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Nod to fife and food (Posted on 2012-06-26) Difficulty: 3 of 5
Substitute each of the capital letters by a different digit from 0 to 9 to satisfy this set of alphametic relationships. None of the numbers can contain any leading zero.

The arithmetic mean of NOD and FIFE is FERN, and:

The geometric mean of NOD and FIFE is FOND, and:

The harmonic mean of NOD and FIFE is FOOD.

Note: For an extra challenge, solve this puzzle without using a computer program.

No Solution Yet Submitted by K Sengupta    
Rating: 4.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: Surprised Comment 5 of 5 |
(In reply to Surprised by brianjn)

My program was a slovenly creation that took the shortcut of permuting all 10 digits though only 7 would be needed, and therefore produced the solution six times over--one for each permutation of the remaining three digits. It took longer to run, but was shorter to write. It did indeed, though, test only the arithmetic and geometric means, and that was sufficient to get the unique (albeit repeated six times) solution.

DECLARE SUB permute (a$)
CLS
a$ = "0123456789": h$ = a$
DO
  n = VAL(MID$(a$, 3, 1))
  o = VAL(MID$(a$, 4, 1))
  d = VAL(MID$(a$, 5, 1))
  f = VAL(MID$(a$, 6, 1))
  i = VAL(MID$(a$, 7, 1))
  e = VAL(MID$(a$, 8, 1))
  r = VAL(MID$(a$, 9, 1))
  nod = 100 * n + 10 * o + d
  fife = 1010 * f + 100 * i + e
  fern = 1000 * f + 100 * e + 10 * r + n
  fond = 1000 * f + 100 * o + 10 * n + d
  food = 1000 * f + 110 * o + d
  IF fern - nod = fife - fern THEN
     IF fond * fond = nod * fife THEN
         PRINT nod; fife; fern; fond; food
     END IF
  END IF
  permute a$
LOOP UNTIL a$ = h$


  Posted by Charlie on 2012-06-27 10:25:45
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