# 2022 AIME II Problems/Problem 14

## Problem

For positive integers $a$, $b$, and $c$ with $a < b < c$, consider collections of postage stamps in denominations $a$, $b$, and $c$ cents that contain at least one stamp of each denomination. If there exists such a collection that contains sub-collections worth every whole number of cents up to $1000$ cents, let $f(a, b, c)$ be the minimum number of stamps in such a collection. Find the sum of the three least values of $c$ such that $f(a, b, c) = 97$ for some choice of $a$ and $b$.

## Solution

Notice that we must have $a = 1$, otherwise $1$ cent stamp cannot be represented. At least $b-1$ numbers of $1$ cent stamps are needed to represent the values less than $b$. Using at most $c-1$ stamps of value $1$ and $b$, it can have all the values from $1$ to $c-1$ cents. Plus $\lfloor \frac{999}{c} \rfloor$ stamps of value $c$, every value up to $1000$ can be represented. Therefore using $\lfloor \frac{999}{c} \rfloor$ stamps of value $c$, $\lfloor \frac{c-1}{b} \rfloor$ stamps of value $b$, and $b-1$ stamps of value $1$, all values up to $1000$ can be represented in sub-collections, while minimizing the number of stamps.

So, $f(a, b, c) = \lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1$. $\lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1 = 97$. We can get the answer by solving this equation. $c > \lfloor \frac{c-1}{b} \rfloor + b-1$ $\frac{999}{c} + c > \lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1 = 97$ $c^2 - 97c + 999 > 0$, $c > 85.3$ or $c < 11.7$ $\lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1 > \frac{999}{c}$ $97 > \frac{999}{c}$, $c>10.3$ $\text{Case } 1:$ $10.3 < c < 11.7 \rightarrow c = 11 \rightarrow \lfloor \frac{999}{11} \rfloor + \lfloor \frac{10}{b} \rfloor + b-1 = 97$ $\lfloor \frac{10}{b} \rfloor + b = 8$, $b=7$ $\text{Case } 2: c>85.3$ $c = 86 \rightarrow$ $\lfloor \frac{999}{86} \rfloor + \lfloor \frac{85}{b} \rfloor + b-1 = 97$ $\lfloor \frac{85}{b} \rfloor + b = 87$, $b=87 > c \rightarrow \text{no solution}$ $c = 87 \rightarrow$ $\lfloor \frac{999}{87} \rfloor + \lfloor \frac{86}{b} \rfloor + b-1 = 97$ $\lfloor \frac{86}{b} \rfloor + b = 87$, $b=86$ or $1$. $\text{We cannot have }b=1 \text{ since it doesn't satisfy } a $\text{of } a \text{ for a total of 96 coins and still be able to make every value from 1 to 1000. Thus } c=87 \text{ yields no solution.}$ $c = 88 \rightarrow$ $\lfloor \frac{999}{88} \rfloor + \lfloor \frac{87}{b} \rfloor + b-1 = 97$ $\lfloor \frac{87}{b} \rfloor + b = 87$, $b=86$ $c = 89 \rightarrow$ $\lfloor \frac{999}{89} \rfloor + \lfloor \frac{88}{b} \rfloor + b-1 = 97$ $\lfloor \frac{88}{b} \rfloor + b = 87$, $b=86$


The $3$ least values of $c$ are $11$, $88$, $89$. $11 + 88+ 89 = \boxed{\textbf{188}}$

~isabelchen ~edited by bobjoebilly

## Video Solution

~MathProblemSolvingSkills.com

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