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2002 AMC 10A Problem 15

2002 AMC 10A Problem 15AMC 10 Step by Step

Name: Date: September 13, 2026

Primes · difficulty 2 of 5 · about 2 min · position P11-15

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?

A)
$150$
B)
$160$
C)
$170$
D)
$180$
E)
$190$

Answer

Key insight

Two-digit primes cannot end in 2, 4, 5 or 6, so those are the tens digits and 1, 3, 7, 9 the units digits; the sum is forced.

Solution

A two-digit prime is odd and not a multiple of $5$ , so its units digit must be $1$ , $3$ , $7$ or $9$ . The eight given digits contain exactly four of these, namely $1, 3, 7, 9$ , and four primes need four units digits, so these are the units digits. The other four digits, $2, 4, 5, 6$ , must all be tens digits.

Whatever the pairing, the sum of the four numbers is
$10\,(2 + 4 + 5 + 6) + (1 + 3 + 7 + 9) = 170 + 20 = 190.$

(Such a pairing does exist, for example $23, 41, 59, 67$ , so the problem is consistent.)

The answer is $\boxed{\textbf{(E)}\ 190}$ .

Why this works

The sum of several two-digit numbers depends only on which digits sit in the tens place and which in the units place, not on how they are matched. Divisibility by $2$ and $5$ decides the units digits completely, so the sum is forced before any prime is written down. Look for what is invariant under the unknown choices.

The trap

Trying to actually build the four primes by trial and error, which is slow and unnecessary since the sum does not depend on the pairing.

Common mistakes

Trying to actually build the four primes by trial and error, which is slow and unnecessary since the sum does not depend on the pairing.
Allowing $5$ as a units digit (thinking of the prime $5$ ) or $1$ as a tens digit, which breaks the forced split.

Techniques

Set up the equation/formula and compute; no special trick needed · Use an invariant, parity, or coloring argument