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A simple logical question (Posted on 2007-07-27) |
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If d(x) denotes the no. positive divisors of x and d(d(d(x))), d(d(x)), d(x) and x can be arranged in 12 different ways. Find possible values of n(x), where n(x) is no. of prime factors of x. Also find possible values of x less than 100.
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Submitted by Praneeth
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Rating: 3.3333 (3 votes)
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Solution:
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The given terms can be arranged in 12 different ways. So, 2 of the terms should be equal.
y=d(y) has only two solutions 1,2.
If one of the four terms x,d(x),d(d(x))and d(d(d(x))) is 1, it implies that all others are 1, but there is only 1 way to arrange these terms. So, x can not be 1.
If you take x=2, then d(x)=2,d(d(x))=2 and so on. From this we can say that the last 2 terms must be equal for them to be arranged in 12 different ways.
So, d(d(x))=2, this implies that d(x) is prime, which in turn implies that x has only 1 distinct prime factor.
If n(x): no. of prime factors, then n(x)=p-1. (where p is odd prime)
If n(x): no. of distinct prime factors, then n(x)=1.
Possible values of x: 4,9,25,49,16,81,64
Note:If d(n,x)=d(d(...ntimes(x))), If n->infinity, Then d(n,x)=2 (if x is a positive integer other than 1) |
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