(PEEM)base 5 + (PEEM)base 7 = (PEEM)base 8

in this equation each of the letters represents a numer.

Like e is 3 and m is 6 and p is 5.

So what number does "Peem" represent?

Results 1 to 3 of 3

- Sep 6th 2007, 02:08 PM #1

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- Sep 2007
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- Sep 6th 2007, 03:36 PM #2
The setup is

$\displaystyle PEEM_5 = P \times 5^3 + E \times 5^2 + E \times 5^1 + M \times 5^0 = 125P + 30E + M$

$\displaystyle PEEM_7 = P \times 7^3 + E \times 7^2 + E \times 7^1 + M \times 7^0 = 343P + 56E + M$

$\displaystyle PEEM_8 = P \times 8^3 + E \times 8^2 + E \times 8^1 + M \times 8^0 = 512P + 72E + M$

So the problem reads:

$\displaystyle (125P + 30E + M) + (343P + 56E + M) = (512P + 72E + M)$

$\displaystyle 44P - 14E - M = 0$

Now, we know that P, E, and M are all less than 5 since PEEM is a number in base 5.

So we can start by postulating P = 1:

$\displaystyle 14E + M = 44$

M is at most 4, so we need E to be at least 3. (And, as it turns out, less than 4.) So try E = 3:

M = 2.

So PEEM = 1332 works.

I can find no other solutions.

-Dan

- Oct 4th 2007, 02:37 PM #3
Since E and M is

__<__4, 14 E + M is__<__15*4 = 60. Any P__>__2 would make 44 P__>__88, hence the equation doesn't meet in both ends. So P__<__1. P = 0 makes 14 E + M = 0, which only solution is E = M = P = 0.

So PEEM is either 1332 as topsquark pointed out or 0000.