The number of pairs of integers $a$ and $b$ with $a<b$ and $a+b<100$ satisfying the equation $\frac{a}{4}+\frac{b}{10}=7$ is
(A) 10
(B) 19
(C) 9
(D) 20
(E) 15
Solution 1

We begin by multiplying the given equation through by 20 to get $20 \times \frac{a}{4}+20 \times \frac{b}{10}=20 \times 7$, or $5 a+2 b=140$.

Since $5 a=140-2 b$ and both 140 and $2 b$ are even, then $5 a$ is even, which means that $a$ is even. We start by trying $a=20$ and $b=20$ which is a solution, since $5 \times 20+2 \times 20=140$.

This pair for $a$ and $b$ satisfies all of the conditions except $ab$ and every time we add two 5 s and subtract five 2 s, $a$ becomes greater and $b$ becomes smaller.

Thus, we need to go in the other direction.

Consider $a=20-2=18$ and $b=20+5=25$.

This is a solution since $5 \times 18+2 \times 25=140$.

Here, $a<b$ and $a+b=43$, so this pair satisfies all of the conditions.

Next, consider $a=18-2=16$ and $b=25+5=30$.

This is a solution since $5 \times 16+2 \times 30=140$.

Here, $a<b$ and $a+b=46$, so this pair satisfies all of the conditions.

Notice that the sum $a+b$ increases by 3 on each of these steps.

This means that doing this 17 more times gets us to $a=16-17 \times 2=-18$ and $b=30+17 \times 5=115$.

This is a solution since $5 \times(-18)+2 \times 115=140$.

Notice that it is still the case that $a<b$ and $a+b<100$.

Repeating this process one more time, we get $a=-20$ and $b=120$, which gives $a+b=100$ and so there are no more pairs that work.

Since we know that $a$ has to be even and we are considering all possible even values for $a$, there can be no other pairs that work.

In total, there are $1+1+17=19$ pairs of integers $a$ and $b$ that satisfy each of the given conditions and the given equation.

## Solution 2

We begin by rearranging the given equation to isolate $b$.

Doing so, we get

$$
\begin{aligned}
\frac{a}{4}+\frac{b}{10} & =7 \\
\frac{b}{10} & =7-\frac{a}{4} \\
10 \times \frac{b}{10} & \left.=10 \times 7-10 \times \frac{a}{4} \quad \text { (multiplying each term by } 10\right) \\
b & =70-\frac{10 a}{4} \\
b & =70-\frac{5 a}{2}
\end{aligned}
$$

Since $b$ is an integer, then $70-\frac{5 a}{2}$ is an integer, which means that $\frac{5 a}{2}$ must be an integer.

Since 2 does not divide 5, then 2 must divide $a$ and so $a$ is even.

Since $a<b$ and $b=70-\frac{5 a}{2}$, then

$$
\begin{aligned}
a & <70-\frac{5 a}{2} \\
2 \times a & \left.<2 \times 70-2 \times \frac{5 a}{2} \quad \text { (multiplying each term by } 2\right) \\
2 a & <140-5 a \\
7 a & <140 \\
a & <20
\end{aligned}
$$

Further, since $a+b<100$ and $b=70-\frac{5 a}{2}$, then

$$
\begin{aligned}
a+70-\frac{5 a}{2} & <100 \\
2 \times a+2 \times 70-2 \times \frac{5 a}{2} & <2 \times 100 \quad(\text { multiplying each term by } 2) \\
2 a+140-5 a & <200 \\
-60 & <3 a \\
-20 & <a
\end{aligned}
$$

Thus $a$ is an even integer that is greater than -20 and less than 20 .

Since there are 19 even integers from -18 to 18 inclusive, we suspect that there are 19 pairs of integers $a$ and $b$ that satisfy the given equation.

It is a good idea (and good practice) to at least check that the largest and smallest of these values of $a$ do indeed satisfy each of the given conditions.

When $a=-18$, we get $b=70-\frac{5(-18)}{2}$ or $b=70-5(-9)$, and so $b=115$.

This pair satisfies the given conditions that $a<b$ and $a+b<100$.

Substituting $a=-18$ and $b=115$ into the given equation, we get

$$
\frac{a}{4}+\frac{b}{10}=\frac{-18}{4}+\frac{115}{10}=\frac{-9}{2}+\frac{23}{2}=\frac{14}{2}=7
$$

and thus $a=-18$ and $b=115$ is a solution.

When $a=18$, we get $b=70-\frac{5(18)}{2}$ or $b=70-5(9)$, and so $b=25$.

This satisfies the given conditions that $a<b$ and $a+b<100$.

Substituting $a=18$ and $b=25$ into the given equation, we get

$$
\frac{a}{4}+\frac{b}{10}=\frac{18}{4}+\frac{25}{10}=\frac{9}{2}+\frac{5}{2}=\frac{14}{2}=7
$$

and thus $a=18$ and $b=25$ is also a solution.

At this point we can be confident that for each of the 19 even integer values of $a$ from -18 to 18 inclusive, there is an integer $b$ for which the pair of integers $a$ and $b$ satisfy each of the given conditions and the given equation.