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Problem 563

AMC 10/12, early questions

Number theory Difficulty 3.6 Multiple choice CEMC Cayley · Canada · 2023

A factory makes chocolate bars. Five boxes, labelled $V$ , $W$ , $X$ , $Y$ , $Z$ , are each packed with 20 bars. Each of
the bars in three of the boxes has a mass of 100 g. Each of the bars in
the other two boxes has a mass of 90 g. One bar is taken from box $V$ , two bars are taken from box $W$ , four bars are taken from box $X$ , eight bars are taken from box $Y$ , and sixteen bars are taken from box
$Z$ . The total mass of these bars
taken from the boxes is 2920 g. The boxes containing the 90 g bars are
labelled

Pick one

Official solution

The number of bars taken from the boxes is $1+2+4+8+16=31$ .

If these bars all had mass 100 g, their total mass would be 3100
g.

Since their total mass is 2920 g, they are $\$3100\text{}$ g} - $2920\text{}$ g} = $180\text{}$
g}$ lighter.

Since all of the bars have a mass of 100 g or of 90 g, then it must be
the case that 18 of the bars are each 10 g lighter (that is, have a mass
of 90 g).

Thus, we want to write 18 as the sum of two of 1, 2, 4, 8, 16 in order
to determine the boxes from which the 90 g bars were taken.

We note that $18 = 2 + 16$ and so the
90 g bars must have been taken from box $W$ and box $Z$ . Can you see why this is unique?

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.