# 2022 AMC 10B Problems/Problem 4

## Problem

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?

$\textbf{(A) }15 \text{ seconds after } 4:58$

$\textbf{(B) }20 \text{ seconds after } 4:58$

$\textbf{(C) }25 \text{ seconds after } 4:58$

$\textbf{(D) }30 \text{ seconds after } 4:58$

$\textbf{(E) }35 \text{ seconds after } 4:58$

## Solution 1

Since the donkey hiccupped the 1st hiccup at $4:00$, he hiccupped for $5 \cdot (700-1) = 3495$ seconds, which is $58$ minutes and $15$ seconds, so the answer is $\boxed{\textbf{(A) }15 \text{ seconds after } 4:58}$.

~MrThinker

## Solution 2 (Faster)

We see that the minute has already been determined. The donkey hiccups once every 5 seconds, or 12 times a minute. $700\equiv 4 \pmod{12}$, so the 700th hiccup happened on the same second as the 4th, which occurred on the $5(4-1)=15$th second. $\boxed{\textbf{(A) }15 \text{ seconds after } 4:58}$.

~not_slay and HIPHOPFROG1

## Solution 3 (Another fast way)

We can add $7\cdot500=3500$ and then minus $5$ seconds. $$4:00+3500\text{ seconds}$$ $$=4:00+3600\text{ seconds }-100\text{ seconds}$$ $$=4:00+1\text{ hour }-100\text{ seconds}$$ $$=5:00-100\text{ seconds}$$ $$=20 \text{ seconds after } 4:58$$ Finally, we minus $5$ seconds giving $\boxed{\textbf{(A) }15 \text{ seconds after } 4:58}$.

~Pancakerunner2