2022 AMC 10B Problem 4

P1-10~2 minPrint

Clocks, Calendars & Time Sequences & Series Set up the equation/formula and compute; no special trick needed

A donkey suffers an attack of hiccups and the first hiccup happens at $4:00$ one afternoon. Suppose that the donkey hiccups regularly every $5$ seconds. At what time does the donkey’s $700$ th hiccup occur?

Show solution without answering

Key insight

From the 1st hiccup to the 700th there are only 699 gaps of 5 seconds, so it happens 3495 seconds (58 minutes 15 seconds) after 4:00.

Solution

Start small to see how many gaps there are.

1. Small cases. The 1st hiccup is at $4{:}00$ exactly. The 2nd is $5$ seconds later (one gap). The 3rd is $10$ seconds later (two gaps). So the $n$ th hiccup comes after $n - 1$ gaps.
2. The 700th hiccup comes after $700 - 1 = 699$ gaps, which is
$699 \cdot 5 = 3495 \text{ seconds after } 4{:}00.$
3. Change to minutes. $58 \cdot 60 = 3480$ , and $3495 - 3480 = 15$ . So $3495$ seconds is $58$ minutes and $15$ seconds.
4. The time is $15$ seconds after $4{:}58$ .

The answer is $\boxed{\textbf{(A)}\ 15 \text{ seconds after } 4{:}58}$ .

Why this works

This is the "fence post" idea: $n$ posts in a row have only $n - 1$ gaps between them. The hiccups are the posts and the $5$ -second waits are the gaps. The answer choices are only $5$ seconds apart, which is exactly one gap, so they are designed to catch the off-by-one mistake. Checking the 2nd and 3rd hiccup first makes the right count clear.

Concise solution

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

The first hiccup is at $4{:}00$ exactly. Between the first and the $700$ th hiccup there are $700 - 1 = 699$ intervals of $5$ seconds each, so the $700$ th hiccup happens
$699 \cdot 5 = 3495 \text{ seconds}$
after $4{:}00$ .

Convert: $3495 = 58 \cdot 60 + 15$ , so that is $58$ minutes and $15$ seconds. The time is $15$ seconds after $4{:}58$ .

The answer is $\boxed{\textbf{(A)}\ 15 \text{ seconds after } 4{:}58}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 21: Sequences & Series (§21.1, §21.2, §21.3, §21.4)
Intermediate Algebra, Chapter 17: More Sequences and Series (§17.1, §17.2)