Problem 25 · 2024 Math Kangaroo
Stretch
Number Theory
digit-sumplace-value
The sum of the digits of N is three times the sum of the digits of \(N+1\). What is the smallest possible sum of the digits of N?
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Answer: C — 12
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Hint 1 of 2
Adding 1 only changes the digit sum a lot when N ends in some nines (each trailing 9 turns into a 0).
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Hint 2 of 2
If N ends in exactly k nines, write the digit sum of N+1 in terms of the digit sum of N and solve.
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Approach: use trailing nines to relate the two digit sums, then minimise
- If N ends in exactly \(k\) nines, then \(N+1\) turns those \(k\) nines into zeros and raises the next digit by 1, so \(S(N+1)=S(N)-9k+1\).
- The condition \(S(N)=3\,S(N+1)\) becomes \(S(N)=3(S(N)-9k+1)\), which simplifies to \(2S(N)=27k-3\); the smallest valid case is \(k=1\), giving \(S(N)=12\).
- This is achievable, e.g. \(N=39\) has digit sum 12 and \(N+1=40\) has digit sum 4, with \(12=3\times4\); so the smallest digit sum of N is 12, answer C.
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