Quick way to solve this?

[sanjay21]

Folks,

This is a problem that I came across in powerprep. Could not find a quick way to solve it. Anyone has any ideas?

Question:
The inverse of a 5 digit number is defined as the digits reversed. E.g 12345 is inverse of 54321. The difference between a number and its inverse is divisible by which of the following?

a) 2
b) 4
c) 6
d) 8
e) 9

Only way I could think of is using sample numbers.

Sanjay

[GRE_Kickboxer]

Try it this way:

Assume abcde are all digits.

abcde = 10,000a + 1,000b + 100c + 10d + e

edcba = 10,000e + 1,000d + 100c + 10b + a

abcde - edcba = 9,999a + 9,990b + etc.

Clearly you are getting multiples of nine because all the digits add up to to a multiple of 9.

[tanmoy]

use the "Divisibility Rule of 9".

if u use 12345, the answer would be 2,4,6,9 so try to use random number for example 13472 or else, then you'll got the number, 9.

[tanmoy]

by divisiblity rule any number subtracted with its reflection will be divisible by 9
because if you write say a 5 digit number it could be as below
number = 10000a+1000b+100c+10d+1e
reflection is =10000e+1000d+100c+10b+a
number -reflection=9999a+990b-0c-990d-9999e
so we know that any number (be any number of digit) subtracted by reflection will be divisible by 9

GRE_kickboxer thy absolutely right

[Mazar760]

An other way _could_ be:

Its easy to recognize, that many differences are odd numers.
-> odd numbers are not divisible by even numbers..
So.. Just the 9 left!