Brazil 1986/4 solution

8th Brazil 1986

Problem 4

Find all 10 digit numbers a₀a₁...a₉ such that for each k, a_k is the number of times that the digit k appears in the number.

Answer

6210001000

Solution

Note that a₀ + a₁ + ... + a₉ = no. of 0s + no. of 1s + ... + no. of 9s = total no. of digits = 10. In other words the sum of the digits is 10. In particular, if i > 5, then a_i = 0 or 1 (because if a_i ≥ 2, then we have at least 2 digits > 5 with sum > 10).

Suppose some a_i ≥ 6. Then a₆ ≥ 1. But a₆ ≤ 1, so a₆ = 1. Hence a₁ > 0. We cannot have a₁ = 1, because then there are at least two 1s, so a₁ ≥ 2. Contradiction. So a₁ ≥ 2. We cannot have a₁ = 6, because then there are six 1s and one 6, total 12, whereas all the digits only total 10. So a₁ must be some other digit k > 1. But we now have at least six 0s, at least two 1s, a k and a 6, which is at least 10. So it must be exactly 10. Hence a₀ = 6 and a₁ = 2. So the number must be 6210001000.

So suppose all a_i ≤ 5. Then in particular a0 ≤ 5, so there are at least 5 non-zero digits, and hence at least 4 of a₁, a₂, ... , a₉ are non-zero. But that means that a₁ + 2a₂ + 3a₃ + ... + 9a₉ ≥ 1 + 2 + 3 + 4 with equality iff a₁ = a₂ = a₃ = a₄ = 1 and a₅ = a₆ = a₇ = a₈ = a₉ = 0. But we must have equality because a₁ + 2a₂ + 3a₃ + ... + 9a₉ is the sum of the digits which is 10. However it is not possible to have a₁ = a₂ = a₃ = a₄ = 1 and a₅ = a₆ = a₇ = a₈ = a₉ = 0 because then there are four 1s, but a₄ = 1.

8th Brazil 1986