Competition problem — Algebra — Maths Olympiad Prep

Algebra Difficulty 4.3 AIME Prove it Japan

Two people, $A$ and $B$ , go up a staircase with a certain number of steps. $A$ goes up 2 steps in one stride, while $B$ goes up 5 steps in one stride, except if $A$ finds only 1 step before the last stride, then he will go up just 1 step to finish, and if $B$ finds 4 or less steps remaining before the last stride, then $B$ will go up whatever steps remaining to climb to finish. $B$ took 6 strides less than $A$ to finish climbing this staircase. Determine all possible numbers of steps for this staircase can have.

Solution

$[19, 20, 21, 22]$

Let $n$ be the number of steps in the staircase, $a$ and $b$ be the number of strides $A$ and $B$ have taken, respectively. Then, we have
$a - b = 6, \quad \frac{n}{2} \le a < \frac{n}{2} + 1, \quad \frac{n}{5} \le b < \frac{n}{5} + 1.$
From these we obtain $\frac{n}{2} - (\frac{n}{5} + 1) < a - b < (\frac{n}{2} + 1) - \frac{n}{5}$ , which yields $\frac{50}{3} < n < \frac{70}{3}$ .

Thus the possibilities for $n$ are 17, 18, 19, 20, 21, 23, but since corresponding values of $a, b, a-b$ are as in the table below, we conclude that the answers we desire are $n = 19, 20, 21, 22$ .

| $n$ | 17 | 18 | 19 | 20 | 21 | 22 | 23 |
|-----|----|----|----|----|----|----|----|
| $a$ | 9 | 9 | 10 | 10 | 11 | 11 | 12 |
| $b$ | 4 | 4 | 4 | 4 | 5 | 5 | 5 |
| $a-b$ | 5 | 5 | 6 | 6 | 6 | 6 | 7 |

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