LCM of two co-prime numbers is their product.

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7^4 = 7*7*7*7 = 2401

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Positive integer n is divided by 5, the remainder is 1 --> n=5q+1n=5q+1, where qq is the quotient --> 1, 6, 11, 16, 21, 26, 31, ...

Positive integer n is divided by 7, the remainder is 3 --> n=7p+3n=7p+3, where pp is the quotient --> 3, 10, 17, 24, 31, ....

There is a way to derive general formula for nn (of a type n=mx+rn=mx+r, where xx is divisor and rr is a remainder) based on above two statements:

Divisor xx would be the least common multiple of above two divisors 5 and 7, hence x=35x=35.

Remainder rr would be the first common integer in above two patterns, hence r=31r=31.

Therefore general formula based on both statements is n=35m+31n=35m+31. Thus the smallest positive integer k such that k+n is a multiple of 35 is 4 --> n+4=35k+31+4=35(k+1)n+4=35k+31+4=35(k+1).

let n = 3

C+n^2 is divisible by 5; so c+9 is divisible by 5. Thus, C = 6

Using other condition; 6+7 = 13 which is not divisible by 5. So, this case does not exist.

Same is for all other options.

3B + 2G, 4B + 1G, 5B

5C2*5C3 + 5C4*5C1 + 5C5 = 126