Problem 256

There are five odd digits: 1, 3, 5, 7, 9.

We consider the positive integers less than 1000 in three sets: those with one digit, those with two digits, and those with three digits.

There are 5 positive one-digit integers with one odd digit (namely 1, 3, 5, 7, 9).

Consider the two-digit positive integers with only odd digits.

Such an integer has the form $XY$ where $X$ and $Y$ are digits.

There are five possibilities for each of $X$ and $Y$ (since each must be odd).

Therefore, there are $5\times 5 = 25$ two-digit positive integers with only odd digits.

Consider the three-digit positive integers with only odd digits.

Such an integer has the form $XYZ$ where $X$, $Y$ and $Z$ are digits.

There are five possibilities for each of $X$, $Y$ and $Z$ (since each must be odd).

Therefore, there are $5\times 5 \times 5= 125$ three-digit positive integers with only odd digits.

In total, there are $5+25+125=155$ positive integers less than 1000 with only odd digits.
Combining the two terms on the right side of the second equation, we obtain $\dfrac{4}{7} = \dfrac{b+a}{ab}$.

Since $a+b=16$, then $\dfrac{4}{7} = \dfrac{16}{ab}$ or $ab = \dfrac{16(7)}{4}=28$.

Therefore, we have $a+b=16$ and $ab=28$.

From the first equation, $b=16-a$.

Substituting into the second equation, we obtain $a(16-a)=28$ or $16a-a^2=28$, which gives $a^2-16a+28=0$.

Factoring, we obtain $(a-14)(a-2)=0$.

Therefore, $a=14$ or $a=2$.

If $a=14$, then $b=16-a=2$.

If $a=2$, then $b=16-a=14$.

Therefore, the two solutions are $(a,b)=(14,2),(2,14)$.

(We note that since $\dfrac{1}{2}+\dfrac{1}{14}=\dfrac{7}{14}+\dfrac{1}{14} = \dfrac{8}{14}=\dfrac{4}{7}$, then both of these pairs are actually solutions to the original system of equations.)