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

Question

Positive integer m is defined such that  m=10^n−36 . Positive integer k is defined such that k equals the sum of the digits of integer m. For which of the following values of n is k a multiple of 5?

A. 107
B. 108
C. 109
D. 110
E. 111

rohit8865 · Answer

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

Kurtosis · Answer

m = 10^n - 36

k = sum of digits of m.

n = 2 --> m = 64
n = 3 --> m = 964
n = 4 --> m = 9964
n = 5 --> m = 99964

The last 2 digits of m is 6 and 4 and 6 + 4 = 10.
The number of 9's in m = value of n - 2

For k to be a multiple of 9, the number of 9's in 'm' must be a multiple of multiple of 5.

A. 107 --> 105 9's

Answer: A

RMD007 · Answer

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.

onamarif · Answer

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

JeffTargetTestPrep · Answer

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

hellosanthosh2k2 · Answer

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)

Kinshook · Answer

Given: Positive integer m is defined such that  m=10^n−36 . Positive integer k is defined such that k equals the sum of the digits of integer m.

Asked: For which of the following values of n is k a multiple of 5?

m=10^n−36  
If n = 2; m = 64
If n = 3; m = 964
If n = 4; m = 9964

If n = n; m = 999.... (n-2)times64
k = Sum of digits of m = (n-2)9+6+4 = 10+9(n-2) = 10 + 9n -18 = 9n - 8 ; k is a multiple of 5

A. 107; k = 9*107-8 = 963 - 8 = 955; k is a multiple of 5
B. 108; k = 9*108-8 = 972 - 8 = 964; k is NOT a multiple of 5
C. 109; k = 9*109-8 = 981 - 8 = 973; k is NOT a multiple of 5
D. 110; k = 9*110-8 = 990 - 8 = 982; k is NOT a multiple of 5
E. 111; k = 9*111-8 = 999 - 8 = 991; k is NOT a multiple of 5

IMO A

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

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