Problem 24 · 2016 Math Kangaroo
Stretch
Number Theory
place-valuecasework
Two three-digit numbers are made up of six different digits. The first digit of the second number is twice as big as the last digit of the first number. (Note: 0 is also a digit, but cannot be the first digit of a number.) What is the smallest possible sum of the two numbers?
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Answer: C — 537
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Hint 1 of 3
The hundreds digits matter most, so make those two as small as you can.
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Hint 2 of 3
The second number's first digit is twice the first number's last digit, so test small even leading digits like 2 or 4.
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Hint 3 of 3
Once the hundreds and that linked pair are fixed, fill the leftover spots with the smallest unused digits.
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Approach: make the hundreds digits smallest, respecting the doubling rule
- The two hundreds digits drive the sum, so we want them tiny; the smallest nonzero hundreds digit is 1 for the first number.
- The second number's hundreds digit must be double the first number's units digit, and the smallest workable pair is units 2 with hundreds 4, so the numbers look like 1?2 and 4?? .
- Fill the remaining slots with the smallest unused digits 0, 3, 5: that gives 102 and 435, all six digits different.
- Their sum \(102 + 435 = 537\) is the smallest possible, choice (C).
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