Eve is a loquacious chatterbox.
Two same characters indicate the same number, and two different characters indicate different numbers. Every character is an integer from 0 to 9, and .
Given that the left side of the equation is an irreducible fraction, find the value of
This problem is a part of <Cryptarithms> series .
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
0 . TALKTALKTALKTALK ⋯ = 9 9 9 9 TALK .
Therefore DID = 1 0 1 , 3 0 3 , 9 0 9 .
i ) DID = 1 0 1
9 9 × EVE = TALK and since E ≥ 2 , the left side is impossible to be a four-digit number.
i i ) DID = 3 0 3
3 3 × EVE = TALK .
Since 3 3 × 4 1 4 = 1 3 6 6 2 > 9 9 9 9 , E = 1 or E = 2 .
If E = 1 , K = 3 . But then D = K , which doesn't satisfy the problem.
So E = 2 . Since DID and EVE are coprime integers, there are 5 values for EVE .
3 3 × 2 1 2 = 6 9 9 6 3 3 × 2 4 2 = 7 9 8 6 3 3 × 2 6 2 = 8 6 4 6 3 3 × 2 7 2 = 8 9 7 6 3 3 × 2 9 2 = 9 6 3 6
So we know that EVE = 2 4 2 and TALK = 7 9 8 6 .
i i i ) DID = 9 0 9
1 1 × EVE = TALK .
Since E = K needs to be satisfied in order for the equation above to hold, it contradicts the problem.
Therefore, the original expression was 3 0 3 2 4 2 = 0 . 7 9 8 6 7 9 8 6 7 9 8 6 7 9 8 6 ⋯ .
∴ L+E+V+I+T+A+T+E+D = 8 + 2 + 4 + 0 + 7 + 9 + 7 + 2 + 3 = 4 2 .