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An Exponent Product And 98 Zeroes Puzzle (Posted on 2007-10-27) Difficulty: 3 of 5
Find the positive integer pair (m, n) such that m*n is the minimum and the last 98 digits in mm*nn are all zeroes.

Note: 99th digit from the right in mm*nn is not zero.

See The Solution Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

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Solution Basic programming language confirms | Comment 9 of 12 |

The lowest I get is 75*98=7350.

CLS

FOR b = 0 TO 13
 FOR y = 0 TO 13
  FOR c = 0 TO 5
   FOR z = 0 TO 5
    m0 = INT(2 ^ b * 5 ^ c + .5)
    n0 = INT(2 ^ y * 5 ^ z + .5)
    IF m0 < 20000 AND n0 < 20000 THEN
     a = 1
     DO
      IF a MOD 5 > 0 THEN
        x = 1
        DO
         IF x MOD 5 > 0 THEN
            m = a * m0
            n = x * n0
            p2 = b * m + y * n
            p5 = c * m + z * n
            IF p2 = 98 AND p5 >= 98 OR p5 = 98 AND p2 >= 98 THEN
             IF m * n < 10300 THEN
               PRINT m, n, m * n
             END IF
            END IF
          END IF
          x = x + 2
        LOOP UNTIL x * n0 > 20000 OR x > 100
      END IF
      a = a + 2
     LOOP UNTIL a * m0 > 20000 OR a > 250
    END IF
   NEXT z
  NEXT c
 NEXT y
NEXT b

produced

105           98            10290
75            98            7350
98            105           10290
98            75            7350

in under 13 seconds.


  Posted by Charlie on 2007-10-28 11:43:48
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