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

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Positive integer m is defined such that m=10n−36. Positive integer k  [#permalink]

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12 Feb 2017, 13:27
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Question Stats:

58% (02:33) correct 42% (02:39) wrong based on 161 sessions

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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

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Re: Positive integer m is defined such that m=10n−36. Positive integer k  [#permalink]

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12 Feb 2017, 20:20
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Bunuel wrote:
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

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
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Re: Positive integer m is defined such that m=10n−36. Positive integer k  [#permalink]

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12 Feb 2017, 20:33
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

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Re: Positive integer m is defined such that m=10n−36. Positive integer k  [#permalink]

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12 Feb 2017, 20:39
Bunuel wrote:
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

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.
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Re: Positive integer m is defined such that m=10n−36. Positive integer k  [#permalink]

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30 Jul 2017, 21:12
$$10^n -36$$ 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
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Re: Positive integer m is defined such that m=10n−36. Positive integer k  [#permalink]

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01 Aug 2017, 16:13
Bunuel wrote:
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

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.

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Re: Positive integer m is defined such that m=10n−36. Positive integer k  [#permalink]

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11 Jan 2018, 05:41
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)
Re: Positive integer m is defined such that m=10n−36. Positive integer k   [#permalink] 11 Jan 2018, 05:41
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