# Problem of the Week Problem C and Solution And the Numbers Are...

## Problem

John and Betty each choose a positive integer that is greater than $$1$$. Betty increases her number by $$1$$. John then takes this new number and multiplies it by his number. This product is equal to $$260$$.

If Betty’s number is larger than John’s number, determine all possible pairs of integers that John and Betty could have chosen.

## Solution

Let John’s integer be $$j$$ and Betty’s integer be $$b$$. We’re given $$j \times (b+1) = 260$$.

In considering the equation $$j\times (b+1)=260$$, we are looking for two integers, each greater than $$1$$, that multiply to $$260$$. We also want Betty’s integer $$b$$ to be greater than John’s integer $$j$$.

We generate the following list of ways to factor $$260$$ as a product of two integers:

$$1\times 260$$, $$2\times 130$$, $$4\times 65$$, $$5\times 52$$, $$10\times 26$$, $$13\times 20$$

We can exclude $$260=1\times 260$$ because both integers must be greater than $$1$$.

Since Betty’s integer is larger than John’s integer, we get the following possibilities:

• $$j=2$$ and $$b+1=130$$. Thus, $$j=2$$ and $$b=129$$.

• $$j=4$$ and $$b+1=65$$. Thus, $$j=4$$ and $$b=64$$.

• $$j=5$$ and $$b+1=52$$. Thus, $$j=5$$ and $$b=51$$.

• $$j=10$$ and $$b+1=26$$. Thus, $$j=10$$ and $$b=25$$.

• $$j=13$$ and $$b+1=20$$. Thus, $$j=13$$ and $$b=19$$.

Therefore, there are five pairs of integers that John and Betty could have chosen. John could have chosen $$2$$ and Betty chose $$129$$, John could have chosen $$4$$ and Betty chose $$64$$, John could have chosen $$5$$ and Betty chose $$51$$, John could have chosen $$10$$ and Betty chose $$25$$, or John could have chosen $$13$$ and Betty chose $$19$$.