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
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Submitted by Praneeth
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Rating: 4.0000 (1 votes)
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Solution:
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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. |
