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 A problem on factors (Posted on 2007-07-16)
Let P(X,A) be the number of positive divisors of A not greater than X; for example, P(1,12)=1, P(2,12)=2, and P(12,12)=6.

1)If P(12,A)=7 and P(A,A)=10, what's A?
2)If x*y is a divisor of A, then show that P(x*y,A) ≥ P(x,A)+P(y,A)-1

 Submitted by Praneeth Rating: 4.0000 (1 votes) Solution: (Hide) 1)If x*y =A, then P(x,A)+P(y,A)=d(A)+1.----(1) So, here P(12,A)=7 So, P(A/12,A)=10+1-7=4. As 12 is a factor of A, all the factors of 12 are factors of A.1,2,3,4,6,12 are factors of A. So, 1st factor of 48 must be 1 and so on 4th factor of 48 must be 4. So, A/12=4 which implies A =12*4=48. Note: If the factors of a number(A) are arranged in ascending order, then (d(A)-r)th factor*(r+1)th factor =A. 2)Without loss of generality, take x < y < xy. No. of positive divisors of A but not xy less than x = P(x,A)-P(x,xy). These divisors when multiplied by y, will be divisors of A between y and xy, but not divisors of xy. No. of positive divisors of A but not xy less than y = P(y,A)-P(y,xy). P(xy,A)≥ d(xy)+P(x,A)-P(x,xy)+P(y,A)-P(y,xy) From eq(1) P(xy,A)≥P(x,A)+P(y,A)-1.

 Subject Author Date re: Part 1 Praneeth Yalavarthi 2007-07-16 12:11:36 Part 1 Jer 2007-07-16 10:57:36

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