When does addition equal multiplication?

Number Theory Level 1

If $A$ and $B$ are both positive integers such that $A + B = A\times B,$ what is the value of $A - B?$

The answer is 0.

6 solutions

Abdur Rehman Zahid
Feb 1, 2016

$\begin{aligned} A+B&=AB\\ A+B-AB&=0\\ A+B(1-A)&=0\\ (A-1)-B(A-1)&=-1\\ (1-B)(A-1)&=-1\\ (B-1)(A-1)&=1\\ \end{aligned}\\ \implies\begin{cases}B-1=1\\ A-1=1\end{cases}\text{or} \begin{cases}B-1=-1\\ A-1=-1\end{cases}$ After solving,we get $(A,B)=(2,2),(0,0)$ .However,since A and B are positive integers,so $(A,B)=(2,2)\implies A-B=\boxed{0}$

Quite good. Indeed, the trick on the combination of AB, A and B is often seen in many contexts.

It's always beneficial to recognise AB - B - A = (A-1)(B-1) - 1, or AB - B + A = (A-1)(B+1) + 1, or AB + A + B = (A + 1)(B+1) -1, etc., so that your argument can be shortened by a bit. You may also see a similar trick used in matrices and junior high-school maths competitions.

By the way, you may apply the exact same technique to the following problem .

Wing Tang - 5 years, 4 months ago

Weiting Hong
Feb 3, 2016

$A \times B$ = A + B

$A \times B$ - A = B

$A \times (B-1)$ = B

A = $\frac{B}{(B-1)}$

In order to have A as an positive integer, B must be 2 (since two adjacent natural numbers are coprime, only something next to 1 may satisfy this condition)

Plug B back to the equation

2A = A+2

A = 2

Therefore: A - B = 2 - 2 = $\boxed{0}$