We define f(n) as the highest non-negative integer power of 7 that

We define f(n) as the highest non-negative integer power of 7 that divides n. What is the minimum value of k such that \(f(1)+f(2)+....+f(k)\) is a positive multiple of 7 ?

(A) 7
(B) 49
(C) 56
(D) 91
(E) 98

f(n) is the highest non-negative integer power of 7 that divides n. This means that f(n) is the highest value of x (where x is a non-negative integer) for which n/7^x is an integer. Or simply put f(n) is the value of the power of 7 in prime factorisation of n. For example:

If n = 14, then f(14) = 1 because the highest power of 7 that divides 14 is 1: 14 = 2*7^1;
If n = 15, then f(14) = 0 because the highest power of 7 that divides 14 is 0: 14 = 3*5*7^0;
...

You can notice that if n is NOT a multiple of 7, then f(n) = 0.

The question asks to find the value of k such that \(f(1)+f(2)+....+f(k)\) is a positive multiple of 7. As discussed all values which are not multiples of 7 will give the value of f as 0. So, we should concentrate on the multiple of 7.

f(7) = 1;
f(14) = 1;
f(21) = 1;
f(28) = 1;
f(35) = 1;
f(42) = 1;
f(49) = 2 (because 49 = 7^2);

At this point the sum is \(f(1)+f(2)+....+f(49) = 1+1+1+1+1+1+2=8\), which is not a multiple of 7.

f(56) = 1 (sum = 9);
f(63) = 1 (sum = 10);
f(70) = 1 (sum = 11);
f(77) = 1 (sum = 12);
f(84) = 1 (sum = 13);
f(91) = 1 (sum = 14 = multiple of 7).

So, basically we need thirteen multiples of 7, out of which twelve give the value of f as 1 and one gives the value of f as 2 (f(49) = 2):

\(f(1)+f(2)+....+f(91) = 1+1+1+1+1+1+2+1+1+1+1+1+1=14\).

Answer: D.

Hope it's clear.