Problem 26 · 2025 Math Kangaroo
Stretch
Number Theory
divisibilitycasework
The six-digit number ABCDEF consists of the digits 1, 2, 3, 4, 5 and 6, with each digit occurring exactly once. The number AB consisting of the first two digits is a multiple of 2. The number ABC is a multiple of 3. The number ABCD is a multiple of 4. The number ABCDE is a multiple of 5, and the entire number ABCDEF is a multiple of 6. What values can the digit F take?
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Answer: B — only 4
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Hint 1 of 3
Apply the rules in order: \(AB\) even, \(ABC\) a multiple of 3, \(ABCD\) of 4, \(ABCDE\) of 5, \(ABCDEF\) of 6.
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Hint 2 of 3
Divisible-by-5 with digits 1–6 forces \(E=5\); divisible-by-2 and divisible-by-6 force \(B\), \(D\), \(F\) to be the three even digits.
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Hint 3 of 3
So \(\{B,D,F\}=\{2,4,6\}\) and \(\{A,C\}=\{1,3\}\); pin them down with the multiple-of-3 and multiple-of-4 rules.
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Approach: apply the divisibility conditions step by step
- Since \(ABCDE\) is a multiple of 5, \(E=5\); \(B\) (and \(F\)) are even and \(D\) is even (for the 4-rule), so the even digits 2, 4, 6 fill \(B,D,F\) and the odd digits 1, 3 fill \(A,C\).
- \(ABC\) divisible by 3 needs \(A+B+C\) divisible by 3; with \(A+C=1+3=4\), \(B\) must be 2, leaving \(\{D,F\}=\{4,6\}\). The 4-rule on \(\overline{CD}\) (with \(C\in\{1,3\}\)) forces \(D=6\).
- That leaves \(F=4\), and indeed 123654 and 321654 satisfy every rule, so \(F\) can only be 4, answer B.
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