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