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A positive integer is divisible by 9 if and only if the sum of its

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A positive integer is divisible by 9 if and only if the sum of its [#permalink]

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A positive integer is divisible by 9 if and only if the sum of its digits is divisible by 9. If n is a positive integer, for which of the following values of k is \(25*10^n + k*10^2n\) divisible by 9?

(A) 9
(B) 16
(C) 23
(D) 35
(E) 47
[Reveal] Spoiler: OA
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Re: A positive integer is divisible by 9 if and only if the sum of its [#permalink]

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New post 14 Mar 2016, 14:06
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For any value of N, when you multiply by 10ˆn or 10ˆ2n you will be only adding zeros. What you only need to do is check if the sum of 2+5+the other possible values for k in the question add up to 9. The only possible answer is E.

Last edited by marcelonac on 15 Mar 2016, 12:11, edited 1 time in total.
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Re: A positive integer is divisible by 9 if and only if the sum of its [#permalink]

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New post 15 Mar 2016, 07:29
marcelonac wrote:
For any value of N, the sum of the digits of 10ˆn and 10ˆ2n will always be 1. Since they're being multiplied by 25 and k, what you only need to do is check if the sum of 2+5+the other possible valued for k in the question add up to 9. The only possible answer is E.


really neat explainantion +1Kudo
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Re: A positive integer is divisible by 9 if and only if the sum of its [#permalink]

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New post 15 Mar 2016, 23:55
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Excellent Question
Here Taking out 10^n common => 100k+25 must be divisible by 9
checking values (try and start with last in such questions of plugging in values ) we get => E
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Re: A positive integer is divisible by 9 if and only if the sum of its [#permalink]

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New post 27 Dec 2016, 06:50
Here is my approach:

It tells us that 25 × 10^n + k × 10^(2n) is divisible by 9 and we know that 10^whatever is not divisible by 9. So we just plug in numbers to find a number that satisfies that the sum of its digits (25 + k) is divisible by 9.
Starting with option E: 25 + 47 = 72 and 7 + 2 = 9. Hence, when k=47 the number is divisible by 9.
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Re: A positive integer is divisible by 9 if and only if the sum of its [#permalink]

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bigdady wrote:
A positive integer is divisible by 9 if and only if the sum of its digits is divisible by 9. If n is a positive integer, for which of the following values of k is 25*10^n + k*10^2n divisible by 9?

(A) 9
(B) 16
(C) 23
(D) 35
(E) 47


25 = 7 (mod 9) = -2 (mod 9)

\(10^n = 10^{2n}\) = 1 (mod 9)

-2 * 1 + k*1 = k - 2

k should have remainder 2 when divided by 9 to give us total remainder 0.

Only option which leaves remainder 2 upon division by 9 is 47 (E).
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Re: A positive integer is divisible by 9 if and only if the sum of its [#permalink]

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New post 27 Jan 2017, 05:18
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This looks like >620 imho

Solution:

25*10^n + k*10^2n = (25 * 10^n) + (k* 10^n * 10^n) {you should be able to see that 10^2n breaks into 10^n * 10^n else work your exponents roots in algebra}

then you can do 10^n * [ 25 +(k * 10^n) ] And here the logic/critical thinking begins

a) The only thing that the very first 10^n does to the number within the square brackets is simply padding it with zero at the end. (as other users suggested before). So it doesn't play any role to the divisibility hence can be fully ignored (N.B. n>0).

b) now 25 + (k * 10^n) simply is K * mul(10) + 25 therefore you care only about the sum of the digits K, 2 and 5. By using "back-solving" you can find E

Therefore E.

good luck!
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Re: A positive integer is divisible by 9 if and only if the sum of its [#permalink]

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New post 28 Jan 2017, 11:32
stonecold wrote:
Excellent Question
Here Taking out 10^n common => 100k+25 must be divisible by 9
checking values (try and start with last in such questions of plugging in values ) we get => E


nice explanation stonecold, thank you :)
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Re: A positive integer is divisible by 9 if and only if the sum of its [#permalink]

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New post 17 Feb 2017, 04:53
Korhand wrote:
stonecold wrote:
Excellent Question
Here Taking out 10^n common => 100k+25 must be divisible by 9
checking values (try and start with last in such questions of plugging in values ) we get => E


nice explanation stonecold, thank you :)



I think the above way misses a crucial point.

Since n>0 then 10^n can be 10 and not 100 (i.e. n = 1 since n>0). Therefore the original relationship deduces to k * 10. So in order to be on the safe side I recommend to solve fro two values (more than two is an overkill imho)

10* k
100 * k
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Re: A positive integer is divisible by 9 if and only if the sum of its [#permalink]

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bigdady wrote:
A positive integer is divisible by 9 if and only if the sum of its digits is divisible by 9. If n is a positive integer, for which of the following values of k is 25*10^n + k*10^2n divisible by 9?

(A) 9
(B) 16
(C) 23
(D) 35
(E) 47


We need to determine for which value of k 25*10^n + k*10^2n is divisible by 9.

We see that 10^n and 10^2n will always have a digit of 1 and then zeros. So, excluding k, the sum of the digits in our expression is 2 + 5 = 7 (since (25)(10^n) has a 2, 5, and zeros).

We need to determine, of our answer choices, which when added to 7 will produce a sum that is divisible by 9. Scanning our answer choices, we see that 47 is the correct answer.

2 + 5 + 4 + 7 = 18, which is divisible by 9.

Answer: E
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Re: A positive integer is divisible by 9 if and only if the sum of its [#permalink]

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I'm just using ridiculous numbers for these questions


so let n be= 10000000000000000

then we get something like:

250000000000.....+k00000000000000000000........

from this it becomes clear that we just have to look for 2+5 + k divisible by 9

7+47=54 which is divisible by 9
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Re: A positive integer is divisible by 9 if and only if the sum of its [#permalink]

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Hi GMATters,

Here's my video explanation of this question:



Enjoy!

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Last edited by Bunuel on 28 Sep 2017, 07:39, edited 3 times in total.
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Re: A positive integer is divisible by 9 if and only if the sum of its [#permalink]

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Thanks for the explanation, very helpful. The youtube link doesn't work, just wanted to let you know ygcrowanhand
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Re: A positive integer is divisible by 9 if and only if the sum of its [#permalink]

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Re: A positive integer is divisible by 9 if and only if the sum of its [#permalink]

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New post 10 Sep 2017, 13:48
ScottTargetTestPrep Bunuel mikemcgarry IanStewart shashankism Engr2012

Quote:
We see that 10^n and 10^2n will always have a digit of 1 and then zeros. So, excluding k, the sum of the digits in our expression is 2 + 5 = 7 (since (25)(10^n) has a 2, 5, and zeros).


Why do we not add for 10^x here and only consider 2 + 5 = 7
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Re: A positive integer is divisible by 9 if and only if the sum of its [#permalink]

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adkikani wrote:
ScottTargetTestPrep Bunuel mikemcgarry IanStewart shashankism Engr2012

Quote:
We see that 10^n and 10^2n will always have a digit of 1 and then zeros. So, excluding k, the sum of the digits in our expression is 2 + 5 = 7 (since (25)(10^n) has a 2, 5, and zeros).


Why do we not add for 10^x here and only consider 2 + 5 = 7



Try to see it this way: 25*10 = 250 (sum of digits = 2+5); similarly 25*10^6 = 25000000 (sum of the digits = 2+5) etc. So in all cases, the sum of the digits for 25*(10^n) will always be 2+5 = 7.

Hope this helps.

P.S. the easiest/most straightforward way for this would have been to assume n=1 and then play with smaller numbers.
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Re: A positive integer is divisible by 9 if and only if the sum of its [#permalink]

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New post 11 Sep 2017, 07:45
What's the level of this question?

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Re: A positive integer is divisible by 9 if and only if the sum of its [#permalink]

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Re: A positive integer is divisible by 9 if and only if the sum of its   [#permalink] 11 Sep 2017, 07:48
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