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Intern  Joined: 23 Dec 2009
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A symmetric number of an another one is a number where the  [#permalink]

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Difficulty:   25% (medium)

Question Stats: 72% (01:19) correct 28% (01:32) wrong based on 594 sessions

HideShow timer Statistics A symmetric number of an another one is a number where the digit are reversed. For instance, 123 is the symmetric of one of 321. Thus the difference of a number and its symmetrical must be divisible by which of the following?

A. 4
B. 5
C. 6
D. 7
E. 9
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Re: NUMBER PROPERTIES HARD QUESTION! HELPP!!!  [#permalink]

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lucalelli88 wrote:
A symmetric number of an another one is a number where the digit are reversed. For instance, 123 is the symmetric of one of 321. Thus the difference 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;)

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.

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Posts: 170
Re: NUMBER PROPERTIES HARD QUESTION! HELPP!!!  [#permalink]

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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
Director  Joined: 03 Sep 2006
Posts: 744
Re: NUMBER PROPERTIES HARD QUESTION! HELPP!!!  [#permalink]

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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.
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Posts: 49
Re: NUMBER PROPERTIES HARD QUESTION! HELPP!!!  [#permalink]

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Bunuel wrote:
lucalelli88 wrote:
A symmetric number of an another one is a number where the digit are reversed. For instance, 123 is the symmetric of one of 321. Thus the difference 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;)

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.

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.
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Re: NUMBER PROPERTIES HARD QUESTION! HELPP!!!  [#permalink]

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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...
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Re: NUMBER PROPERTIES HARD QUESTION! HELPP!!!  [#permalink]

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forsaken11 wrote:
Bunuel wrote:
lucalelli88 wrote:
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;)

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.

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.

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..

[/quote]Can we not use the divisibility property of nine here? [/quote]
that we cannot assume it that way
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Regards,
Harsha

Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat Satyameva Jayate - Truth alone triumphs
Manager  Joined: 14 Apr 2011
Posts: 165
Re: NUMBER PROPERTIES HARD QUESTION! HELPP!!!  [#permalink]

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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.
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Re: NUMBER PROPERTIES HARD QUESTION! HELPP!!!  [#permalink]

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Bunuel wrote:
lucalelli88 wrote:
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;)

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.

i am totally agree with your solution but the question is asking about "the number and difference must be divisible".
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Chauahan Gaurav
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Posts: 55804
Re: NUMBER PROPERTIES HARD QUESTION! HELPP!!!  [#permalink]

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ichauhan.gaurav wrote:
Bunuel wrote:
lucalelli88 wrote:
A symmetric number of an another one is a number where the digit are reversed. For instance, 123 is the symmetric of one of 321. Thus the difference 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;)

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.

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

The questions asks: the difference of a number and its symmetrical must be divisible by which of the following...
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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: A symmetric number of an another one is a number where the  [#permalink]

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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.

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Rich
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Re: A symmetric number of an another one is a number where the  [#permalink]

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lucalelli88 wrote:
A symmetric number of an another one is a number where the digit are reversed. For instance, 123 is the symmetric of one of 321. Thus the difference of a number and its symmetrical must be divisible by which of the following?

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

Dear Moderator,
Found this " Arithmetic" question under " work/rate problems ", hope you will look into this. Thank you.
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- Stne
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Posts: 55804
Re: A symmetric number of an another one is a number where the  [#permalink]

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stne wrote:
lucalelli88 wrote:
A symmetric number of an another one is a number where the digit are reversed. For instance, 123 is the symmetric of one of 321. Thus the difference of a number and its symmetrical must be divisible by which of the following?

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

Dear Moderator,
Found this " Arithmetic" question under " work/rate problems ", hope you will look into this. Thank you.

_______________
Fixed. Thank you.
_________________ Re: A symmetric number of an another one is a number where the   [#permalink] 17 Sep 2018, 06:14
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