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Balti Tee 2011 · Valikvooru ülesanne

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Nonnegative integers aa and bb have the following property: d(na)≥d(nb)d(na) \ge d(nb) for each positive integer nn (where d(k)d(k) is the number of divisors of kk). Prove that aa is divisible by bb.

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Let a=p1α1…pmαma = p_1^{\alpha_1} \dots p_m^{\alpha_m}, b=p1β1…pmβmb = p_1^{\beta_1} \dots p_m^{\beta_m} be the prime decompositions of these numbers (we assume that some αk,βk\alpha_k, \beta_k can be equal to 00). Let us check that for each kk αk≥βk\alpha_k \ge \beta_k. Indeed, if the inequality does not hold for some kk, say, α1<β1\alpha_1 < \beta_1, then for n=p2s…pmsn = p_2^s \dots p_m^s we have

1≤d(na)d(nb)=α1(s+α2)…(s+αm)β1(s+β2)…(s+βm)1 \le \frac{d(na)}{d(nb)} = \frac{\alpha_1(s + \alpha_2) \dots (s + \alpha_m)}{\beta_1(s + \beta_2) \dots (s + \beta_m)}

For big ss this fraction is close to α1β1<1\frac{\alpha_1}{\beta_1} < 1. A contradiction.