Track / Stage 5 / 375 of 400 #975 of 1964

Problem 975

AIME late

Number theory Difficulty 5.9 Prove it

SUBIECTUL 2

a) Show that the number $\frac{3^{10}+7 \cdot 3^{11}}{8 \cdot 3^{10}+3^{11}}$ is a natural number.

b) Show that if $a$ and $b$ are natural numbers and $2a + 3b$ is divisible by 11, then the fraction $\frac{a+7b}{8a+b}$ is reducible.

R.M.T

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

## SUBJECT 2

a) Show that $\frac{3^{10}+7 \cdot 3^{11}}{8 \cdot 3^{10}+3^{11}}$ is a natural number.

b) Show that if $a$ and $b$ are natural numbers and $2a + 3b$ is divisible by 11, then the fraction $\frac{a+7b}{8a+b}$ is reducible.
a)

$\frac{3^{10}+7 \cdot 3^{11}}{8 \cdot 3^{10}+3^{11}}=\frac{3^{10}(1+7 \cdot 3)}{3^{10}(8+3)}=$	$1 p$
$=\frac{3^{10} \cdot 22}{3^{10} \cdot 11}=2 \in \mathbb{N}$ .	$1 p$

$2a + 3b = \mathbf{M}_{11} \Rightarrow 6(2a + 3b) = \mathbf{M}_{11} \Rightarrow 12a + 18b = \mathbf{M}_{11} \Rightarrow a + 7b = \mathbf{M}_{11}(1)$ .	$2 p$
Then $2a + 3b = \mathbf{M}_{11} \Rightarrow 4(2a + 3b) = \mathbf{M}_{11} \Rightarrow 8a + 12b = \mathbf{M}_{11} \Rightarrow 8a + b = \mathcal{M}_{11}(2)$ .	$2 p$
From (1) and (2), it follows that the fraction $\frac{a+7b}{8a+b}$ is reducible.	$1 p$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.