A symmetric number of an another one is a number where the

Question

A symmetric number of an another one is a number where the digit are reversed. for istance 123 is the symmetric of one of 321. thus the different of a number and its symmetrical must be divisible by which of the following?

A 4
B 5
C 6
D 7
E 9

I really dont know the answer... can you explain me step by step... thank you for your help;)

A. 4
B. 5 
C. 6
D. 7
E. 9

Bunuel · Accepted Answer

Let's consider the example of three digit symmetric numbers {abc} and {cba}. Three digit number can be represented as: {abc}=100a+10b+c and {cba}=100c+10b+a. The difference would be:

{abc}-{cba}=100a+10b+c-(100c+10b+a)=99a-99c=99(a-c).

Two digit: {ab} and {ba}. {ab}-{ba}=10a+b-(10b+a)=9a-9b=9(a-b)

Hence the difference of two symmetric numbers (2 digit, 3 digit, ...) will always be divisible by 9.

Answer: E.

mrblack · Answer

You can also try plugging in some numbers.

2 digits: 34 and 43 -> the difference is 9, which is divisible by 9
3 digits: 123 and 321 -> the difference is 198, which is divisible by 9
3 digits: 111 and 111 -> the difference is 0, which is divisible by 9

LM · Answer

Should be 9 because the difference will always be a multiple of 9.

For example,

xyz = 100x+10y+z
zyx= 100z+10y+x

(zyx) - (xyz) = 99(z-x)

which si always divisible by 9 or 11. Therefore the answer choice is E.

forsaken11 · Answer

I like your solution as it taught me a new way of doing this problem. But I tried another way and was wondering if I am correct in my approach.

Can we not use the divisibility property of nine here? Any number is divisible by nine if the sum of its digits is equal to nine. A number symmetrical to the original number will still have the sum of its digit equal to nine. Hence, 9 is the answer.

harshavmrg · Answer

Hi Forsaken...i guess you are assuming here that the digits taken as example are delebrately taken which are already divisble by 9...

As usual Brunel's explanation is really awasome and teaches new ways...

harshavmrg · Answer

For example lets take 21 and its reverse is 12....and we are asking if the difference between them is divisble by 9....indeed it is..but if you observe the numbers 21 and 12 are not divisble by 9..

hopefully this answers your question 
 Can we not use the divisibility property of nine here?  
that we cannot assume it that way

vinayrsm · Answer

for this question, we could also quickly pick 2 numbers and see what we get. 100 & 1 => diff of 99, 20 & 2=> diff of 18, 9 is the largest common number.

ichauhan.gaurav · Answer

i am totally agree with your solution but the question is asking about "the number and difference must be divisible".
can you please expleain

Bunuel · Answer

The questions asks: the  difference  of a number and its symmetrical must be divisible by which of the following...

EMPOWERgmatRichC · Answer

Hi All,

This question is actually based on a 'math truism' of a sort. If you've ever taken an accounting class, then you might have learned about "transpositional errors" - errors that happen when you put the same digits in the wrong "order." While this question focuses on symmetrical numbers, the issue is exactly the same and works in any variation.

Putting a set of digits in a different order will ALWAYS lead to a difference that is divisible by 9.

23 and 32 is a difference of 9, which is divisible by 9
147 and 714 is a difference of 567, which is divisible by 9
34567 and 56374 is a difference of 21,807 which is divisible by 9

It goes to show that some of the concepts that you're going to learn for the GMAT have some value even after you're done with the Test.

GMAT assassins aren't born, they're made,
Rich

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