2002 AMC 10A Problems/Problem 15

Problem

Using the digits 1, 2, 3, 4, 5, 6, 7, and 9, form 4 two-digit prime numbers, using each digit only once. What is the sum of the 4 prime numbers?

$\text{(A)}\ 150 \qquad \text{(B)}\ 160 \qquad \text{(C)}\ 170 \qquad \text{(D)}\ 180 \qquad \text{(E)}\ 190$

Solution

Since a multiple-digit prime number is not divisible by either 2 or 5, it must end with 1, 3, 7, or 9 in the units place. The remaining digits given must therefore appear in the tens place. Hence our answer is $20 + 40 + 50 + 60 + 1 + 3 + 7 + 9 = 190\Rightarrow\boxed{(E)= 190}$.

(Note that we did not need to actually construct the primes. If we had to, one way to match the tens and ones digits to form four primes is $23$, $41$, $59$, and $67$.)

(You could also guess the numbers, if you have a lot of spare time.)

Video Solution by Daily Dose of Math

https://youtu.be/4dr5Mk4-hOI

~Thesmartgreekmathdude

See Also

2002 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png