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The "reflection" of a positive integer is obtained by reversing its digits. For example, 321 is the reflection of 123. The difference between a five-digit integer and its reflection must be divisible by which of the following? a. 2 b. 4 c. 5 d. 6 e. 9

I would appreciate if you provide explanation along with an answer

Re: Number and its reflection PowerPrep [#permalink]

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15 Mar 2009, 05:09

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E? wat is OA? I use POE here. The difference in the last digit of the 5-digit number and its reflection can be odd or even, depending upon the last two digits are odd or even. We are asked to find an answer that is always (must) true...so 2,4,6 all gone ( as they are multiples of 2 ) 5 has a restriction that the difference should end in 0 or 5 which is not always true, so 5 is gone..

9 remains, I tried with 3-4 combinations and always got the diff as multiple of 9

May be someone else can have a better way..

kbulse wrote:

The "reflection" of a positive integer is obtained by reversing its digits. For example, 321 is the reflection of 123. The difference between a five-digit integer and its reflection must be divisible by which of the following? a. 2 b. 4 c. 5 d. 6 e. 9

I would appreciate if you provide explanation along with an answer

Re: Number and its reflection PowerPrep [#permalink]

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15 Mar 2009, 05:20

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I got the answer as e. 9.

Explanation: Take any 5 digit integer as 10000a+1000b+100c+10d+e. Hence its reflection would be 10000e+1000d+100c+10b+a. Difference between the two would be:

9999a+990b-990d-9999e = 9(1111a+110b-110d-1111e). hence this number is divisible by 9.

Process of elimination is a great approach here, since as Economist did above, you can quite quickly rule out wrong answers here. You can also see why 9 is the correct answer if you know one elementary fact from number theory:

-to find the remainder when you divide a number by 9, you can add the digits, and find the remainder of that sum.

Because we can find remainders when dividing by 9 by adding digits (and the same works for 3), the numbers ABCDE and EDCBA must have the same remainder when we divide them by 9. That is:

ABCDE = 9q + r EDCBA = 9k + r

So ABCDE - EDCBA = 9q + r - (9k + r) = 9(q-k)

and their difference is a multiple of 9.
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Re: Number and its reflection PowerPrep [#permalink]

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18 Mar 2009, 16:42

IanStewart wrote:

Process of elimination is a great approach here, since as Economist did above, you can quite quickly rule out wrong answers here. You can also see why 9 is the correct answer if you know one elementary fact from number theory:

-to find the remainder when you divide a number by 9, you can add the digits, and find the remainder of that sum.

Because we can find remainders when dividing by 9 by adding digits (and the same works for 3), the numbers ABCDE and EDCBA must have the same remainder when we divide them by 9. That is:

ABCDE = 9q + r EDCBA = 9k + r

So ABCDE - EDCBA = 9q + r - (9k + r) = 9(q-k)

and their difference is a multiple of 9.

Wow......that is ONE excellent explanation Ian. Thank you for giving another view on solving the problem.

Re: Number and its reflection PowerPrep [#permalink]

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18 Mar 2009, 17:53

my solution is similar to abhishek's @ economist can you please explain "The difference in the last digit of the 5-digit number and its reflection can be odd or even, depending upon the last two digits are odd or even."

Re: Number and its reflection PowerPrep [#permalink]

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18 Mar 2009, 18:57

lav wrote:

my solution is similar to abhishek's @ economist can you please explain "The difference in the last digit of the 5-digit number and its reflection can be odd or even, depending upon the last two digits are odd or even."

Infact the high lighted part itself is enough to knock of ans 2 / 4 / 6. (Since the difference of 5 digit number or for that matter different of any two numbers can either be off / even and we cannot confirmly say that this difference is divisible by 2 / 4 /6)