There is actually another way to solve this puzzle, figuiring out the answer digit by digit, backwards.
Can someone figure out what it is?
Yes, I did it this way. It is slower, but perhaps more fun.
First you write the equation like a long multiplication question:
1 a b c d e
x_______3
a b c d e 1
This is the logic:
e x 3 ends in 1. e must be 7 (7x3=21, carry the 2)
d x 3 ends in 7-2. d must be 5 (5x3=15, carry the 1)
c x 3 ends in 5-1. c must be 8 (8x3=24, carry the 2)
b x 3 ends in 8-2. b must be 2 (2x3=6, no carry)
a x 3 ends in 2. a must be 4 (4x3=12, carry the 1)
1 x 3 ends in 4-1. this is true.
______21 a b c d
7x_______3
a b c d
7 1
_____11 a b c
5 7
x_______3
a b c
5 7 1
___21 a b
8 5 7
x_______3
a b
8 5 7 1
1 a
2 8 5 7
x_______3
a
2 8 5 7 1
11
4 2 8 5 7
x_______3
4 2 8 5 7 1
(carries in yellow)