Multiplication. Addition. What's The Difference?

Number Theory Level 1

$\begin{aligned} A \times B \times C &=& X \\ A + B + C &=& X \end{aligned}$

$A, B,$ and $C$ are positive integers satisfying the system above. Find $X.$

The answer is 6.

2 solutions

Kay Xspre
Feb 4, 2016

Let $abc = a+b+c$ , we will get that $c(ab-1) = a+b$ or $c = \frac{a+b}{ab-1}$

From here, since C is a whole number, the denominator in fractional form may only be 1 or -1, which means

$|ab-1|=1\Rightarrow ab = 0, 2$

However, we cannot use $ab = 0$ as A and/or B shall be zero, which does not satisfy the positive integer condition. Hence, we can only use $ab = 2$ . Regardless of $(a, b) = (1, 2), (2, 1), c = 3$ . The answer is then $abc = 6$ and for each integer (order is disregarded)

$(a, b, c) = (1, 2, 3)$

Why only 1 or -1. For example it could be 8/4=2

kattos kattos - 5 years, 4 months ago

I agree that further proof is necessary....

Without loss of generality, assume that $0 \lt a \le b \le c$ . Then $a + b + c \le c + c + c = 3c$ .

Now suppose $ab \gt 3$ . Then since $c$ is positive we have that $abc \gt 3c$ , which combined with the previous result implies that $abc \gt 3c \ge a + b + c \Longrightarrow abc \gt a + b + c$ . So since we wish to have $abc = a + b + c$ we are forced to conclude that $ab \le 3$ . This leaves us with the following cases to examine:

$ab = 1 \Longrightarrow c = a + b + c \Longrightarrow a + b = 0$ , which is not valid as $a,b \gt 0$ . .
$ab = 2 \Longrightarrow 2c = a + b + c \Longrightarrow a + b = c$ . Now as $ab = 2$ and $0 \lt a \le b$ we must have $a = 1, b = 2$ , and thus $c = 3$ . Thus $(a,b,c) = (1,2,3)$ is a valid solution. .
$ab = 3 \Longrightarrow 3c = a + b + c \Longrightarrow 2c = a + b$ . Now as $ab = 3$ and $0 \lt a \le b$ we must have $a = 1, b = 3$ , and thus $2c = 4 \Longrightarrow c = 2$ . But this makes $c \lt b$ , and besides, simply duplicates the solution found in the $ab = 2$ case.

Thus, having examined all possible scenarios, we can conclude that $x = a + b + c = abc = 1*2*3 = \boxed{6}$ is the unique solution.

Brian Charlesworth - 5 years, 4 months ago

isn't 0 a positive integer?

Mardokay Mosazghi - 5 years, 4 months ago

@Mardokay Mosazghi – $0$ is neither positive nor negative, and we can refer to it as neutral. By the same token, it is the only element in the set of integers that are both non-negative and non-positive.

Brian Charlesworth - 5 years, 4 months ago

why not 0, 0+0+0 is 0, also, 0^3=0,

Kyle Prest - 5 years, 4 months ago

Because a,b,c are positive integers

kattos kattos - 5 years, 4 months ago

when i first looked at the question, i thought 0 had to be the answer, but u have to think about it long enough

Bryan Gonzalez - 5 years, 4 months ago

It says A, B and C are positive integers. 0 is neutral.

Matheus Rezende - 5 years, 4 months ago

@Matheus Rezende – oh, i see now, It makes sense. 1+2+3, 1x2x3,

Kyle Prest - 5 years, 4 months ago

Lance Fernando
Jul 2, 2016

Perfect numbers had this property that all factors except itself will add up to the number itself. However, since it had three factors, the very first perfect number is the answer: six.

Multiplication. Addition. What's The Difference?

The answer is 6.

2 solutions

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