Cryptograms (Problem 4, Version 2)

First note that the number $3 \times B$ must end in the digit $B$ . We're told $B$ is not zero; so $B=5$ . Writing out the rest as an equation, $300R+30G+15=1000W+555$

Subtracting $15$ from both sides and dividing by $10$ : $30R+3G=100W+54$

We see that $W$ is a multiple of $3$ . So the smallest the right-hand side can be is $354$ . But the largest the left-hand side could be (obtained when $R=9$ and $G=8$ ) is $294$ .

Therefore there are no solutions.

This is a very easy problem! $WBBB=3RGB$ , therefore $3B+W\;mod\;3=0$ . So W must be divisible by 3. But $W<3$ , because $W_{max}(=987*3)=\boxed{2}961$ . Thus $W=0$ , but the problem says that $W$ is a non-zero digit.

I can prove that there is no solution to this problem . Why then the answer $0$ is incorrect? Following is the proof :

We are given that $3(100R+10G+B)=1000W+111B$

Hence $W$ must be divisible by $3$ . Since it can't be zero, it's possible values are $3,6,9$ . We will dwell with $W=3$ here, the other alternatives can be proved in exactly the similar fashion.

For $W=3, 100R+10G=1000+36B$ .

So $B$ must be $5$ . Then

$10R+G=118$ , which is impossible, since the maximum possible value of $10R+G$ is $99$ .