likesachallenge wrote:In base four, each digit in a number represents the number of copies of that power of four....
The above was an uncredited portion of
this article. If you read the article, then you may have a better understanding of the various solutions which have been posted. :wink:
The 'Lenny Conundrum' authors, in Round 230, wrote:In the following equation, M, E, and P represent specific digits.
(MEEP)base 5 + (MEEP)base 7 = (MEEP)base 8
What is the number represented by MEEP?
Using the method explained in the article (above), we first note that, since the smallest base is base-5, none of the digits can be a 5 or greater. We have to work with 0, 1, 2, 3, and 4. Also, M cannot reasonably be zero. We will assume that "specific" implies "distinct"; that is, M, E, and P represent three
different digits from among our choices.
Converting the base-x expressions into base-10 (that is, "regular") values, we have:
. . . . .(125M + 25E + 5E + P) + (343M + 49E + 7E + P) = 512M + 64E + 8E + P
Simplify:
. . . . .468M + 86E + 2P = 512M + 72E + P
Combine "like" terms, and simplify some more:
. . . . .14E + P = 44M
Note that, since 14 is even and 44 is even, then 14E and 44M will be even. The only way for the last equation above to work is for P then to be even. That is, P must be one of 0, 2, and 4.
Let's work by cases, since we have only five digits from which to choose.
If M = 0, then the leading digit in each number is a zero, which isn't how we write numbers. So M cannot be zero.
If M = 1, then:
. . . . .14E + P = 44
If P = 0, then:
. . . . .14E + 0 = 44
. . . . .14E = 44
. . . . .E = 44/14 = 22/7
This won't work. On to the next sub-case: If P = 2, then:
. . . . .14E + 2 = 44
. . . . .14E = 42
. . . . .E = 42/14 = 3
This will work! M = 1, P = 2, E = 3. Continue checking for any other solutions.
If P = 4, then:
. . . . .14E + 4 = 44
. . . . .14E = 40
. . . . .E = 40/14 = 20/7
This also won't work. Check the next case for M.
If M = 2, then:
. . . . .14E + P = 88
If P = 0, then:
. . . . .14E + 0 = 88
. . . . .14E = 88
. . . . .E = 88/14 = 44/7
No. If P = 2, then:
. . . . .14E + 2 = 88
. . . . .14E = 86
. . . . .E = 86/14 = 43/7
No. If P = 4, then:
. . . . .14E + 4 = 88
. . . . .14E = 84
. . . . .E = 84/14 = 6
But 6 is too large a digit. We've exhausted the sub-cases, so M cannot be 2.
If M = 3, then:
. . . . .14E + P = 132
If P = 0, then:
. . . . .14E + 0 = 132
. . . . .14E = 132
. . . . .E = 132/14 = 66/7
No. If P = 2, then:
. . . . .14E + 2 = 132
. . . . .14E = 130
. . . . .E = 130/14 = 65/7
No. If P = 4, then:
. . . . .14E + 4 = 132
. . . . .14E = 128
. . . . .E = 128/14 = 64/7
No. So M cannot be 3.
If M = 4, then:
. . . . .14E + P = 176
Consider the sub-cases: If P = 0, then:
. . . . .14E + 0 = 176
. . . . .14E = 176
. . . . .E = 176/14 = 88/7
No. If P = 2, then:
. . . . .14E + 2 = 176
. . . . .14E = 174
. . . . .E = 174/14 = 87/7
No. IF P = 4, then:
. . . . .14E + 4 = 176
. . . . .14E = 172
. . . . .E = 172/14 = 86/7
No. So M cannot be 4. There remains only one value to check for M:
If M = 5, then:
. . . . .14E + P = 220
If P = 0, then:
. . . . .14E + 0 = 220
. . . . .14E = 220
. . . . .E = 220/14 = 110/7
No. If P = 2, then:
. . . . .14E + 2 = 220
. . . . .14E = 218
. . . . .E = 218/14 = 109/7
No. If P = 4, then:
. . . . .14E + 4 = 220
. . . . .14E = 216
. . . . .E = 216/14 = 108/7
No.
This exhausts all possible cases, with only one solution found:
. . . . .********************
. . . . .** M = 1, E = 3, P = 2 **
. . . . .********************
Verifying:
. . .1332(base 5) = 1(5^3) + 3(5^2) + 3(5^1) + 2(5^0)
. . . . .= 1(125) + 3(25) + 3(5) + 2(1)
. . . . .= 125 + 75 + 15 + 2 = 217(base 10)
. . .1332(base 7) = 1(7^3) + 3(7^2) + 3(7^1) + 2(7^0)
. . . . .= 1(343) + 3(49) + 3(7) + 2(1)
. . . . .= 343 + 147 + 21 + 2 = 513(base 10)
. . .1332(base 8) = 1(8^3) + 3(8^2) + 3(8^1) + 2(8^0)
. . . . .= 1(512) + 3(64) + 3(8) + 2(1)
. . . . .= 512 + 192 + 24 + 2 = 730(base 10)
. . . . .217 + 513 = 730
So the only solution we found is now confirmed valid.
Please check my work.
Eliz.