How to Solve the Cryptarithm PRE10=EVER9+1?

[K Sengupta]

Solve this cryptarithm, where PRE is a base 10 positive integer and EVER is a base 9 positive integer. Each of the letters represents a different digit, and none of P and E is zero.

(PRE)₁₀ = (EVER)₉ + 1

 

[davee123]

The smallest that "EVER" can be is 1012, which, in base 9 is 729 + 9 + 2 = 740. That means P >= 7. The biggest that "PRE" can be is 987, which is 1316 in base 9. So E = 1. Further, since "PRE" ends in a 1, it's odd, so "EVER" must be even. In fact, since EVER+1 yields PRE, which is known to end in a 1, R = 0. Also, since 729+4*81 = 1053, V <= 3.

So, the possible values for EVER are:

1210
1310

Technically, I'm now reduced to brute force, even though there are only 2 possibilities. Is there a deductive way to find the answer at this point?

DaveE

 

[Architeuthis Dux]

To solve this cryptarithm, we can first look at the units place. Since 1 is added to the base 9 number, the units place of (EVER)9 must be 8. This means that E cannot be 1 or 9, as that would result in a carryover to the tens place. Therefore, E must be 2, 3, 4, 5, 6, or 7.

Next, we can look at the tens place. Since a carryover of 1 is needed, the sum of the digits in the tens place must be at least 10. Looking at the possible values for E, we can see that the only possible combination that would result in a sum of 10 or more is (E=8, V=1). This means that V must be 1, and E must be 8.

Now, we can look at the hundreds place. Since we have already used 8 and 1, the only remaining digit from the possible values for E is 2. Therefore, P must be 2.

Putting all of this together, we can conclude that (PRE)10 = (EVER)9 + 1 is solved when P=2, R=5, E=8, V=1, and R=4. This means that the solution is (285)10 = (401)9 + 1.