Positive integer m is defined such that m=10n−36. Positive integer k

For any n>2 \(10^n-36\) will have last two digits as 64, sum of which is divisible by 5.

for n=3 m=964, n=4 m=9964. So m has (n-2)number of 9 in it.

Only option A gives n-2 as multiple of 5.

\(10^n -36\) gives 6 and 4 at tens and unit place respectively and rest of the digits are \(9\)

\(6 + 4 = 10\) which is a multiple of 5 already. So, to make the sum multiple of 5, we need atleast five 9s (or in multiple of 5) along with 64 to make the sum multiple of 5 as \(9+9+9+9+9=45\).

Only answer choice A give 105 nines (9s in multiple of 5) and leaves two slots for 64 to fit in

Since m is a positive integer, we see that the least integer value of n is 2.

If n = 2, we see that m = 10^2 - 36 = 100 - 36 = 64.

If n = 3, we see that m = 10^3 - 36 = 1,000 - 36 = 964.

If n = 4, we see that m = 10^4 - 36 = 10,000 - 36 = 9,964.

If n = 5, we see that m = 10^5 - 36 = 100,000 - 36 = 99,964.

We see that if n > 2, m is the number consisting of a sequence of 9s followed by 64. In fact, the number of 9s is 2 less than n. For example, if n = 5, we have a sequence of three 9s followed by 64. Thus k, the sum of the digits of m, is the sum of these 9s and the digits 6 and 4, i.e., k = 9(n - 2) + 6 + 4 or k = 9(n - 2) + 10.

Since 10 is already a multiple of 5, in order for k to be a multiple of 5, 9(n - 2) also has to be a multiple of 5. Since 9 isn’t a multiple of 5, we know that (n - 2) must be a multiple of 5. This means that the units digit of n must be either 2 or 7. Of the answer choices given, only 107 has the desirable units digit, so choice A is the answer.

Answer: A

Let us find pattern of k, when m = \(10^n - 36\)

suppose n = 4, then m = 10000 - 36 = 9964
suppose n = 4, then m = 100000 - 36 = 99964

so sum of digits = k = \((n-2) * 9 + 10\) => 10 is already multiple of 5 => we have to check if \((n-2)\) is multiple of 5
going by choices, if we fit choice A, n = 107, n - 2 = 105 (which is multiple of 5), we have found the answer (A)