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2006 AMC 10B Problem 22

2006 AMC 10B Problem 22AMC 10 Step by Step

Name: Date: October 28, 2026

Diophantine Equations · difficulty 4 of 5 · about 5 min · position P21-25

Elmo makes $N$ sandwiches for a fundraiser. For each sandwich he uses $B$ globs of peanut butter at $4\cent$ per glob and $J$ blobs of jam at $5\cent$ per blob. The cost of the peanut butter and jam to make all the sandwiches is $\$ 2.53$ . Assume that $B$ , $J$ , and $N$ are positive integers with $N>1$ . What is the cost of the jam Elmo uses to make the sandwiches?

A)$ 1.05
B)$ 1.25
C)$ 1.45
D)$ 1.65
E)$ 1.85

Answer

Key insight

In cents, sandwiches times cost per sandwich is 253 = 11 times 23. Only 11 sandwiches at 23 cents each works: 2 globs, 3 blobs.

Solution

1. Work in cents so everything is a whole number. One sandwich costs $4B + 5J$ cents, and $N$ sandwiches cost
$N \times (4B + 5J) = 253.$
2. Break $253$ into factors. $253 = 11 \times 23$ , and both $11$ and $23$ are prime. So the number of sandwiches $N$ must be $1$ , $11$ , $23$ or $253$ .
3. Rule out the easy ones. $N = 1$ is not allowed ( $N > 1$ ). Each sandwich has at least $1$ glob and $1$ blob, so it costs at least $4 + 5 = 9$ cents; $N = 253$ would make each sandwich cost $1$ cent, which is too little.
4. Try $N = 23$ . Each sandwich would cost $11$ cents, so $4B + 5J = 11$ . Try $J = 1$ : $4B = 6$ , not a whole number of globs. Try $J = 2$ : $4B = 1$ , no. $J = 3$ is already too big. No answer.
5. Try $N = 11$ . Each sandwich costs $23$ cents, so $4B + 5J = 23$ . Try each $J$ :

$J$	$1$	$2$	$3$	$4$
$4B = 23 - 5J$	$18$	$13$	$8$	$3$
$B$	no	no	$2$	no

Only $J = 3$ , $B = 2$ works.
6. Jam cost. $11$ sandwiches, $3$ blobs each, $5$ cents a blob: $11 \times 3 \times 5 = 165$ cents, which is $\$1.65$ .

The answer is $\boxed{\textbf{(D)}\ \$1.65}$ .

Why this works

When a product of unknown whole numbers equals a known number, factor that number: there are only a few ways to split it. Facts like "each sandwich uses at least one of each" and " $N > 1$ " cut the list down further. What remains is small enough to check by trying values.

The trap

Stopping too early and answering the jam cost for one sandwich (15 cents) or the peanut butter total (88 cents), instead of the jam for all 11 sandwiches.

Concise solution

Work in cents. One sandwich costs $4B + 5J$ cents, so all of them cost
$N(4B + 5J) = 253.$
Now $253 = 11 \cdot 23$ , and both factors are prime, so $N$ must be one of $1, 11, 23, 253$ . We are told $N > 1$ , and since $B, J \ge 1$ the per-sandwich cost is at least $4 + 5 = 9$ , which rules out $N = 253$ (per-sandwich cost $1$ ).

$N = 23$ would need $4B + 5J = 11$ . With $J = 1$ , $4B = 6$ ; with $J = 2$ , $4B = 1$ . No solution.
$N = 11$ needs $4B + 5J = 23$ . With $J = 1$ , $4B = 18$ ; $J = 2$ , $4B = 13$ ; $J = 3$ , $4B = 8$ , so $B = 2$ ; $J = 4$ makes $5J = 20$ and $4B = 3$ . The unique solution is $B = 2$ , $J = 3$ .

So Elmo makes $11$ sandwiches with $3$ blobs of jam each: $11 \cdot 3 \cdot 5 = 165$ cents.

The answer is $\boxed{\textbf{(D)}\ \$1.65}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 7: Algebra With Integers
Introduction to Number Theory, Chapter 2: Primes and Composites
Introduction to Number Theory, Chapter 12: Introduction to Modular Arithmetic
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.2)

Techniques

Bound the quantity above/below or estimate to pin it down · Organized listing / direct enumeration