Printable

2015 AMC 10B Problem 3

2015 AMC 10B Problem 3AMC 10 Step by Step

Name: Date: September 13, 2026

Linear Equations & Word Problems · difficulty 1 of 5 · about 2 min · position P1-10

Isaac has written down one integer two times and another integer three times. The sum of the five numbers is $100$ , and one of the numbers is $28.$ What is the other number?

A)
$8$
B)
$11$
C)
$14$
D)
$15$
E)
$18$

Answer

Key insight

If 28 were written twice the other number would be 44/3, not an integer; so 28 is written three times and the other is 8.

Solution

Let $a$ be the integer written twice and $b$ the integer written three times, so $2a + 3b = 100$ . The number $28$ is either $a$ or $b$ .

If $a = 28$ : then $3b = 100 - 56 = 44$ , so $b = \tfrac{44}{3}$ , which is not an integer. Rejected.

If $b = 28$ : then $2a = 100 - 84 = 16$ , so $a = 8$ . This is an integer, and $2 \cdot 8 + 3 \cdot 28 = 16 + 84 = 100$ checks.

The answer is $\boxed{\textbf{(A)}\ 8}$ .

Why this works

The statement is deliberately ambiguous about which role $28$ plays, so test both. The word "integer" is the deciding constraint: one case produces a fraction and dies. Whenever a word problem says "integer," expect divisibility to eliminate a case.

The trap

Assuming 28 is the number written twice, getting 44/3 = 14.67, and choosing 14 or 15 as 'close enough'.

Common mistakes

Assuming 28 is the number written twice, getting 44/3 = 14.67, and choosing 14 or 15 as 'close enough'.
Writing $3a + 2b = 100$ with the roles swapped and not noticing that both cases then fail, or forgetting to check the arithmetic $3 \cdot 28 = 84$ .

Techniques

Test small/specific values or special cases to find or verify the answer