15
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I hope it's not too easy.

As CAKE I'm worth 573,566.

As PIE I'm worth 33,062.

What am I worth as a TART?

Note : It's not a word game.

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  • 5
    $\begingroup$ I can't work out if the double entendre in the title question is deliberate... $\endgroup$ – David Richerby Sep 14 '17 at 17:36
20
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You would be worth

1,366,985

as a TART.

Because ...

They are base 36 numbers, with the capital letters standing as digits 10 to 35 in lexical order.
$CAKE_{36} = 12*36^3 + 10*36^2 + 20*36 + 14 = 573,566$
$PIE_{36} = 25*36^2 + 18*36 + 14 = 33,062$
and $TART_{36} = 29*36^3 + 10*36^2 + 27*36 + 29 = 1,366,985$

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  • 3
    $\begingroup$ Welcome to Puzzling! (Take the Tour!) We generally put Spoiler tags around answers to avoid revealing the answers to someone who wants to try to solve the puzzle on their own. I've done that for you. $\endgroup$ – Rubio Sep 14 '17 at 11:50
3
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Incomplete Answer

Assuming that all of these are just variables multipled together, let's start with CAKE.

The prime factorization of CAKE is 2, 7, 53, 773. That tells us what our variables are, but not the order they are in.

Next, I noticed that the prime factorization of PIE is 2, 61, 271, which shares the common factor 2 with CAKE. Because of this, we can determine that E is 2.

Because E is 2, the letter A must be either 7, 53, or 773.

However, I'm stuck from here.

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  • 3
    $\begingroup$ How can you eliminate E=1? $\endgroup$ – boboquack Sep 14 '17 at 10:21
  • $\begingroup$ @boboquack, as 1 is not a prime number, possibly ! $\endgroup$ – Mea Culpa Nay Sep 14 '17 at 10:25
  • $\begingroup$ @rsp My answer IS incomplete. I'm just trying to help any other solvers. $\endgroup$ – Julian Lachniet Sep 14 '17 at 10:26
  • $\begingroup$ @MeaCulpaNay So? There was no assumption that all the letters were prime numbers, I could have C=573566, P=33062, A=K=E=I=1 $\endgroup$ – boboquack Sep 14 '17 at 10:30
  • $\begingroup$ @boboquack - In which case, the puzzle would be unsolvable, as there would be no way to determine what R was (we could guess T was also 1). Generally, one can presume that the puzzles presented to us by our fellows here are solvable (with the possible exception of ones they're sharing because they couldn't solve them). $\endgroup$ – RDFozz Sep 14 '17 at 23:44

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