C50 (Robert Beezer) A three-digit number has two properties. The tens-digit and the ones-digit add up to 5. If the number is written with the digits in the reverse order, and then subtracted from the original number, the result is 792 . Use a system of equations to find all of the three-digit numbers with these properties.

Solution (Robert Beezer) Let $a$ be the hundreds digit, $b$ the tens digit, and $c$ the ones digit. Then the first condition says that $b+c=5$. The original number is $100 a+10 b+c$, while the reversed number is $100 c+10 b+a$. So the second condition is
$$792=(100 a+10 b+c)-(100 c+10 b+a)=99 a-99 c$$
So we arrive at the system of equations
$$\begin{gathered} b+c=5 \ 99 a-99 c=792 \end{gathered}$$
Using equation operations, we arrive at the equivalent system
\begin{aligned} & a-c=8 \ & b+c=5 \end{aligned}
We can vary $c$ and obtain infinitely many solutions. However, $c$ must be a digit, restricting us to ten values $(0-9)$. Furthermore, if $c>1$, then the first equation forces $a>9$, an impossibility. Setting $c=0$, yields 850 as a solution, and setting $c=1$ yields 941 as another solution.

C51 (Robert Beezer) Find all of the six-digit numbers in which the first digit is one less than the second, the third digit is half the second, the fourth digit is three times the third and the last two digits form a number that equals the sum of the fourth and fifth. The sum of all the digits is 24. (From The MENSA Puzzle Calendar for January 9, 2006.)

Solution (Robert Beezer) Let abcdef denote any such six-digit number and convert each requirement in the problem statement into an equation.

\begin{aligned} c & =\frac{1}{2} b \ d & =3 c \ 10 e+f & =d+e \ 24 & =a+b+c+d+e+f \end{aligned}
In a more standard form this becomes
\begin{aligned} a-b & =-1 \ -b+2 c & =0 \ -3 c+d & =0 \ -d+9 e+f & =0 \ a+b+c+d+e+f & =24 \end{aligned}
Using equation operations (or the techniques of the upcoming Section RREF), this system can be converted to the equivalent system
\begin{aligned} & a+\frac{16}{75} f=5 \ & b+\frac{16}{75} f=6 \ & c+\frac{8}{75} f=3 \ & d+\frac{8}{25} f=9 \ & e+\frac{11}{75} f=1 \end{aligned}
Clearly, choosing $f=0$ will yield the solution $a b c d e=563910$. Furthermore, to have the variables result in single-digit numbers, none of the other choices for $f(1,2, \ldots, 9)$ will yield a solution.

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