Remainder - tens digit

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Remainder - tens digit

by mgm » Sun Jul 21, 2013 11:08 am
What is the value of the two-digit positive integer n?

(1) When n is divided by 5, the remainder is equal to the tens digit of n.

(2) When n is divided by 9, the remainder is equal to the tens digit of n.

OA C

I guess this can be solved by listing out possible values but is there a better approach ?
Source: — Data Sufficiency |

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by [email protected] » Sun Jul 21, 2013 11:42 am
Hi mgm,

Listing the possibilities is a great way to PROVE what the correct answer is. DS questions have no "safety net", so you'll have very little chance to catch a mistake if you make one. The only way to know that you're correct is to have proof (and listing the possibilities is how you get that proof).

In this question, there are some Number Property shortcuts that you might find helpful (and they'll help you to save some time at the end of this question). Here's the big rule:

When dealing with remainders, what you're dividing by will limit the possibilities. In Fact 1, by dividing by 5, your remainder can only be 0, 1, 2, 3 or 4. It CAN'T be 5 and it CAN'T be greater than 5. This means that the 2-digit number CAN'T be > 49 because the 10s digit wouldn't match the remainder.

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Rich
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by Brent@GMATPrepNow » Sun Jul 21, 2013 12:04 pm
mgm wrote:What is the value of the two-digit positive integer n?

(1) When n is divided by 5, the remainder is equal to the tens digit of n.

(2) When n is divided by 9, the remainder is equal to the tens digit of n.

OA C

I guess this can be solved by listing out possible values but is there a better approach ?
I don't believe there's an approach that doesn't involve listing. But, as Rich demonstrates, we can apply some logic.

Target question: What is the value of the two-digit positive integer n?

Statement 1: When n is divided by 5, the remainder is equal to the tens digit of n
If the tens digit of n is 1, then n could be 11 or 15
If the tens digit of n is 2, then n could be 22 or 27
If the tens digit of n is 3, then n could be 33 or 38
Etc.
So, n could equal 11, 16, 22, 27, 33, 38, 44, or 49
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: When n is divided by 9, the remainder is equal to the tens digit of n
If the tens digit of n is 1, then n could be 10 or 19
If the tens digit of n is 2, then n could be 20 or 29
If the tens digit of n is 3, then n could be 30 or 39
Etc.
So, n could equal 10, 19, 20, 29, 30, 39, 40, 49, 50, 59 . . . 90 or 99
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined:
There's only one value of n that both lists have in common.
So n must = 49
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer = C

Cheers,
Brent
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by Matt@VeritasPrep » Sun Jul 21, 2013 10:10 pm
We can do it without listing, I think.

Let's say our two digit number is ab, where a is the tens digit and b is the units digit. One way to represent this number is 10a + b, so let's do that.

S1 tells us 10a + b = 5x + a. (The number is equal to a multiple of 5 plus the remainder of the number's tens digit.)

S2 tells us 10a + b = 9y + a. (Same idea as above.)

Cleaning up our equations, we have 9a + b = 5x and 9a + b = 9y.

Working with the second equation, if 9a + b = 9y, b is either 0 or 9, but b cannot be 0, as if the number ended in 0 it would leave no remainder when divided by 5. So b is 9.

Now 9a + 9 = 5x, or 9(a+1) = 5x. That means x is a multiple of 9, and since we're dealing with a two-digit number that doesn't end in 0, x must be 9. Hence 9(a+1) = 5*9, or (a+1) = 5, or a = 4.

So ab = 49. This is probably how I would do it on the test, but it does require a little number theory.