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The Magic Cafe Forum Index » » Puzzle me this... » » A matter of factorization (0 Likes) Printer Friendly Version

landmark
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After a little bit of pencil pushing, you find that your favorite number can be expressed in the prime factored form of (a^x) * (b^y)* (c^z) . . . etc. , where a, b, c . . . are distinct primes, and x, y, z . . . are integers.

How many factors does your number have?

How can you show that your answer must be correct?

e.g. 16200 can be expressed as (2^3) (3 ^4) (5^2) . How many factors (prime and non-prime) does 16200 contain?

Jack Shalom
Psy-Kosh
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There're a couple ways of looking at that.

The simplest is this: consider 2^x * 3^y * 5^z

now, let x, y, and z range between 0 and the actual correct exponent for 16200

Each of those possibilities will correspond to a unique number that divides 16200. (if you don't want do count 1 as a factor, just subtract 1 from the answer..)

so we have 4*5*3 = 60 possibilities. or 59 not counting 1. or 58 not counting either 1 or 16200.
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