Number theory - Find the number satisfying the condition.

Find a 3 digit number which equals to 4 times the product of its digits.

My approach:

I considered the Number to be $\overline{ABC}$ then wrote the relation $$100A+10B+C=4A\times B\times C$$ But i don't know how to proceed further. Please help.

• Are you at programming or mathematics competition? If programming, try brute-force algorithm. – yW0K5o Aug 14 '18 at 18:17
• Cuz a stupid program could easily find this. E.g. my finding 384 in about 2 seconds of coding. If its a math competition, you have to be more careful – Don Thousand Aug 14 '18 at 18:18
• If this is a math problem, my trick would be to first try to prove that $A\cdot B\cdot C\leq100$ – Don Thousand Aug 14 '18 at 18:20
• also ABC must be divisible by 4 – yW0K5o Aug 14 '18 at 18:22

Long solution without programming.

$100A+10B+C=4\times A\times B\times C (1)$

Divide both parts of (1) by 4

$25A+\frac{5}{2}B+\frac{C}{4}=A\times B\times C (2)$

From (2)

Based on Daniel Fischer suggestion

$C=2$ (2) goes to $25A+\frac{5}{2}B+\frac{2}{4}=2\times A\times B => \frac{50A+5B+1}{4}=A\times B$ and $50A+5B+1$ can not be divided by 4

$C=6$ (2) goes to $25A+\frac{5}{2}B+\frac{3}{2}=6\times A\times B => \frac{50A+5B+3}{12}=A\times B$ and $50A+5B+3$ can not be divided by 2

$C$ must be 4 or 8 and $B$ must be even.

If $c=8$ (2) can be rewritten as $25A+\frac{5}{2}B+2=8 \times A\times B (3')$

or

$\frac{25}{8}A+\frac{5}{16}B+\frac{1}{4}=A\times B (3'')$

(3'') can not be solved for for integers A and B because of sum plus $\frac{1}{4}$ and B must be divisible by 16 (wrong).

Try $C=4$

$25A+\frac{5}{2}B+1=4 \times A\times B (4')$ or

$\frac{25}{4}A+\frac{5}{8}B+\frac{1}{4}=A\times B (4'')$

Try (4'') with $B=2,4,6,8,0$ $B=0$ (4'') impossible

$B=2$ (4'') goes to $\frac{25}{4}A+\frac{5}{4}+\frac{1}{4}=2\times A (4''')$ or $25A+6=8A => 17A= -6$- impossible

$B=4$ (4'') goes to $\frac{25}{4}A+\frac{10}{4}+\frac{1}{4}=4\times A (4''')$ or $25A+11=16A => 9A= -11$- impossible

$B=6$ (4'') goes to $\frac{25}{4}A+\frac{5}{8}6+\frac{1}{4}=6\times A (4''')$ or $25A+16=24A => A= -16$- impossible

$B=8$ (4'') goes to $\frac{25}{4}A+\frac{5}{8}8+\frac{1}{4}=8\times A (4''')$ or $25A+21=32A => 7A= 21 => A=3$ - possible

384

• In $(2)$, why can you rule out $C = 2$ or $6$ and $B$ odd? – Daniel Fischer Aug 14 '18 at 19:12
• your solution implies in some way that $4 | 10b + c \quad \Rightarrow \quad 4 | 10 b \quad \wedge \quad 4|c$ that is false – JayTuma Aug 14 '18 at 19:26
• @jaytuma, my solution is not implies what you've said. – yW0K5o Aug 14 '18 at 19:35
• the first line of the Daniel Fischer suggestion is false because $50 A + 5 B + 1$ is divided by $4$ with $B = 3$ (we had no hypotesis on $B$ yet). I appreciate your edit, but i still belive it's not a complete solution – JayTuma Aug 14 '18 at 20:01
• @jaytuma I agree. – yW0K5o Aug 14 '18 at 20:19

Hint: One good way to begin with is always to look at congruence class. For example it's easy to see that $4$ divides the RHS, hence it has to divide the LHS and in particular $4$ divides $c + 10b$. Thus $c$ is even which implies even a better condition: $8$ divides $100a + 10b + c$.

Another intresting condition is that if any digits were $5$ then your number is divisible by $10$ and so $c = 0$, but the RHS then implies that your number is $0$.

[from Barry Cipra observation]

Checking the equation mod $3$ we get $$A + B + C = ABC \mod 3$$ Assuming that $A,B,C$ are all non divisible by $3$, then we are left with 4 cases to check, but modulus $3$ they are all not possibile $$\begin{array}{ll} 1 + 1 + 1 = 0 \neq 1 = 1 \cdot 1 \cdot 1 & \quad 1 + 1 + 2 = 1 \neq 2 = 1 \cdot 1 \cdot 2 \\ 1 + 2 + 2 = 2 \neq 1 = 1 \cdot 2 \cdot 2 & \quad 2 + 2 + 2 = 0 \neq 2 = 2 \cdot 2 \cdot 2 \end{array}$$

I encourage you to find more conditions, until eventually you are left with a feasible number of cases that can be checked by hand

• From $A+B+C\equiv ABC$ mod $3$ you can argue that one of the digits (at least) is a multiple of $3$ and the other two sum to a multiple of $3$. – Barry Cipra Aug 14 '18 at 23:45
• Thank you, i added your observation – JayTuma Aug 15 '18 at 12:25