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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, 03:57
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Question Stats:

31% (02:29) correct 69% (02:34) wrong based on 236 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

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

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New post Updated on: 17 Dec 2017, 02:49
4
5
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 number
c max = 6 because 6+3 =9 =a max
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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Originally posted by sahilvijay on 07 Sep 2017, 04:17.
Last edited by sahilvijay on 17 Dec 2017, 02:49, edited 2 times in total.
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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, 04:55
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, 05:39
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, 03:44
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, 00:00
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, 16:56
1
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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If a three-digit positive integer has its digits reversed, the  [#permalink]

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New post 18 Sep 2018, 19:59
1
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


xyz-zyx=297
297/99=3=x-z
6 possible x-z combinations: 9-6,8-5,7-4,6-3,5-2,4-1
10 possible y values: 0-9
6*10=60 possible pairs
D
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If a three-digit positive integer has its digits reversed, the &nbs [#permalink] 18 Sep 2018, 19:59
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