Question

An eccentric casino owner decides that his casino should only use chips in $\$5$ and $\$7$ denominations. Which of the following amounts cannot be paid out using these chips?

• A. 31$
• B. 29$
• C. 26$
• D. 23$
• E. 21$

Answer 1

The correct option is D 23$

The payouts will have to be in the sum of some integer number of $5chipsandsomeintegernumberof$ $7$ chips. Which of the answer choices cannot be the sum? One efficient way to eliminate choices is first to cross off any multiples of $7$ and/ or $5$, which eliminates (E). Now, any other possible sums must have at least one $5$ and one $7$ in them. So you can subtract off $5^{\prime }s$ one at a time until your each a multiple of $7$. (It is easier to subtract $5^{\prime }s$ than Z's, because our number system is $base-10$.) Thus:

Answer choice (A): $31-5=26;26-5=21$, a multiple of $7$; this eliminates (A). (In other words, $31=3\times 7+2\times 5.)$

Answer choice (B): $29-5=24;24-5=19;19-5=14$, a multiple of $7$; this eliminates (B).

Answer choice (C): $26-5=21$, a multiple of $7$; this eliminates (C).

So the answer must be (D), $23$. You check by successively subtracting $5$ and looking for multiples of $7:23-5=18$, not a multiple of $7;18-5=13$, also not a multiple of $7;13-5=8$, not a multiple of $7$; and no smaller result will be a multiple of $7$ either.