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If a three-digit positive integer has its digits reversed, the

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If a three-digit positive integer has its digits reversed, the [#permalink]

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New post 07 Sep 2017, 04:57
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A
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D
E

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Question Stats:

38% (01:32) correct 62% (01:58) wrong based on 105 sessions

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Fresh GMAT Club Tests' Challenge Question:



If a three-digit positive integer has its digits reversed, the resulting three-digit positive integer is less than the original integer by 297. How many such pairs are possible?

A. 3
B. 6
C. 7
D. 60
E. 70
[Reveal] Spoiler: OA

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Re: If a three-digit positive integer has its digits reversed, the [#permalink]

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New post 07 Sep 2017, 05:17
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IMO D

abc -cba =297
100a+10b+c - (100c+10b+a) = 297
100(a-c) +c-a = 297
99(a-c) =297
a-c =3
b can take 0-9 => 10 values
but c can not be 0 since reverse number is also 3 digit humber
so c can take 1,2,3,4,5,6 => 6 values
Hence total numbers possible
6x10=60

D is the Answer
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Re: If a three-digit positive integer has its digits reversed, the [#permalink]

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New post 07 Sep 2017, 05:55
Bunuel wrote:

Fresh GMAT Club Tests' Challenge Question:



If a three-digit positive integer has its digits reversed, the resulting three-digit positive integer is less than the original integer by 297. How many such pairs are possible?

A. 3
B. 6
C. 7
D. 60
E. 70

100x+10y+z-100z+10y+x=297
99(x-z)=297
x-z = 3
there are 6 possible numbers for x. for example: 9y6-6y9=297 and so on till x=4
as for y. there are 10 possibilities: from 0 to 9
so 6*10=60 pairs possible.
so answer is D

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Re: If a three-digit positive integer has its digits reversed, the [#permalink]

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New post 07 Sep 2017, 06:39
Ans: D

N=100x+10y+z
=> (100x+10y+z) - (100x+10y+z) =297
=> x-z =3
(1) x <=9, z <=6 <=> Total 6 pairs z=1,2,3,4,5 & 6
(2) y can be 0 to 9(Total 10)

Total possibilities = 6*10=60

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Re: If a three-digit positive integer has its digits reversed, the [#permalink]

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New post 07 Oct 2017, 04:44
sahilvijay wrote:
IMO D

abc -cba =297
100a+10b+c - (100c+10b+a) = 297
100(a-c) +c-a = 297
99(a-c) =297
a-c =3
b can take 0-9 => 10 values
but c can not be 0 since reverse number is also 3 digit humber
so c can take 1,2,3,4,5,6 => 6 values
Hence total numbers possible
6x10=60

D is the Answer


What about a?
b*c= 10 * 6= 60
so what about a? where have it gone?

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Re: If a three-digit positive integer has its digits reversed, the [#permalink]

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New post 10 Oct 2017, 01:00
lor12345 wrote:
sahilvijay wrote:
IMO D

abc -cba =297
100a+10b+c - (100c+10b+a) = 297
100(a-c) +c-a = 297
99(a-c) =297
a-c =3
b can take 0-9 => 10 values
but c can not be 0 since reverse number is also 3 digit humber
so c can take 1,2,3,4,5,6 => 6 values
Hence total numbers possible
6x10=60

D is the Answer




What about a?
b*c= 10 * 6= 60
so what about a? where have it gone?




No need to consider a - above solution is self explanatory
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Re: If a three-digit positive integer has its digits reversed, the [#permalink]

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New post 11 Oct 2017, 17:56
Answer is D
Possible combinations are 4b1, 5b2, 6b3,7b4,8b5, 9b6
B can be any dogit from 0-9 , hence 10 possiblities. and the above combination are only 6 , therefore answer is 6*10 = 60

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Re: If a three-digit positive integer has its digits reversed, the   [#permalink] 11 Oct 2017, 17:56
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