Problem 21 · 2014 Math Kangaroo
Stretch
Counting & Probability
factorizationcareful-counting
In the equation \(N \times U \times (M + B + E + R) = 33\) each letter stands for a different digit (0, 1, 2, …, 9). In how many different ways can the letters be replaced by different digits?
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Answer: D — 48
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Hint 1 of 2
33 factors very few ways into three positive integers using single digits.
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Hint 2 of 2
Fix which factor is the bracket sum, then count digit arrangements.
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Approach: factor 33, pin the bracket sum, then count digit arrangements
- The bracket \(M+B+E+R\) is a sum of four different digits, so it lies between \(0+1+2+3=6\) and 30; the only divisor of 33 in that range is 11, forcing \(N\times U=3\), i.e. \(\{N,U\}=\{1,3\}\).
- The four bracket digits must be different and avoid 1 and 3; the smallest four-digit total from the remaining digits is \(0+2+4+5=11\), so \(\{M,B,E,R\}=\{0,2,4,5\}\) is the only possibility.
- Arrangements: \(2\) ways to assign \(N,U\) times \(4!=24\) ways to assign the bracket digits give \(2\times24=\) 48 ways.
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