# Three-Digit numbers divisbile by 3

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?

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

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.

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 $a=c=9$.

• Why can b be only 3 and 6? It can be 9 12 15 etc – Aspiring Mathlete Apr 24 '14 at 12:04
• 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. – gnasher729 Apr 24 '14 at 12:21

$$a+b+c=4b$$ $$a+c=3b$$ $$3|a+b+c$$ $$3|4b$$ 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 $10$.