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Remainder With Square And Cube (Posted on 2008-10-05) Difficulty: 2 of 5
M is the smallest positive integer such that 2008*M is a perfect square and 2009*M is a perfect cube.

Analytically determine the remainder when M is divided by 25.

See The Solution Submitted by K Sengupta    
Rating: 4.0000 (2 votes)

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Solution Solution | Comment 2 of 3 |

The factorization of 2008 is 23 x 251, thus a perfect square of 2008*K, where K is a positive integer, requires another factor of 2 and 251, i.e. K = 24 x 2512.

The factorization of 2009 is 72 x 41, thus a perfect cube of 2009*L, where L is a positive integer, requires another factor of 7 and two more of 41, i.e., L = 73 x 413.

Now if M must have the factors 2 x 7 x 251 x 412 and 2008*M is a perfect square and 2009*M is a perfect cube, then two more factors of 2 and 251 and three more factors of 7 are needed with the composite factors of both K and L, thus, the smallest value of M would need to have the factors 23 x 74 x 412 x 2513. These factors equate to 510588495274648.

To find the remainder of M divided by 25, as any number being a multiple of 100 is also a multiple of 25, we only need to take the last two digits, 48, modulo 25. (48 - 25 = 23) The result being 23, thus the remainder of M/25 is 23/25.

 

Edited on October 6, 2008, 11:46 pm
  Posted by Dej Mar on 2008-10-06 01:12:35

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