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2015 AMC 10B Problem 21

2015 AMC 10B Problem 21AMC 10 Step by Step

Name: Date: September 13, 2026

Number Properties · difficulty 4 of 5 · about 6 min · position P21-25

Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than 5 steps left). Suppose Dash takes 19 fewer jumps than Cozy to reach the top of the staircase. Let $s$ denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of $s$ ?

A)
$9$
B)
$11$
C)
$12$
D)
$13$
E)
$15$

Answer

Key insight

Jumps are ceilings, and ceil(n/2) - ceil(n/5) stays within 1 of 3n/10, so only n = 62 to 66 need checking.

Solution

For a staircase of $n$ steps, Cozy needs $\lceil n/2 \rceil$ jumps and Dash needs $\lceil n/5 \rceil$ jumps. We need
$\left\lceil \frac{n}{2} \right\rceil - \left\lceil \frac{n}{5} \right\rceil = 19.$

Bound the ceilings: $\frac{n}{2} \le \lceil n/2 \rceil \le \frac{n}{2} + \frac12$ and $\frac{n}{5} \le \lceil n/5 \rceil \le \frac{n}{5} + \frac45$ . Subtracting,
$\frac{3n}{10} - \frac45 \;\le\; 19 \;\le\; \frac{3n}{10} + \frac12,$
so $18.5 \le \frac{3n}{10} \le 19.8$ , i.e. $61.7 \le n \le 66$ . Only $n = 62, 63, 64, 65, 66$ can work.

Check each:
- $n = 62$ : $31 - 13 = 18$ . No.
- $n = 63$ : $32 - 13 = 19$ . Yes.
- $n = 64$ : $32 - 13 = 19$ . Yes.
- $n = 65$ : $33 - 13 = 20$ . No.
- $n = 66$ : $33 - 14 = 19$ . Yes.

So $s = 63 + 64 + 66 = 193$ , and the sum of its digits is $1 + 9 + 3 = 13$ .

The answer is $\boxed{\textbf{(D)}\ 13}$ .

Why this works

"Jump the last few steps if fewer remain" is the ceiling function. Ceilings are awkward to solve exactly, but each one is trapped within a known distance of the fraction it rounds, so the equation pins $n$ to a short interval; from there a table finishes the job. Whenever a problem mixes floors or ceilings, bound first, enumerate second.

The trap

Treating the jump counts as n/2 and n/5 exactly (giving the single value n = 63.33) or forgetting that a partial final jump still counts as one jump.

Common mistakes

Treating the jump counts as n/2 and n/5 exactly (giving the single value n = 63.33) or forgetting that a partial final jump still counts as one jump.
Reporting $s = 193$ or the number of solutions ( $3$ ) instead of the digit sum, or missing $n = 66$ by stopping the search at $65$ .

Techniques

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