2
$\begingroup$

How many three digit numbers are divisible by 3 and have an additional property that the sum of of their digits is 4 times the middle digit? My approach: let the number be $abc$ so $$abc \equiv 0\pmod{3}$$ and $$a + b+ c= 4b$$ I'm stuck now. Any help?

$\endgroup$
  • $\begingroup$ Actually, you have $0\equiv100a+10b+c\equiv a+b+c\pmod{3}$. $\endgroup$ – robjohn Apr 24 '14 at 11:50
3
$\begingroup$

Note that $a+c=3b$

Hence, choose $a,c$ such that their sum is multiple of three. You will automatically get a $b$ free with each case.

$\endgroup$
1
$\begingroup$

First, $b$ should be dividable by $3$, since $a+c=3b$, we have $0<3b<18$. So $b$ can be $3$ and $6$.

For $b=3$, we have $a+c=9$, there are 9 choices;

for $b=6$, we have $a+c=18$, then the only choice is $a=c=9$.

$\endgroup$
  • $\begingroup$ Why can b be only 3 and 6? It can be 9 12 15 etc $\endgroup$ – Aspiring Mathlete Apr 24 '14 at 12:04
  • 1
    $\begingroup$ b = 9 makes the sum of digits 36 which is too big for a 3 digit number. b = 12 or 15 is impossible since it is a single digit. $\endgroup$ – gnasher729 Apr 24 '14 at 12:21
0
$\begingroup$

$$a+b+c=4b$$ $$a+c=3b$$ $$3|a+b+c$$ $$3|4b$$ since, $b=0,1,2, \ldots, 9$

therefore $b=0,3,6,9$

now,make different cases- $a3c,a6c,a9c$

the third case is not possible or doesn't have any solution.

first and second case have solutions-

case $1$- $(a,c)=(1,8),(2,7),(3,6),(4,5)$ and their revese too like-$(a,b)=(8,1), \ldots$ and one more solution is $(9,0)$ but reverse not possilble and for case $2$- one solution $(a,b)=(9,9)$ therefore total numbers are $10$.

$\endgroup$

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.