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Newsless Alphametics (Posted on 2009-04-15) Difficulty: 2 of 5
x, y and z (in this order), with x < y < z, are three positive integers in arithmetic sequence that satisfy this set of alphametic equations.

y4 - x4 = NEWS, and

z4 - y4 = LESS.

where, each of the capital letters in bold represents a different decimal (base 10) digit from 0 to 9.

Determine all possible value(s) of the triplet (x, y, z).

What are the possible four digit number(s) that represent WELL?

Note: Neither L nor N is zero.

No Solution Yet Submitted by K Sengupta    
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Solution computer solution | Comment 3 of 5 |

The following program determines that z can be no larger than 14, as, if it were, then even only a 1 unit difference with y would make z^4-y^4 be longer than 4 digits:

   10   for Z=10 to 10000
   20   Diff=Z^4-(Z-1)^4
   25   print Z,Diff
   30   if Diff>9999 then print Z:end
   40   next
 10      3439
 11      4641
 12      6095
 13      7825
 14      9855
 15      12209
No for(next)

That leads to the following program and its run:

   10   for X=1 to 12
   20   for Y=X+1 to 13
   30   for Z=Y+1 to 14
   40      News=cutspc(str(Y^4-X^4))
   50      Less=cutspc(str(Z^4-Y^4))
   60      Good=1
   70      if len(News)<>4 or len(Less)<>4 then Good=0
   80      if Good then
   90        :for I=1 to 3
  100         :if instr(I+1,News,mid(News,I,1))>0 then Good=0:endif
  110        :next
  120        :if mid(Less,3,1)<>mid(Less,4,1) then Good=0:endif
  130        :if mid(Less,3,1)<>mid(News,4,1) then Good=0:endif
  140        :if mid(Less,2,1)<>mid(News,2,1) then Good=0:endif
  150        :if instr(2,Less,mid(Less,1,1))>0 then Good=0:endif
  160        :if instr(News,mid(Less,1,1))>0 then Good=0:endif
  170        :if Good then print X;Y;Z,News;" ";Less
  180   next
  190   next
  200   next
 12  13  14     7825 9855

So (x,y,z) can only be (12,13,14). WELL is 2899.

The program did not check for the integers x, y and z to be in arithmetic sequence, but the only set that fit the other criteria were in fact in arithmetic sequence.

In fact, even if we allowed x, y and z to be any digits between 0 and 14 without regard to which is largest or smallest, there are no further solutions.

  Posted by Charlie on 2009-04-15 13:48:49
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