Harry, Tom and I each replaced each asterisk in the diagram below with a single digit, such that the two horizontal 4-digit numbers, the horizontal 3-digit number and the vertical 5-digit number were all perfect cubes:

****
   *
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We each found a different solution that used exactly nine of the ten digits 0-9. Harry's unused digit was the same as Tom's.

What was my solution?

The cubes can only be:

3-digit cubes: 125 / 216 / 343 / 512 / 729.

4-digit cubes: 1000 / 1331 / 1728 / 2197 / 2744 / 3375 / 4096 / 4913 / 5832 / 6859 / 8000 / 9261.

5-digit cubes: 10648 / 12167 / 13824 / 15625 / 17576 / 19583 / 21952 / 24389 / 27000 / 29791 / 32768 / 35937 / 39304 / 42875 / 46656 / 50653 / 54872 / 59319 / 64000 / 68921 / 74088 / 79507 / 85184 / 91125 / 97336.

It seems that there is a lot of possibilities, but in fact there are only very few to be analyzed. The number of 3-digit cubes (5) invites us to approach the problem from them.

Letīs call the 3-digit middle cube, MC; the 5-digit vertical cube, VC; the 4-digit upper cube, UC; the 4-digit bottom cube, BC; and the missing digits, MD. We have:

      MC        VC       UC        BC         MD  -----------------------------------------------------  a)   125      17576     1331      4096      (8)  b)                      9261      4096      (3,8)   c)            79507     2197      2197      (3,4,6,8)  d)   216      15625     1331      3375      (0,4,8,9)  e)                      9261      3375      (0,4,8)  f)            19683     1331      4913      (0,5,7)  g)                      9261      4913      (0,4,5,7)  h)            46656     2744      4096      (3,8)  i)            50653     3375      4913      (8)  j)   343      24389     5832      6859      (0,1,7)  k)            39304     4913      2744      (5,6,8)   l)            59319     3375      6859      (0,2)  m)            97336     6859      4096      (1,2)  n)   729      21952     5832      5832      (0,4,6)   o)            35937     4913      2197      (0,6,8)  p)            68921     4096      1331      (5)  q)                                9261      (3,5)
Observe that we didnīt use the 3-digit cube 512,   because none of the 5-digit cubes has a "2" as its middle digit.     The unique cases where only one digit is missing are the   cases a(8), i(8), and p(5).    And since the missing digits in your friendīs solutions  are equal, they found (a) and (i), and your solution is (p).
       friend1       friend2             you            1331           3375             4096             7              0                8           125            216              729             7              5                2          4096           4913             1331
    

Edited on November 19, 2005, 6:02 pm

Posted by pcbouhid on 2005-11-19 07:47:34