3
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ Are you at programming or mathematics competition? If programming, try brute-force algorithm. $\endgroup$
    – yW0K5o
    Aug 14, 2018 at 18:17
  • 2
    $\begingroup$ 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 $\endgroup$ Aug 14, 2018 at 18:18
  • 1
    $\begingroup$ If this is a math problem, my trick would be to first try to prove that $A\cdot B\cdot C\leq100$ $\endgroup$ Aug 14, 2018 at 18:20
  • $\begingroup$ also ABC must be divisible by 4 $\endgroup$
    – yW0K5o
    Aug 14, 2018 at 18:22

2 Answers 2

2
$\begingroup$

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

$\endgroup$
2
  • 1
    $\begingroup$ 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 $3$. $\endgroup$ Aug 14, 2018 at 23:45
  • $\begingroup$ Thank you, i added your observation $\endgroup$
    – JayTuma
    Aug 15, 2018 at 12:25
1
$\begingroup$

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

Question, comments, edits?

$\endgroup$
5
  • 1
    $\begingroup$ In $(2)$, why can you rule out $C = 2$ or $6$ and $B$ odd? $\endgroup$ Aug 14, 2018 at 19:12
  • $\begingroup$ your solution implies in some way that $4 | 10b + c \quad \Rightarrow \quad 4 | 10 b \quad \wedge \quad 4|c$ that is false $\endgroup$
    – JayTuma
    Aug 14, 2018 at 19:26
  • $\begingroup$ @jaytuma, my solution is not implies what you've said. $\endgroup$
    – yW0K5o
    Aug 14, 2018 at 19:35
  • 1
    $\begingroup$ 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 $\endgroup$
    – JayTuma
    Aug 14, 2018 at 20:01
  • $\begingroup$ @jaytuma I agree. $\endgroup$
    – yW0K5o
    Aug 14, 2018 at 20:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.