Alex has $\$ 2.65$. He has only dimes (worth $\$ 0.10$ each) and quarters (worth $\$ 0.25$ each). He has more dimes than quarters. What is the smallest number of coins that Alex could have?
(A) 25
(B) 16
(C) 13
(D) 19
(E) 22
Solution 1

Alex has 265 cents in total.

Since 265 is not divisible by 10, Alex cannot have only dimes, so must have at least 1 quarter. If Alex has 1 quarter, then he has $265-25=240$ cents in dimes, so 24 dimes. Alex cannot have 2 quarters, since $265-2(25)=215$ is not divisible by 10 . If Alex has 3 quarters, then he has $265-3(25)=190$ cents in dimes, so 19 dimes. Continuing this argument, we can see that Alex cannot have an even number of quarters, since the total value in cents of these quarters would end in a 0 , making the total value of the dimes end in a 5 , which is not possible.

If Alex has 5 quarters, then he has $265-5(25)=140$ cents in dimes, so 14 dimes.

If Alex has 7 quarters, then he has $265-7(25)=90$ cents in dimes, so 9 dimes.

If Alex has 9 quarters, then he has $265-9(25)=40$ cents in dimes, so 4 dimes.

If Alex has more than 9 quarters, then he will have even fewer than 4 dimes, so we do not need to investigate any more possibilities since we are told that Alex has more dimes than quarters. So the possibilities for the total number of coins that Alex has are $1+24=25,3+19=22$, $5+14=19$, and $7+9=16$.

Therefore, the smallest number of coins that Alex could have is 16 .

(Notice that each time we increase the number of quarters above, we are in effect exchanging 2 quarters (worth 50 cents) for 5 dimes (also worth 50 cents).)

## Solution 2

Suppose that Alex has $d$ dimes and $q$ quarters, where $d$ and $q$ are non-negative integers.

Since Alex has $\$ 2.65$, then $10 d+25 q=265$ or $2 d+5 q=53$.

Since the right side is odd, then the left side must be odd, so $5 q$ must be odd, so $q$ must be odd.

If $q \geq 11$, then $5 q \geq 55$, which is too large.

Therefore, $qq$ and $d+q$ smallest is $q=7$ and $d=9$, giving 16 coins in total.

ANSWER: (B)