Solving a Cryptarithm

You are given the following cryptarithm and asked to decode what 24687 95310 decodes to:

	0	1	2	3	4	5	6	7	8	9
S
O
L
A
R
T
M
P
N
E

SOCIAL    social  _____ _____

+ SOLAR     solar  24687 95310

VEHICLE   vehicle

Immediately we know that V must be 1 because you can only carry a single digit from the previous column addition.  Furthermore, since there is only one digit in the previous column, it must be a 9 in order to carry from the column before that which means that the first two digits of the final result must be 10 telling us the mappings of V and E. We can mark that in the table.

	0	1	2	3	4	5	6	7	8	9
S										ü
O
C
I
A
L
R
V		ü
E	ü
H

SOCIAL  9ocial  _____ S__V_

+ SOLAR   9olar  24687 95310

VEHICLE 10hicl0

Some quick observations we can learn from what we have filled in so far:

In the first column we have: L+R=10 which because of the numbers already mapped can only be 2+8, 3+7 or 4+6 in either order.

In the second column we add the carry from the first column to A+A  giving us L which must be odd.  Based on what we learned in the first column, we know that L must be either 3 or 7 which means A must be one of 1, 3, 6 or 8.  We can quickly try all 4 options

A=1 won’t work since V=1 already

For A=3 we end up with L=7, but L+R=10 means that R would also be 3 so we can’t use that.

For A=6 we have L=3 which forces R=7 which means it is a possibility.

For A=8 we get L=7 which forces R=3 leaving it as a possibility. 

Either way we know that either L or R is 3 and the other is 7 and that A must be either 6 or 8, so we mark it in the table.  We also know that since A > 5 there is a carry into the next column

	0	1	2	3	4	5	6	7	8	9
S										ü
O
C
I
A
L
R
V		ü
E	ü
H

 SOCIAL     9ocial  _____ S__V_

+ SOLAR   +  9olar  24687 95310

VEHICLE    10hicl0

With the next column, we know that (carry from previous column) 1+I+L=C.  Since L must be either 3 or 7 and I and C are both limited to only five possible values, we look at the ten possible combinations to see which work.

L	I	1+I+L=C	Notes
3	2	6
3	4	8
3	5	9	S=9
3	6	(carry)0	E=0
3	8	(carry)2
7	2	(carry)0	E=0
7	4	(carry)2
7	5	(carry)3	C≠3
7	6	(carry)4
7	8	(carry)6

Immediately this eliminates C=5 and I=5 leaving only H or O to be 5.

With H and O in mind, we notice that (possible carry)+O+9=H (with a carry).  This tells us that O>H and that either O-1=H or O-2=H depending on the carry from the previous column.  Since one of them must be 5 we either have O=5 and H=4 or H=5 and O=6.  This means that there can not be a carry from C+O and C+O<9. 

With this information in hand, we fill in our table and eliminate quite a few options:

	0	1	2	3	4	5	6	7	8	9
S										ü
O
C
I
A
L
R
V		ü
E	ü
H

 SOCIAL     9ocial  _____ S__V_

+ SOLAR   +  9olar  24687 95310

VEHICLE    10hicl0

The only column we haven’t looked at is the (possible carry from previous column+)C+O=I.  Taking into account what we learned with the I+L column and knowing that there are only two possible values for I and 4 possible values for C or I we can test them out quickly in a table.

C	O	Carry+C+O=I	Notes
2	5	8
2	6	9	S=9
4	5	(carry)0	E=0
4	6	(carry)1	V=1
6	5	(carry)2
6	5	(carry)3	I≠3
8	5	(carry)3	I≠3
8	6	(carry)4

Since we previously determined that C+O<9 this tells us that the only possible answer is that C=2, O=5 and I=8.  Since we know what when O=5, H=4 we can fill that in too.

This leaves A=6 as the only option.  Previously we also determined that for A=6, L=3 and R=7 which gives us the final table and we can fill in the letters for the answer.

	0	1	2	3	4	5	6	7	8	9
S										ü
O						ü
C			ü
I									ü
A							ü
L				ü
R								ü
V		ü
E	ü
H					ü

 SOCIAL     952863 CHAIR SOLVE

+ SOLAR   +  95367  24687 95310

VEHICLE    1048230