Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.1 AIME, harder Find the answer

9. Factory A bought a machine, which produces a certain product every day. After several days, Factory B also bought the same machine and started producing the same product at the same speed.

When the total number of products produced by Factory A was 6 times that of Factory B, both factories had produced for an integer number of days, at which point they each bought another identical machine and put it into production; when the total number of products produced by Factory A was 4 times that of Factory B, both factories had produced for an integer number of days, at which point they each bought two more identical machines and put them into production; when the total number of products produced by Factory A was 2 times that of Factory B, both factories had produced for an integer number of days, at which point Factory A had been producing the product for more than four months.

If Factory A started producing the product on January 1, 2009, and each machine was used every day with a constant production efficiency, then, when the total number of products produced by Factory A was 2 times that of Factory B, the date was $\qquad$ month day in 2009.

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

9.5 9th May.

Let when the total number of products produced by Factory A is 6 times that of Factory B, Factory A has produced for $x$ days, and Factory B has produced for $\frac{x}{6}$ days. After another $y$ days, the total number of products produced by Factory A is 4 times that of Factory B. Then
$x+2 y=4\left(\frac{x}{6}+2 y\right) \text {, }$

which means $x=18 y$ .
Let after another $z$ days, the total number of products produced by Factory A is 2 times that of Factory B. Then
$x+2 y+4 z=2\left(\frac{x}{6}+2 y+4 z\right),$

which means $\frac{2}{3} x=2 y+4 z$ .
From equation (1) we get
$5 y=2 z$ .
Thus, $y$ is a multiple of 2.
Let $y=2 n\left(n \in \mathbf{N}_{+}\right)$ . Then
$x+y+z=18 y+y+\frac{5}{2} y=43 n \text {. }$

From the given information,
$\begin{array}{l} 120=31+28+31+30 \\ \leqslant 43 n \leqslant 31+28+31+30+31=151 . \end{array}$

Therefore, $x+y+z=43 n=129$ .
So, $129-31-28-31-30=9$ .
Thus, when the total number of products produced by Factory A is 2 times that of Factory B, the date is May 9, 2009.

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