Problem 17 · 2018 Math Kangaroo
Hard
Number Theory
place-valuedivisibility
How many three-digit numbers have the property that the two-digit number obtained by deleting the middle digit is exactly one ninth of the original number?
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Answer: D — 4
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Hint 1 of 2
Write the number as \(100a+10b+c\); deleting the middle digit gives \(10a+c\).
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Hint 2 of 2
The condition \(100a+10b+c = 9(10a+c)\) simplifies to a relation between the digits.
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Approach: set up the place-value equation and count solutions
- Let the number be \(100a+10b+c\); deleting the middle digit gives \(10a+c\).
- The condition \(100a+10b+c = 9(10a+c)\) becomes \(10a+10b = 8c\), i.e. \(5(a+b)=4c\).
- So \(c\) is a multiple of 5, and \(c\neq 0\) forces \(c=5\), giving \(a+b=4\).
- With \(a\ge 1\): \((a,b)=(1,3),(2,2),(3,1),(4,0)\) — 4 numbers (135, 225, 315, 405).
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