Last two digits are 64

and the rest digits of m will be 9's
so how many 9's...its n-2 nos. of 9
suppose n=4 or 10^n= 10^n =10000
then 10000-36 = 9964
thus sum will be 2*9 +6+4= 28
Lets check options
(a) n=107 thus sum = 9*105+6+4 =945+10=955
(b) n=108 thus sum = 9*106+6+4 =954+10=964
(c) n=109 thus sum = 9*107+6+4 =963+10=973
(d) n=110 thus sum = 9*108+6+4 =972+10=982
(e) n=111 thus sum = 9*109+6+4 =981+10=991

only option (a) is resulting as 955 ,divisible by 5

Ans A

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.

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
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