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2006 AMC 10B Problem 22

2006 AMC 10B Problem 22AMC 10 Step by Step

Name: Date: September 13, 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, N(4B+5J) = 253 = 11*23; the per-sandwich cost is at least 9 and N > 1, so N = 11, B = 2, J = 3.

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 $\boxed{\textbf{(D)}\ \$1.65}$ .

Why this works

The disguised statement is an integer factorization: a product of unknown positive integers equals a known number, so factor it and test the few divisor splits. Bounds from "positive integers" (per-sandwich cost at least $9$ ) and the given $N > 1$ prune the cases. The leftover two-variable equation is tiny enough to solve by trying $J$ values.

Alternative approach

Mod $5$ shortcut: $4B + 5J = 23$ means $4B \equiv 3 \pmod 5$ , so $B \equiv 2$ ; only $B = 2$ keeps $4B < 23$ . Similarly $4B + 5J = 11$ needs $B \equiv 4 \pmod 5$ , impossible since $4 \cdot 4 > 11$ .

The trap

Reporting the jam cost per sandwich (15 cents) or the peanut butter total ($0.88), instead of jam for all 11 sandwiches.

Common mistakes

Reporting the jam cost per sandwich (15 cents) or the peanut butter total ( $0.88), instead of jam for all 11 sandwiches. - Not converting to cents and failing to see$ 253 = 11 \cdot 23 $, or forgetting to reject$ N = 1$.

Techniques

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