2006 AMC 10B Problem 22

P21-25~5 minPrint

Diophantine Equations Modular Arithmetic Primes Bound the quantity above/below or estimate to pin it down Organized listing / direct enumeration

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?

Show solution without answering

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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)