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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
M37-46
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Updated on: 17 Dec 2017, 03:49
Originally posted by sahilvijay on 07 Sep 2017, 05:17.
Last edited by sahilvijay on 17 Dec 2017, 03:49, edited 2 times in total.
Last edited by sahilvijay on 17 Dec 2017, 03:49, edited 2 times in total.
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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 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
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
26 Nov 2021, 01:42
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Official Solution:
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. \(70\)
Say the original three-digit number is \(abc\) and the reversed number is \(cba\).
Given: \(abc-cba=297\);
\((100a + 10b + c ) - (100c + 10b + c) = 297\);
\(a-c=3\).
\((a, c)\) can be (9, 6), (8, 5), (7, 4), (6, 3), (5, 2), and (4, 1) (6 pairs). Notice that (6, 0) pair is not possible: \(c\) cannot be 0 because in this case \(cba\) would be a two-digit number, not a three-digit number as given in the stem.
Finally, don't forget about \(b\). It can take 10 values: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.
So, there are a total of \(6*10=60\) numbers are possible.
Answer: D
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. \(70\)
Say the original three-digit number is \(abc\) and the reversed number is \(cba\).
Given: \(abc-cba=297\);
\((100a + 10b + c ) - (100c + 10b + c) = 297\);
\(a-c=3\).
\((a, c)\) can be (9, 6), (8, 5), (7, 4), (6, 3), (5, 2), and (4, 1) (6 pairs). Notice that (6, 0) pair is not possible: \(c\) cannot be 0 because in this case \(cba\) would be a two-digit number, not a three-digit number as given in the stem.
Finally, don't forget about \(b\). It can take 10 values: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.
So, there are a total of \(6*10=60\) numbers are possible.
Answer: D
