Problem 25 · 2025 Math Kangaroo
Stretch
Number Theory
caseworkdivisibility
A four-digit number ABCD is multiplied by its units digit D. The result is a different four-digit number DXYA, whose units and thousands digits are swapped compared to the original number (that is, ABCD × D = DXYA). Different letters can stand for the same digits. How many four-digit numbers ABCD have this property?
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Answer: E — 11
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Hint 1 of 3
The product is still a 4-digit number whose thousands digit is the original units digit \(D\) and whose units digit is the original thousands digit \(A\).
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Hint 2 of 3
First nail down \(A\) and \(D\) from the size and the last-digit rules, then see how free the middle digits are.
Still stuck? Show hint 3 →
Hint 3 of 3
Since multiplying by \(D\) only just stays 4-digit, \(A\) must be very small; the units digit of \(A\times D\) must equal \(A\).
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Approach: pin the end digits, then count the free middle
- The result \(DXYA\) starts with \(D\) and ends with \(A\); for \(ABCD\times D\) to stay 4-digit while its leading digit jumps to \(D\), the only fit is \(A=1\) and \(D=9\) (then \(1BC9\times9\) lands in the 9000s and ends in \(\dots1\), since \(9\times9=81\)).
- With \(A=1,\,D=9\) the number is \(\overline{1BC9}\); checking shows every such number works because \(\overline{1BC9}\times9\) automatically becomes \(\overline{9XY1}\).
- So \(B\) and \(C\) are free over \(1009,1019,\dots,1109\) — that is 11 numbers, answer E.
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