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Hellenic Mathematical Olympiad

Algebra Difficulty 5.7 AIME, harder Prove it Greece

Determine all positive integers which are equal to 1313 times the sum of their digits.

Solution

Let κ\kappa be the number of digits of the integer AA which is equal to 1313 times the sum of its digits. The least possible AA is 10κ−110^{\kappa-1}, while the maximal possible sum of the digits is 9κ9\kappa. Therefore we need to have:
10κ−1≤13⋅9κ=117κ.(1) 10^{\kappa-1} \leq 13 \cdot 9\kappa = 117\kappa. \qquad (1)
For κ≥4\kappa \ge 4, we will prove using induction that: 10κ−1>117κ10^{\kappa-1} > 117\kappa, that is, relation (1) is not valid. In fact, for κ=4\kappa = 4 we have: 104−1=103>117⋅4=46810^{4-1} = 10^3 > 117 \cdot 4 = 468. If 10κ−1>117κ10^{\kappa-1} > 117\kappa, for the arbitrary κ>4\kappa > 4, then we get:
10(κ+1)−1=10κ=10⋅10κ−1>10⋅117κ=117⋅10κ>117⋅(κ+1). 10^{(\kappa+1)-1} = 10^{\kappa} = 10 \cdot 10^{\kappa-1} > 10 \cdot 117\kappa = 117 \cdot 10\kappa > 117 \cdot (\kappa+1).
Therefore the number κ\kappa must be less or equal to 33.

* We have to reject the case κ=1\kappa = 1, since A=α<13αA = \alpha < 13\alpha, with 0<α≤90 < \alpha \le 9.
* Similarly we reject the case with κ=2\kappa = 2, since A=10α+β<13(α+β)A = 10\alpha + \beta < 13(\alpha + \beta).
* Let κ=3\kappa = 3 and A=αβγ‾=100α+10β+γA = \overline{\alpha\beta\gamma} = 100\alpha + 10\beta + \gamma, 0<α≤90 < \alpha \le 9, 0≤β,γ≤90 \le \beta, \gamma \le 9.

Then we have:
100α+10β+γ=13⋅(α+β+γ),0<α≤9,0≤β,γ≤9 100\alpha + 10\beta + \gamma = 13 \cdot (\alpha + \beta + \gamma), \quad 0 < \alpha \le 9, 0 \le \beta, \gamma \le 9
⇔87α=3β+12γ,0<α≤9,0≤β,γ≤9 \Leftrightarrow 87\alpha = 3\beta + 12\gamma, \quad 0 < \alpha \le 9, 0 \le \beta, \gamma \le 9
⇔29α=β+4γ,0<α≤9,0≤β,γ≤9 \Leftrightarrow 29\alpha = \beta + 4\gamma, \quad 0 < \alpha \le 9, 0 \le \beta, \gamma \le 9
Since 0≤β+4γ≤45⇒29α≤45⇒α≤10 \le \beta + 4\gamma \le 45 \Rightarrow 29\alpha \le 45 \Rightarrow \alpha \le 1 (since α≠0\alpha \ne 0), and hence:
β+4γ=29⇒β=29−4γ≥0⇒0≤β=29−4γ≤9 \beta + 4\gamma = 29 \Rightarrow \beta = 29 - 4\gamma \ge 0 \Rightarrow 0 \le \beta = 29 - 4\gamma \le 9
⇒−29≤−4γ≤−20⇒5≤γ≤294⇒γ∈{5,6,7} \Rightarrow -29 \le -4\gamma \le -20 \Rightarrow 5 \le \gamma \le \frac{29}{4} \Rightarrow \gamma \in \{5, 6, 7\}
* For γ=5⇒β=9\gamma = 5 \Rightarrow \beta = 9 and A=195A = 195.
* For γ=6⇒β=5\gamma = 6 \Rightarrow \beta = 5 and A=156A = 156.
* For γ=7⇒β=1\gamma = 7 \Rightarrow \beta = 1 and A=117A = 117.

Therefore, the positive integers which are equal to 1313 times the sum of their digits are 117117, 156156, and 195195.

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