2015 AMC 10B Problem 21

P21-25~6 minPrint

Number Properties Linear Equations & Word Problems Bound the quantity above/below or estimate to pin it down Organized listing / direct enumeration

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 $s$ ?

Show solution without answering

Key insight

Jump counts are steps divided by 2 or 5, rounded up. Their difference stays close to 3/10 of the steps, so only 62 to 66 need checking.

Solution

Step 1: how many jumps for $n$ steps?

Cozy jumps $2$ steps at a time, with one short jump at the end if $n$ is odd. So for $n = 7$ steps he needs $4$ jumps: $n \div 2 = 3.5$ , rounded up. Dash jumps $5$ at a time, so for $n = 7$ he needs $2$ jumps: $7 \div 5 = 1.4$ , rounded up.

So Cozy needs $n \div 2$ rounded up, and Dash needs $n \div 5$ rounded up. We want Cozy's count minus Dash's count to be $19$ .

Step 2: narrow down $n$ .

Rounding up adds less than one jump. More exactly:
- Cozy's count is between $\frac{n}{2}$ and $\frac{n}{2} + \frac12$ .
- Dash's count is between $\frac{n}{5}$ and $\frac{n}{5} + \frac45$ .

Without rounding, the difference would be $\frac{n}{2} - \frac{n}{5} = \frac{3n}{10}$ . The rounding can push the difference up by at most $\frac12$ or down by at most $\frac45$ . So
$\frac{3n}{10} - \frac45 \le 19 \le \frac{3n}{10} + \frac12.$
The right half gives $\frac{3n}{10} \ge 18.5$ , so $n \ge 61.7$ . The left half gives $\frac{3n}{10} \le 19.8$ , so $n \le 66$ . Only $n = 62, 63, 64, 65, 66$ are possible.

Step 3: check each one.

$n$	Cozy's jumps	Dash's jumps	Difference	Works?
$62$	$31$	$13$	$18$	no
$63$	$32$	$13$	$19$	yes
$64$	$32$	$13$	$19$	yes
$65$	$33$	$13$	$20$	no
$66$	$33$	$14$	$19$	yes

Step 4: finish. The possible staircases have $63$ , $64$ or $66$ steps, so $s = 63 + 64 + 66 = 193$ . Its digits add to $1 + 9 + 3 = 13$ .

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

Why this works

"Take a short jump at the end if needed" means dividing and rounding up. Rounding is hard to solve exactly, but it never changes a number by a whole jump, so the equation traps $n$ in a short range. Then a small table finishes the job. When a problem involves rounding up or down, first find a short range, then check every number in it.

Concise solution

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

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 $\boxed{\textbf{(D)}\ 13}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 1: Integers: The Basics
Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.2)
Introduction to Algebra, Chapter 20: Special Functions (§20.3)