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 The box (Posted on 2009-01-02)
A box has integer dimensions, and when each one is increased by 2, its volume doubles. What is the largest possible dimension?

 See The Solution Submitted by pcbouhid No Rating

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 computer exploration (spoiler) | Comment 3 of 7 |

PRINT : PRINT "--"
FOR t = 2 TO 15000
FOR a = 1 TO t / 2
b = t - a
x = 2 * a * b / ((a + 2) * (b + 2))
IF x <> 1 THEN
c = INT(2 / (x - 1) + .5)
IF c >= b THEN
IF (a + 2) * (b + 2) * (c + 2) = 2 * a * b * c THEN
PRINT a; b; c, a + 2; b + 2; c + 2
END IF
END IF
END IF
NEXT
NEXT

finds only the following, after checking all cases where the smaller two dimensions add up to 15,000 or less (23 being the largest total of those two dimensions found):

` 5   5   98     7   7  100 4   7   54     6   9   56 5   6   28     7   8   30 4   8   30     6  10   32 5   7   18     7   9   20 6   6   16     8   8   18 4   9   22     6  11   24 5   8   14     7  10   16 6   7   12     8   9   14 3  11  130     5  13  132 4  10   18     6  12   20 6   8   10     8  10   12 3  12   70     5  14   72 3  13   50     5  15   52 4  12   14     6  14   16 3  14   40     5  16   42 3  15   34     5  17   36 3  16   30     5  18   32 3  18   25     5  20   27 3  20   22     5  22   24`

so it's highly likely that 3x11x130 contains the highest dimension before increasing by 2, and that 132 is the largest dimension after the increase by 2.

3x11x130 = 4290
5x13x132 = 8580

8580/4290 = 2

Edited on January 2, 2009, 6:38 pm
 Posted by Charlie on 2009-01-02 18:33:28

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