💡First, I recognize that triangle DEF is a right triangle with a right angle at point D.Given:- DE = 30 cm (one leg of the triangle)- DF = 24 cm (the other leg of the triangle)To find the area of a right triangle, I can use the formula:[text{Area} = frac{1}{2} times text{base} times text{height}]Here, DE can be considered the base and DF the height.Plugging in the values:[text{Area} = frac{1}{2} times 30 , text{cm} times 24 , text{cm} = frac{1}{2} times 720 , text{cm}^2 = 360 , text{cm}^2]Therefore, the area of triangle DEF is 360 square centimeters.

💡Alright, so I have this problem here: Given that ( b ) is a multiple of 1428, I need to find the greatest common divisor (gcd) of ( b^2 + 11b + 30 ) and ( b + 6 ). Hmm, okay. Let me think about how to approach this.First, I remember that the Euclidean Algorithm is a reliable method for finding the gcd of two numbers. Maybe I can apply that here. The Euclidean Algorithm involves dividing the larger number by the smaller one and then taking the remainder, repeating this process until the remainder is zero. The last non-zero remainder is the gcd.In this case, the two expressions are ( b^2 + 11b + 30 ) and ( b + 6 ). Since ( b^2 + 11b + 30 ) is a quadratic expression and ( b + 6 ) is linear, I can try dividing ( b^2 + 11b + 30 ) by ( b + 6 ) and find the remainder. That should help me simplify the problem.Let me set up the division. I'll write ( b^2 + 11b + 30 ) divided by ( b + 6 ). To perform polynomial division, I can think of it like this:1. Divide ( b^2 ) by ( b ) to get ( b ).2. Multiply ( b + 6 ) by ( b ) to get ( b^2 + 6b ).3. Subtract this from the original polynomial: ( (b^2 + 11b + 30) - (b^2 + 6b) = 5b + 30 ).So, the remainder after dividing ( b^2 + 11b + 30 ) by ( b + 6 ) is ( 5b + 30 ). According to the Euclidean Algorithm, now I need to find the gcd of ( b + 6 ) and ( 5b + 30 ).Wait a second, ( 5b + 30 ) can be factored as ( 5(b + 6) ). That means ( 5b + 30 ) is just 5 times ( b + 6 ). So, if I'm taking the gcd of ( b + 6 ) and ( 5(b + 6) ), it should be ( b + 6 ) itself, right? Because ( b + 6 ) divides ( 5(b + 6) ) exactly.But hold on, the problem states that ( b ) is a multiple of 1428. That might be important. Let me note that 1428 is a specific number. Maybe I should factorize 1428 to understand its properties better.Let me factorize 1428:1428 divided by 2 is 714.714 divided by 2 is 357.357 divided by 3 is 119.119 divided by 7 is 17.17 is a prime number.So, the prime factorization of 1428 is ( 2^2 times 3 times 7 times 17 ).Okay, so ( b ) is a multiple of 1428, which means ( b = 1428k ) for some integer ( k ). Therefore, ( b + 6 = 1428k + 6 ). I can factor out a 6 from this expression: ( 6(238k + 1) ). So, ( b + 6 ) is a multiple of 6.Earlier, I thought that the gcd is ( b + 6 ), but since ( b + 6 ) is a multiple of 6, maybe the gcd is 6? Let me check.Wait, if ( b + 6 ) is 6 times something, and ( 5b + 30 ) is 5 times ( b + 6 ), which is 5 times 6 times something, so 30 times something. So, the gcd of ( b + 6 ) and ( 5b + 30 ) is the same as the gcd of ( b + 6 ) and 5 times ( b + 6 ). Since 5 and ( b + 6 ) might share a common divisor, but 5 is prime. So, unless ( b + 6 ) is a multiple of 5, the gcd would be 1 or 5.But ( b ) is a multiple of 1428, which is 1428. Let me check if 1428 is a multiple of 5. 1428 divided by 5 is 285.6, which is not an integer. So, 1428 is not a multiple of 5, which means ( b = 1428k ) is also not a multiple of 5. Therefore, ( b + 6 ) is ( 1428k + 6 ). Let me see if this can be a multiple of 5.Let me compute ( 1428k + 6 ) modulo 5. Since 1428 divided by 5 is 285 with a remainder of 3, so 1428 ≡ 3 mod 5. Therefore, ( 1428k ≡ 3k ) mod 5. Adding 6, which is 1 mod 5, so ( 1428k + 6 ≡ 3k + 1 ) mod 5.For ( b + 6 ) to be a multiple of 5, ( 3k + 1 ≡ 0 ) mod 5. That is, ( 3k ≡ -1 ≡ 4 ) mod 5. Multiplying both sides by the inverse of 3 mod 5, which is 2, since 3*2=6≡1 mod5. So, k ≡ 4*2=8≡3 mod5. So, if k ≡3 mod5, then ( b +6 ) is a multiple of 5.But since k can be any integer, depending on k, ( b +6 ) might or might not be a multiple of 5. However, the problem doesn't specify any particular k, just that b is a multiple of 1428. So, in general, the gcd could be 5 or 1, depending on k.Wait, but earlier I thought the gcd was ( b +6 ), but that can't be because ( b +6 ) is much larger than 5. So, perhaps I made a mistake earlier.Let me go back. The Euclidean Algorithm says that gcd(a, b) = gcd(b, a mod b). So, in this case, gcd(b² +11b +30, b +6) = gcd(b +6, (b² +11b +30) mod (b +6)).Earlier, I found that (b² +11b +30) mod (b +6) is 5b +30. So, now I need to compute gcd(b +6, 5b +30). But 5b +30 is 5*(b +6). So, gcd(b +6, 5*(b +6)) = |b +6|, since b +6 divides 5*(b +6). But that would mean the gcd is b +6. But that contradicts my earlier thought.Wait, no. The Euclidean Algorithm says that gcd(a, b) = gcd(b, a mod b). So, in this case, a = b² +11b +30, b = b +6. Then, a mod b is 5b +30. So, now we have gcd(b +6, 5b +30). But 5b +30 is 5*(b +6). So, the gcd is gcd(b +6, 5*(b +6)) which is |b +6|, because b +6 divides 5*(b +6). So, the gcd is b +6.But that can't be, because if I plug in a specific value of b, say b =1428, then b +6 =1434. Let's compute gcd(1428² +11*1428 +30, 1434).Wait, let me compute 1428² +11*1428 +30. That's a huge number, but maybe I can compute it modulo 1434 to find the gcd.Alternatively, since we know that gcd(b² +11b +30, b +6) = gcd(b +6, 5b +30) = gcd(b +6, 5*(b +6)) = b +6, but that would mean the gcd is b +6, which is 1434 in this case. But 1434 is 6*239, and 239 is a prime number. So, is 1434 the gcd? That seems too large because the other polynomial is quadratic.Wait, maybe I'm misunderstanding something. Let me try plugging in b =1428 into the two expressions.First, compute b +6 =1428 +6=1434.Next, compute b² +11b +30. That's 1428² +11*1428 +30. Let me compute this modulo 1434 to find the remainder, which should be the same as 5b +30.Wait, 1428 mod 1434 is -6. So, 1428² mod1434 is (-6)²=36. 11*1428 mod1434 is 11*(-6)= -66. 30 mod1434 is 30.So, adding them up: 36 -66 +30=0. So, the remainder is 0. That means 1434 divides b² +11b +30 exactly when b=1428. So, in this case, the gcd is 1434.But wait, that contradicts my earlier thought that the gcd is 6. Hmm.Wait, let me check with another multiple of 1428. Let's take b=2856 (which is 1428*2). Then, b +6=2862.Compute b² +11b +30 mod2862.Again, b=2856, so b mod2862=2856-2862= -6.So, b² mod2862=(-6)^2=36.11b mod2862=11*(-6)= -66.30 mod2862=30.Adding up: 36 -66 +30=0. So, again, the remainder is 0. So, 2862 divides b² +11b +30 exactly. So, the gcd is 2862.Wait, so in both cases, when b=1428 and b=2856, the gcd is b +6. So, is the gcd always b +6? But the problem says "find the greatest common divisor", implying it's a constant, not depending on b.But in my earlier thought process, I thought the gcd might be 6, but that seems incorrect because when I plug in specific values, the gcd is b +6, which is larger than 6.Wait, maybe I made a mistake in the Euclidean Algorithm step. Let me go through it again.We have gcd(b² +11b +30, b +6). Let me denote a = b² +11b +30 and d = b +6.We can write a = q*d + r, where q is the quotient and r is the remainder. As I did before, dividing a by d, I get q = b +5 and r = 5b +30 -5*(b +6)=5b +30 -5b -30=0. Wait, that can't be right because earlier I thought the remainder was 5b +30.Wait, no, actually, when I perform polynomial division, I have:Divide b² +11b +30 by b +6.First term: b² / b = b. Multiply b +6 by b: b² +6b.Subtract: (b² +11b +30) - (b² +6b) =5b +30.Now, divide 5b +30 by b +6. 5b / b =5. Multiply b +6 by5:5b +30.Subtract: (5b +30) - (5b +30)=0.So, the remainder is 0. Therefore, the gcd is the last non-zero remainder, which was 5b +30. But wait, in the Euclidean Algorithm, after finding the remainder, we then compute gcd(d, r). So, in this case, gcd(b +6, 5b +30). But 5b +30 =5*(b +6). So, gcd(b +6,5*(b +6))=b +6, since b +6 divides 5*(b +6). Therefore, the gcd is b +6.But in the problem, b is a multiple of 1428, so b +6 is 1428k +6. Let me factor out 6:6*(238k +1). So, b +6 is a multiple of 6. Therefore, the gcd is b +6, which is a multiple of 6.But the problem asks for the greatest common divisor, which is a specific number, not an expression in terms of b. So, perhaps I need to find the constant gcd, not depending on b.Wait, but if b is any multiple of 1428, then b +6 can be different values, each time the gcd would be b +6, which varies. But the problem says "find the greatest common divisor", implying it's a fixed number. So, maybe I'm misunderstanding something.Alternatively, perhaps the gcd is always 6, regardless of the multiple. Let me check with b=1428: gcd(1428² +11*1428 +30,1434). As I computed earlier, 1434 divides the quadratic exactly, so the gcd is 1434, which is 6*239. So, 6 is a factor, but the gcd is larger.Wait, maybe I need to find the constant divisor that divides both expressions for any multiple b of 1428. So, the gcd must divide both expressions for all b=1428k. Therefore, the gcd is the common divisor of all possible b +6 and b² +11b +30 for b=1428k.So, let me think about it differently. Let me denote d = gcd(b² +11b +30, b +6). Then, d divides both b² +11b +30 and b +6. Therefore, d divides any linear combination of these two. For example, d divides (b² +11b +30) - b*(b +6) = (b² +11b +30) - (b² +6b) =5b +30. So, d divides 5b +30.Similarly, d divides (5b +30) -5*(b +6)=5b +30 -5b -30=0. So, d divides 0, which means d divides any multiple of b +6. But since d divides b +6, and d divides 5b +30, which is 5*(b +6), d must divide the gcd of b +6 and 5*(b +6). But the gcd of b +6 and 5*(b +6) is b +6, as b +6 divides 5*(b +6). Therefore, d divides b +6.But we also have that d divides 5b +30, which is 5*(b +6). So, d divides both b +6 and 5*(b +6). Therefore, d must divide the gcd of b +6 and 5*(b +6), which is b +6. So, d divides b +6.But since d also divides b +6, the only possible divisors are the divisors of b +6. However, since b is a multiple of 1428, b +6 is 1428k +6. Let me factor this:1428k +6=6*(238k +1). So, b +6 is a multiple of 6. Therefore, d must be a divisor of 6*(238k +1). But since d divides b +6, and b +6 is 6*(238k +1), d must be a divisor of 6*(238k +1).But we also have that d divides 5b +30=5*(b +6)=5*6*(238k +1)=30*(238k +1). So, d divides 30*(238k +1). But since d divides both 6*(238k +1) and 30*(238k +1), d must divide the gcd of 6*(238k +1) and 30*(238k +1). The gcd of 6 and 30 is 6. Therefore, d must divide 6.So, the possible values of d are the divisors of 6, which are 1, 2, 3, 6.Now, we need to find the greatest common divisor, so the largest possible d is 6. But we need to check if 6 actually divides both expressions for all b=1428k.Let me check if 6 divides b +6. Since b is a multiple of 1428, which is divisible by 6, b is divisible by 6. Therefore, b +6 is 6m +6=6(m +1), which is divisible by 6. So, 6 divides b +6.Now, does 6 divide b² +11b +30? Let's check modulo 6.Since b is a multiple of 1428, which is divisible by 6, b ≡0 mod6. Therefore, b² ≡0 mod6, 11b ≡0 mod6, and 30≡0 mod6. So, b² +11b +30 ≡0 +0 +0=0 mod6. Therefore, 6 divides b² +11b +30.Therefore, 6 divides both expressions, so the gcd is at least 6. But earlier, I thought the gcd could be larger, like 1434, but that was for a specific b. However, the problem asks for the gcd in general, for any multiple b of 1428. So, the gcd must be a divisor that works for all such b. Since 6 is the largest such divisor, the gcd is 6.Wait, but earlier when I plugged in b=1428, the gcd was 1434, which is a multiple of 6. So, how come the gcd is 6? Because for a specific b, the gcd could be larger, but since the problem asks for the gcd in general, meaning for any b multiple of 1428, the gcd must be a divisor that works for all such b. Therefore, the gcd is 6.To confirm, let me check with another multiple of 1428, say b=1428*2=2856. Then, b +6=2862. Compute gcd(2856² +11*2856 +30,2862).Again, compute 2856² +11*2856 +30 mod2862.Since 2856=2862-6, so 2856≡-6 mod2862.Therefore, 2856²≡(-6)^2=36 mod2862.11*2856≡11*(-6)= -66 mod2862.30≡30 mod2862.Adding up:36 -66 +30=0. So, 2862 divides the quadratic exactly. Therefore, gcd is 2862, which is 6*477. So, again, the gcd is a multiple of 6. But since we need the gcd that works for all b=1428k, the only common divisor is 6.Therefore, the greatest common divisor is 6.

💡Okay, so I have this problem where I need to show that for any strictly positive integer ( q ) and a polynomial ( P ) with integer coefficients, there exists a positive integer ( n ) such that the sum ( P(1) + P(2) + cdots + P(n) ) is divisible by ( q ). Hmm, interesting. Let me try to break this down step by step.First, I know that ( P ) is a polynomial with integer coefficients. So, ( P(x) ) can be written as ( a_k x^k + a_{k-1} x^{k-1} + cdots + a_1 x + a_0 ), where each ( a_i ) is an integer. That might be useful later on.Now, the problem is about finding an ( n ) such that the sum ( S(n) = P(1) + P(2) + cdots + P(n) ) is divisible by ( q ). So, I need to show that there exists some ( n ) where ( S(n) equiv 0 mod q ).I remember that when dealing with modular arithmetic, sometimes considering the problem modulo ( q ) can simplify things. Maybe I can look at the behavior of ( P(x) ) modulo ( q ) and see if that helps.Let me think about the values of ( P(x) ) modulo ( q ). Since ( P ) has integer coefficients, ( P(x) mod q ) will depend on ( x mod q ). That is, if two integers ( x ) and ( y ) are congruent modulo ( q ), then ( P(x) equiv P(y) mod q ). So, the polynomial ( P ) has a periodic behavior modulo ( q ) with period ( q ).Wait, that might be useful. If ( P(x) ) is periodic modulo ( q ) with period ( q ), then the sequence ( P(1), P(2), ldots, P(q) ) modulo ( q ) will repeat every ( q ) terms. So, the sum ( S(n) ) modulo ( q ) might also have some periodic behavior.Let me try to formalize this. Let's denote ( S(n) = sum_{i=1}^n P(i) ). Then, modulo ( q ), this sum can be written as ( sum_{i=1}^n P(i) mod q ).Given the periodicity, if I consider blocks of ( q ) consecutive integers, the sum of ( P(i) ) over each block modulo ( q ) will be the same. So, if I denote ( T = sum_{i=1}^q P(i) mod q ), then for any multiple of ( q ), say ( n = kq ), the sum ( S(n) ) modulo ( q ) would be ( kT mod q ).Hmm, so if I can find a ( k ) such that ( kT equiv 0 mod q ), then ( S(kq) equiv 0 mod q ). That would give me the desired ( n = kq ).But wait, is ( T ) necessarily non-zero? If ( T equiv 0 mod q ), then ( S(q) equiv 0 mod q ), and we can take ( n = q ). If not, then we need to find some ( k ) such that ( kT equiv 0 mod q ).Since ( T ) is an integer, and ( q ) is a positive integer, the equation ( kT equiv 0 mod q ) is equivalent to ( q ) dividing ( kT ). By the properties of modular arithmetic, this will happen if ( k ) is a multiple of ( q / gcd(T, q) ). So, if I take ( k = q / gcd(T, q) ), then ( kT ) will be divisible by ( q ).Therefore, ( n = kq = q times (q / gcd(T, q)) = q^2 / gcd(T, q) ) will satisfy ( S(n) equiv 0 mod q ). So, such an ( n ) exists.Wait, but is ( T ) necessarily an integer? Yes, because ( P(i) ) are integers, so their sum ( T ) is also an integer. Therefore, ( gcd(T, q) ) is well-defined.Alternatively, maybe I can use the pigeonhole principle. Since there are only finitely many possible remainders modulo ( q ) for the partial sums ( S(1), S(2), ldots, S(q) ), by the pigeonhole principle, at least two of them must be congruent modulo ( q ). Then, the difference between these two sums would be divisible by ( q ), giving me a block of consecutive terms whose sum is divisible by ( q ).Let me elaborate. Consider the partial sums ( S(1), S(2), ldots, S(q) ). Each of these is an integer, so when taken modulo ( q ), each can be congruent to one of ( 0, 1, 2, ldots, q-1 ). There are ( q ) partial sums and ( q ) possible remainders. If any ( S(i) equiv 0 mod q ), then we're done with ( n = i ).If not, then by the pigeonhole principle, there must be two partial sums ( S(j) ) and ( S(k) ) with ( j < k ) such that ( S(j) equiv S(k) mod q ). Then, the sum ( S(k) - S(j) = sum_{i=j+1}^k P(i) ) is divisible by ( q ). So, the sum from ( j+1 ) to ( k ) is divisible by ( q ).But the problem asks for the sum from 1 to ( n ), not necessarily a consecutive block. Hmm, so maybe this approach doesn't directly give me the result, but it's related.Wait, perhaps I can combine these ideas. If I consider the partial sums modulo ( q ), either one of them is zero modulo ( q ), or two of them are equal modulo ( q ), giving a block sum divisible by ( q ). But the problem is about the cumulative sum up to ( n ), not a block in the middle.Alternatively, maybe I can use the fact that the polynomial ( P(x) ) has integer coefficients, so the differences ( P(x+1) - P(x) ) also have integer coefficients, and so on. Maybe considering the telescoping sums or something like that.Alternatively, perhaps I can use induction on the degree of the polynomial. For example, if ( P(x) ) is a constant polynomial, say ( P(x) = c ), then the sum ( S(n) = c times n ). So, to have ( c times n equiv 0 mod q ), we can take ( n = q ) if ( c ) and ( q ) are coprime, or a smaller multiple otherwise.If ( P(x) ) is linear, say ( P(x) = ax + b ), then ( S(n) = a times frac{n(n+1)}{2} + b times n ). We need to find ( n ) such that ( a times frac{n(n+1)}{2} + b times n equiv 0 mod q ). This is a quadratic equation in ( n ), and since we're working modulo ( q ), there should be a solution.Similarly, for higher-degree polynomials, the sum ( S(n) ) will be a polynomial of degree ( k+1 ) if ( P(x) ) is degree ( k ). So, perhaps for any polynomial, the sum ( S(n) ) will eventually cover all residues modulo ( q ), or at least hit zero.Wait, but how can I formalize that? Maybe using the fact that the sequence ( S(n) mod q ) must eventually repeat, and within the first ( q ) terms, either one of them is zero, or there's a repetition leading to a block sum divisible by ( q ).But again, the problem is about the cumulative sum up to ( n ), not a block. So, maybe I need a different approach.Let me think about the polynomial ( P(x) ) modulo ( q ). Since ( P(x) ) has integer coefficients, ( P(x) mod q ) is a polynomial with coefficients in ( mathbb{Z}/qmathbb{Z} ). So, ( P(x) ) can be considered as a function from ( mathbb{Z}/qmathbb{Z} ) to ( mathbb{Z}/qmathbb{Z} ).Now, the sum ( S(n) = sum_{i=1}^n P(i) ) can be considered as the sum of ( P(1), P(2), ldots, P(n) ) modulo ( q ). If I can show that this sum can be zero modulo ( q ), then I'm done.Alternatively, maybe I can use the fact that the average of the first ( n ) terms of a polynomial sequence can be made divisible by ( q ) by choosing a suitable ( n ).Wait, another idea: consider the sum ( S(n) ) modulo ( q ). Since there are only finitely many possible residues modulo ( q ), the sequence ( S(1), S(2), ldots ) modulo ( q ) must eventually repeat. So, by the pigeonhole principle, there must be some ( n ) where ( S(n) equiv 0 mod q ).But is that necessarily true? Let me think. The partial sums ( S(n) ) modulo ( q ) can take on ( q ) different values. If none of them is zero, then by the pigeonhole principle, two of them must be equal, say ( S(j) equiv S(k) mod q ) for some ( j < k ). Then, ( S(k) - S(j) = sum_{i=j+1}^k P(i) equiv 0 mod q ). But this gives a block sum divisible by ( q ), not necessarily the cumulative sum up to ( k ).Hmm, so maybe this approach doesn't directly give me the result, but it's related. I need a way to ensure that the cumulative sum itself is zero modulo ( q ).Wait, perhaps I can use the fact that the polynomial ( P(x) ) modulo ( q ) is periodic with period ( q ). So, the sequence ( P(1), P(2), ldots ) modulo ( q ) repeats every ( q ) terms. Therefore, the sum ( S(n) ) modulo ( q ) will have a periodic component as well.Let me denote ( T = sum_{i=1}^q P(i) mod q ). If ( T equiv 0 mod q ), then ( S(q) equiv 0 mod q ), and we can take ( n = q ). If not, then consider the sums ( S(q), S(2q), S(3q), ldots ). Each of these sums is congruent to ( kT mod q ), where ( k ) is the number of blocks of ( q ) terms.So, ( S(kq) equiv kT mod q ). We need to find ( k ) such that ( kT equiv 0 mod q ). Since ( T ) is fixed, this is equivalent to finding ( k ) such that ( q ) divides ( kT ). The smallest such ( k ) is ( k = q / gcd(T, q) ). Therefore, ( n = kq = q^2 / gcd(T, q) ) will satisfy ( S(n) equiv 0 mod q ).This seems promising. So, regardless of the value of ( T ), as long as ( T ) is an integer, which it is, we can find such an ( n ). Therefore, such an ( n ) exists.Wait, but what if ( T ) is zero? Then, ( n = q ) works. If ( T ) is not zero, then ( n = q^2 / gcd(T, q) ) works. So, in either case, such an ( n ) exists.Alternatively, maybe I can use the fact that the sum of a polynomial over a complete residue system modulo ( q ) is divisible by ( q ). Is that a theorem? I think it might be related to the fact that the sum of a polynomial over a complete set of residues modulo ( q ) can be expressed in terms of Bernoulli numbers or something like that, but I'm not sure.Wait, actually, I recall that for any polynomial ( P(x) ) with integer coefficients, the sum ( sum_{x=0}^{q-1} P(x) ) is divisible by ( q ) if ( P(x) ) is a polynomial of degree less than ( q-1 ). But I'm not sure if that's directly applicable here.Alternatively, maybe I can use the fact that the sum ( S(n) ) is a polynomial of degree ( k+1 ), and therefore, its behavior modulo ( q ) must eventually cycle or cover all residues, ensuring that zero is achieved.But I think the earlier approach using the periodicity and the pigeonhole principle is more straightforward. By considering the sum over blocks of ( q ) terms, and noting that the sum over each block is congruent modulo ( q ), we can find a multiple of ( q ) such that the total sum is divisible by ( q ).So, to summarize, here's the plan:1. Recognize that ( P(x) mod q ) is periodic with period ( q ).2. Compute the sum ( T = sum_{i=1}^q P(i) mod q ).3. If ( T equiv 0 mod q ), then ( n = q ) works.4. If not, find the smallest ( k ) such that ( kT equiv 0 mod q ), which is ( k = q / gcd(T, q) ).5. Then, ( n = kq ) will satisfy ( S(n) equiv 0 mod q ).This shows that such an ( n ) exists, specifically ( n = q^2 / gcd(T, q) ).Alternatively, using the pigeonhole principle on the partial sums ( S(1), S(2), ldots, S(q) ), either one of them is zero modulo ( q ), or two are equal, leading to a block sum divisible by ( q ). However, since the problem asks for the cumulative sum up to ( n ), the first approach is more direct.I think this makes sense. I should probably write this up more formally to make sure all the steps are clear.

💡Okay, so I have this problem where I need to prove that for a given prime number ( p ) and positive integers ( m ) and ( c ), there are infinitely many positive integers ( n ) such that ( f(p, n) equiv c pmod{m} ). Here, ( f(p, n) ) is the largest integer ( k ) such that ( p^k ) divides ( n! ). First, I remember that ( f(p, n) ) can be calculated using Legendre's formula, which is:[f(p, n) = sum_{i=1}^{infty} leftlfloor frac{n}{p^i} rightrfloor]This formula adds up the number of multiples of ( p ), ( p^2 ), ( p^3 ), and so on in ( n! ). So, ( f(p, n) ) is essentially the exponent of the highest power of ( p ) dividing ( n! ).Now, the problem is asking about the congruence of this exponent modulo ( m ). Specifically, we need to find infinitely many ( n ) such that ( f(p, n) equiv c pmod{m} ). I think the key here is to analyze the behavior of ( f(p, n) ) modulo ( m ). Since ( m ) is a positive integer, it's finite, so the residues modulo ( m ) are cyclic. Maybe we can find a pattern or periodicity in ( f(p, n) ) modulo ( m ) that allows us to find infinitely many ( n ) satisfying the condition.Let me consider how ( f(p, n) ) changes as ( n ) increases. Each time ( n ) crosses a multiple of ( p ), ( p^2 ), ( p^3 ), etc., the value of ( f(p, n) ) increases by 1, 1, 1, etc., respectively. So, the function ( f(p, n) ) is a step function that increases by 1 at multiples of ( p ), by another 1 at multiples of ( p^2 ), and so on.This suggests that ( f(p, n) ) can be expressed as:[f(p, n) = leftlfloor frac{n}{p} rightrfloor + leftlfloor frac{n}{p^2} rightrfloor + leftlfloor frac{n}{p^3} rightrfloor + cdots]Each term in this sum contributes to the total exponent of ( p ) in ( n! ).Now, if I think about ( f(p, n) ) modulo ( m ), I need to find ( n ) such that the sum of these floor functions is congruent to ( c ) modulo ( m ). Since ( m ) is fixed, maybe I can construct ( n ) in such a way that each term in the sum contributes a specific amount modulo ( m ).Perhaps I can choose ( n ) such that each term ( leftlfloor frac{n}{p^i} rightrfloor ) is congruent to some value modulo ( m ). If I can control these terms, I might be able to make the entire sum congruent to ( c ) modulo ( m ).But how do I ensure that there are infinitely many such ( n )? Maybe by choosing ( n ) in a specific form, like ( n = kp^t ) for some ( t ), or using a sequence where ( n ) increases in a way that the contributions to ( f(p, n) ) cycle through the residues modulo ( m ).Wait, another thought: since ( f(p, n) ) is a sum of terms that each increase at different rates (because each ( p^i ) grows exponentially), the overall behavior of ( f(p, n) ) modulo ( m ) might be periodic or have some repeating pattern. If that's the case, then by the Chinese Remainder Theorem or some similar argument, we can find infinitely many ( n ) where the sum hits the desired residue ( c ).Alternatively, maybe I can use the fact that for sufficiently large ( n ), the higher powers of ( p ) (like ( p^3, p^4 ), etc.) contribute less frequently, so the main contribution to ( f(p, n) ) comes from the first few terms. If I can control the lower terms, perhaps I can adjust ( n ) to get the desired congruence.Let me try to formalize this a bit. Suppose I fix some exponent ( t ) such that ( p^t ) is larger than ( m ). Then, for ( n ) much larger than ( p^t ), the terms ( leftlfloor frac{n}{p^{t+1}} rightrfloor, leftlfloor frac{n}{p^{t+2}} rightrfloor, ldots ) will be relatively small compared to the earlier terms. So, maybe I can focus on the first ( t ) terms of the sum.If I can adjust ( n ) such that the sum of the first ( t ) terms is congruent to ( c ) modulo ( m ), then the remaining terms, being small, won't affect the congruence. But I need to ensure that this is possible for infinitely many ( n ).Perhaps I can construct ( n ) as a sum of powers of ( p ). For example, take ( n = p^{a_1} + p^{a_2} + cdots + p^{a_k} ) where ( a_1 > a_2 > cdots > a_k ). Then, when I compute ( f(p, n) ), each term ( leftlfloor frac{n}{p^i} rightrfloor ) will be the sum of ( p^{a_j - i} ) for each ( a_j geq i ).Wait, let me compute this more carefully. If ( n = p^{a_1} + p^{a_2} + cdots + p^{a_k} ), then:[leftlfloor frac{n}{p^i} rightrfloor = leftlfloor frac{p^{a_1} + p^{a_2} + cdots + p^{a_k}}{p^i} rightrfloor = p^{a_1 - i} + p^{a_2 - i} + cdots + p^{a_k - i}]provided that ( a_j geq i ) for each term. So, the total ( f(p, n) ) becomes:[f(p, n) = sum_{i=1}^{infty} sum_{j=1}^k p^{a_j - i} quad text{where} quad a_j geq i]This can be rewritten as:[f(p, n) = sum_{j=1}^k sum_{i=1}^{a_j} p^{a_j - i} = sum_{j=1}^k frac{p^{a_j} - 1}{p - 1}]Ah, interesting! So, ( f(p, n) ) is the sum of these geometric series for each power of ( p ) in the representation of ( n ).Therefore, if I can choose the exponents ( a_j ) such that each ( frac{p^{a_j} - 1}{p - 1} ) contributes a specific amount modulo ( m ), then the total sum can be made congruent to ( c ) modulo ( m ).But how do I choose these ( a_j ) to make this happen? Maybe I can choose each ( a_j ) such that ( frac{p^{a_j} - 1}{p - 1} equiv c_j pmod{m} ) for some ( c_j ), and then choose the number of terms ( k ) such that the sum of ( c_j ) is congruent to ( c ) modulo ( m ).But I need to ensure that there are infinitely many such ( a_j ). That is, for each ( c_j ), there should be infinitely many exponents ( a_j ) such that ( frac{p^{a_j} - 1}{p - 1} equiv c_j pmod{m} ).Is this possible? Let me think about the sequence ( frac{p^a - 1}{p - 1} ) modulo ( m ). This sequence is periodic because there are only finitely many residues modulo ( m ). So, by the Pigeonhole Principle, some residues must repeat infinitely often.Therefore, for each desired residue ( c_j ), there are infinitely many exponents ( a_j ) such that ( frac{p^{a_j} - 1}{p - 1} equiv c_j pmod{m} ).So, if I can choose ( k ) and ( c_j ) such that ( sum_{j=1}^k c_j equiv c pmod{m} ), then I can construct ( n ) as the sum of ( p^{a_j} ) with these exponents ( a_j ), and ( f(p, n) equiv c pmod{m} ).But I need to ensure that this can be done for infinitely many ( n ). That is, I can choose different sets of exponents ( a_j ) each time, leading to different ( n ), but all satisfying ( f(p, n) equiv c pmod{m} ).Wait, but how do I ensure that the sum ( sum_{j=1}^k c_j equiv c pmod{m} ) can be achieved with different ( k ) and different ( c_j )? Maybe I can fix ( k ) and vary ( c_j ), or vary ( k ) while keeping the sum congruent to ( c ).Alternatively, perhaps I can use the fact that the sequence ( frac{p^a - 1}{p - 1} ) modulo ( m ) is periodic, so there exists some period ( d ) such that ( frac{p^{a + d} - 1}{p - 1} equiv frac{p^a - 1}{p - 1} pmod{m} ) for all sufficiently large ( a ).If that's the case, then for each residue ( c_j ), there are infinitely many ( a_j ) such that ( frac{p^{a_j} - 1}{p - 1} equiv c_j pmod{m} ). Therefore, I can choose as many ( a_j ) as needed to make the sum congruent to ( c ) modulo ( m ).But I need to formalize this argument. Maybe I can use the Chinese Remainder Theorem or some combinatorial argument to show that such ( n ) exist infinitely often.Alternatively, perhaps I can consider the function ( f(p, n) ) modulo ( m ) and show that it is surjective or that it cycles through all residues modulo ( m ), hence hitting ( c ) infinitely often.Wait, another approach: since ( f(p, n) ) is a sum of terms that each increase at different rates, the overall function ( f(p, n) ) modulo ( m ) must eventually cycle through all residues modulo ( m ) infinitely often. Therefore, by the infinite pigeonhole principle, there must be infinitely many ( n ) such that ( f(p, n) equiv c pmod{m} ).But I need to make this more precise. Maybe I can consider the behavior of ( f(p, n) ) as ( n ) increases and show that the residues modulo ( m ) are dense or cover all possibilities infinitely often.Alternatively, perhaps I can construct a sequence ( n_k ) such that ( f(p, n_k) equiv c pmod{m} ) for each ( k ), and show that this sequence can be extended infinitely.Wait, going back to the earlier idea of expressing ( n ) as a sum of powers of ( p ). If I can choose each ( a_j ) such that ( frac{p^{a_j} - 1}{p - 1} equiv c_j pmod{m} ), and then choose ( k ) such that ( sum_{j=1}^k c_j equiv c pmod{m} ), then I can construct infinitely many such ( n ) by choosing different combinations of ( a_j ).But how do I ensure that such ( a_j ) exist for each ( c_j )? Maybe by solving the congruence ( frac{p^{a_j} - 1}{p - 1} equiv c_j pmod{m} ) for ( a_j ).This seems related to solving exponential congruences. Since ( p ) and ( m ) are given, perhaps I can use properties of modular arithmetic to find solutions ( a_j ).Alternatively, since ( frac{p^a - 1}{p - 1} ) is a geometric series, maybe I can analyze its periodicity modulo ( m ). If the sequence ( frac{p^a - 1}{p - 1} ) modulo ( m ) is periodic, then each residue appears infinitely often, which would allow me to choose ( a_j ) such that ( frac{p^{a_j} - 1}{p - 1} equiv c_j pmod{m} ).Therefore, if I can show that for each residue ( c_j ) modulo ( m ), there are infinitely many ( a_j ) such that ( frac{p^{a_j} - 1}{p - 1} equiv c_j pmod{m} ), then I can construct infinitely many ( n ) by summing the corresponding ( p^{a_j} ).But I need to make sure that this is possible. Let me consider the function ( g(a) = frac{p^a - 1}{p - 1} ) modulo ( m ). Since ( g(a) ) is periodic modulo ( m ), there exists some period ( d ) such that ( g(a + d) equiv g(a) pmod{m} ) for all ( a geq t ) for some ( t ).Therefore, the residues of ( g(a) ) modulo ( m ) repeat every ( d ) terms. Hence, each residue that appears in the sequence ( g(a) ) modulo ( m ) must appear infinitely often.Thus, for any desired residue ( c_j ), there are infinitely many ( a_j ) such that ( g(a_j) equiv c_j pmod{m} ).Therefore, I can choose ( k ) and ( c_j ) such that ( sum_{j=1}^k c_j equiv c pmod{m} ). Since there are infinitely many choices for each ( a_j ), I can construct infinitely many ( n ) by choosing different combinations of ( a_j ) leading to different ( n ) but all satisfying ( f(p, n) equiv c pmod{m} ).Wait, but how do I ensure that the sum ( sum_{j=1}^k c_j equiv c pmod{m} ) can be achieved? Maybe by choosing ( k ) such that ( k cdot c_j equiv c pmod{m} ) for some ( c_j ), but that might not always be possible.Alternatively, perhaps I can use the fact that the set of residues ( {c_j} ) modulo ( m ) is closed under addition, so by choosing appropriate ( c_j ), I can reach any desired residue ( c ).But I need to formalize this. Maybe I can use the fact that the additive group modulo ( m ) is cyclic, so any element can be expressed as a sum of generators. Therefore, if I can find a generator ( g ) such that ( g ) modulo ( m ) is a generator, then I can express ( c ) as a multiple of ( g ).But I'm not sure if this directly applies. Maybe a better approach is to consider that since the sequence ( g(a) ) modulo ( m ) is periodic, and each residue appears infinitely often, then by choosing multiple terms ( g(a_j) ), I can combine them to reach any desired residue ( c ) modulo ( m ).For example, if I can find one ( a ) such that ( g(a) equiv 1 pmod{m} ), then I can use multiple such terms to reach any ( c ). But I don't know if ( g(a) equiv 1 pmod{m} ) is always possible.Alternatively, maybe I can use the fact that the set of residues ( {g(a)} ) modulo ( m ) generates the entire additive group modulo ( m ), or at least a subgroup that contains ( c ).But I'm not entirely sure about this. Maybe I need to approach this differently.Let me think about the function ( f(p, n) ) modulo ( m ). Since ( f(p, n) ) is a sum of terms that each increase at different rates, the overall function ( f(p, n) ) modulo ( m ) must eventually cycle through all residues modulo ( m ) infinitely often. Therefore, by the infinite pigeonhole principle, there must be infinitely many ( n ) such that ( f(p, n) equiv c pmod{m} ).But I need to make this more precise. Maybe I can consider the behavior of ( f(p, n) ) as ( n ) increases and show that the residues modulo ( m ) are dense or cover all possibilities infinitely often.Alternatively, perhaps I can construct a sequence ( n_k ) such that ( f(p, n_k) equiv c pmod{m} ) for each ( k ), and show that this sequence can be extended infinitely.Wait, another idea: since ( f(p, n) ) is a non-decreasing function, and it increases by at least 1 each time ( n ) crosses a multiple of ( p ), the function ( f(p, n) ) modulo ( m ) must cycle through all residues modulo ( m ) infinitely often. Therefore, for any ( c ), there are infinitely many ( n ) such that ( f(p, n) equiv c pmod{m} ).But I need to verify this. Let me consider that ( f(p, n) ) increases by 1 at each multiple of ( p ), by another 1 at each multiple of ( p^2 ), and so on. Therefore, the increments to ( f(p, n) ) are not uniform, but they are frequent enough to ensure that the residues modulo ( m ) are hit infinitely often.In other words, since ( f(p, n) ) increases by at least 1 infinitely often, and the increments can be controlled to adjust the residue modulo ( m ), we can ensure that every residue class modulo ( m ) is achieved infinitely often.Therefore, there must be infinitely many ( n ) such that ( f(p, n) equiv c pmod{m} ).But I need to formalize this argument. Maybe I can use the fact that the function ( f(p, n) ) modulo ( m ) is surjective and periodic, hence hits every residue infinitely often.Alternatively, perhaps I can use the fact that the sequence ( f(p, n) ) modulo ( m ) is dense in the integers modulo ( m ), meaning that every residue appears infinitely often.But I'm not entirely sure about the density argument. Maybe a better approach is to consider that since ( f(p, n) ) increases by at least 1 infinitely often, and the increments can be adjusted by choosing ( n ) appropriately, we can ensure that the residues cycle through all possibilities modulo ( m ).Therefore, by the infinite pigeonhole principle, since there are infinitely many ( n ) and only finitely many residues modulo ( m ), each residue must appear infinitely often.Hence, there are infinitely many positive integers ( n ) such that ( f(p, n) equiv c pmod{m} ).I think this line of reasoning makes sense, but I need to make sure it's rigorous. Maybe I can use the fact that the function ( f(p, n) ) modulo ( m ) is not eventually constant, and since it's non-decreasing, it must cycle through all residues modulo ( m ) infinitely often.Alternatively, perhaps I can use a more algebraic approach by considering the periodicity of ( f(p, n) ) modulo ( m ) and showing that the desired residue ( c ) is achieved infinitely often.In conclusion, I believe that by analyzing the behavior of ( f(p, n) ) modulo ( m ) and using properties of periodicity and the infinite pigeonhole principle, we can show that there are indeed infinitely many positive integers ( n ) such that ( f(p, n) equiv c pmod{m} ).

💡Okay, so I have this polynomial: 6x⁴ + b₃x³ + b₂x² + b₁x + 10 = 0. I need to find the number of different possible rational roots. Hmm, I remember something called the Rational Root Theorem from algebra. Let me try to recall how that works.The Rational Root Theorem says that any possible rational root of a polynomial equation with integer coefficients is of the form ±p/q, where p is a factor of the constant term and q is a factor of the leading coefficient. So, in this case, the constant term is 10, and the leading coefficient is 6. Alright, so first, I need to list all the factors of 10. The factors of 10 are ±1, ±2, ±5, and ±10. These are all the possible values for p. Next, I need to list all the factors of 6, which are ±1, ±2, ±3, and ±6. These will be the possible values for q.Now, according to the theorem, the possible rational roots are all combinations of p over q. That means I need to take each factor of 10 and divide it by each factor of 6. Let me write these out:Starting with p = 1:- 1/1 = 1- 1/2 = 0.5- 1/3 ≈ 0.333...- 1/6 ≈ 0.166...Similarly, for p = -1:- -1/1 = -1- -1/2 = -0.5- -1/3 ≈ -0.333...- -1/6 ≈ -0.166...Moving on to p = 2:- 2/1 = 2- 2/2 = 1- 2/3 ≈ 0.666...- 2/6 = 1/3 ≈ 0.333...For p = -2:- -2/1 = -2- -2/2 = -1- -2/3 ≈ -0.666...- -2/6 = -1/3 ≈ -0.333...Next, p = 5:- 5/1 = 5- 5/2 = 2.5- 5/3 ≈ 1.666...- 5/6 ≈ 0.833...And for p = -5:- -5/1 = -5- -5/2 = -2.5- -5/3 ≈ -1.666...- -5/6 ≈ -0.833...Finally, p = 10:- 10/1 = 10- 10/2 = 5- 10/3 ≈ 3.333...- 10/6 = 5/3 ≈ 1.666...And for p = -10:- -10/1 = -10- -10/2 = -5- -10/3 ≈ -3.333...- -10/6 = -5/3 ≈ -1.666...Now, let me list all these possible rational roots:Positive:1, 0.5, 0.333..., 0.166..., 2, 1, 0.666..., 0.333..., 5, 2.5, 1.666..., 0.833..., 10, 5, 3.333..., 1.666...Negative:-1, -0.5, -0.333..., -0.166..., -2, -1, -0.666..., -0.333..., -5, -2.5, -1.666..., -0.833..., -10, -5, -3.333..., -1.666...Wait, I notice some duplicates here. For example, 1 appears twice because both 1/1 and 2/2 give 1. Similarly, 0.333... comes from both 1/3 and 2/6. I need to eliminate these duplicates to find the distinct possible rational roots.Let me list them again, making sure each is unique:Positive:1, 2, 5, 10, 0.5, 0.333..., 0.166..., 2.5, 1.666..., 0.833..., 3.333...Negative:-1, -2, -5, -10, -0.5, -0.333..., -0.166..., -2.5, -1.666..., -0.833..., -3.333...Wait, I think I missed some. Let me make sure I have all the unique fractions:From p = 1: 1, 0.5, 0.333..., 0.166...From p = 2: 2, 1 (duplicate), 0.666..., 0.333... (duplicate)From p = 5: 5, 2.5, 1.666..., 0.833...From p = 10: 10, 5 (duplicate), 3.333..., 1.666... (duplicate)Similarly for the negatives.So, compiling the unique positive possible rational roots:1, 2, 5, 10, 0.5, 0.333..., 0.166..., 2.5, 1.666..., 0.833..., 3.333...And the unique negative ones:-1, -2, -5, -10, -0.5, -0.333..., -0.166..., -2.5, -1.666..., -0.833..., -3.333...Now, let's count them. For each positive, there's a corresponding negative, so I can count the positive ones and double the count.Looking at the positive list:1, 2, 5, 10, 0.5, 0.333..., 0.166..., 2.5, 1.666..., 0.833..., 3.333...That's 11 distinct positive possible rational roots. Therefore, including their negatives, that's 22 distinct possible rational roots.Wait, but let me make sure I didn't miss any. Let me list them all:Positive:1, 2, 5, 10, 1/2, 1/3, 1/6, 5/2, 5/3, 5/6, 2/3Negative:-1, -2, -5, -10, -1/2, -1/3, -1/6, -5/2, -5/3, -5/6, -2/3Yes, that's 11 positive and 11 negative, totaling 22 distinct possible rational roots.So, the number of different possible rational roots of the polynomial is 22.

💡First, I need to determine how much the man earns each day. Let's denote the amount he earns on the first day as ( E ).The man follows a pattern where he earns ( E ) rupees on the first day and spends Rs 15 on the second day. This cycle repeats every two days.Since there are 60 days, there are 30 such earn-spend cycles.At the end of each cycle, the man's net amount is ( E - 15 ) rupees.After 30 cycles, the man has Rs 60. This gives us the equation:[30 times (E - 15) = 60]Solving for ( E ):[30E - 450 = 60][30E = 510][E = frac{510}{30}][E = 17]Therefore, the man earns Rs 17 on the first day and on every alternate day thereafter, including the third day.