Competition problem — Number theory

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Number theory Difficulty 5.7 AIME, harder Prove it Ukraine

Let $[x_1, x_2]$ and $[y_1, y_2, y_3]$ be the least common multiple of $x_1, x_2$ and $y_1, y_2, y_3$ respectively. For any positive integers $a, b, c, d$ let $A$ and $B$ be such that:
$A = [a, b, c] \cdot [a, b, d] \cdot [a, c, d] \cdot [b, c, d] \text{ and } B = [a, b] \cdot [a, c] \cdot [a, d] \cdot [b, c] \cdot [b, d] \cdot [c, d].$

Show that $A^6 \geq B^4$ .

Solution

Take prime $p$ that divides $abcd$ . Without loss of generality, let the prime number be a factor of $a, b, c, d$ of degree $a_1 \geq b_1 \geq c_1 \geq d_1$ respectively. We can find the biggest degree of $p$ — $p_A$ and $p_B$ , that divide $A$ and $B$ respectively.

$p_A = 3a_1 + b_1, \quad p_B = 3a_1 + 2b_1 + c_1.$

Then degrees $p'_A$ and $p'_B$ for numbers $A^6$ and $B^4$ equal:

$p'_A = 6p_A = 18a_1 + 6b_1 \geq p'_B = 4p_B = 12a_1 + 8b_1 + 4c_1.$

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.