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26 Nov 2021, 01:41
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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. \(70\)
A. \(3\)
B. \(6\)
C. \(7\)
D. \(60\)
E. \(70\)
26 Nov 2021, 01:41
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Bookmarks
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: \(abc-cba=297\);
\((100a + 10b + c ) - (100c + 10b + a) = 297\);
\(a-c=3\).
\((a, c)\) can be (9, 6), (8, 5), (7, 4), (6, 3), (5, 2), and (4, 1) (6 pairs). Notice that (3, 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: \(abc-cba=297\);
\((100a + 10b + c ) - (100c + 10b + a) = 297\);
\(a-c=3\).
\((a, c)\) can be (9, 6), (8, 5), (7, 4), (6, 3), (5, 2), and (4, 1) (6 pairs). Notice that (3, 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
