Problem:
There are positive integers x and y that satisfy the system of equations
log10x+2log10(gcd(x,y))=60log10y+2log10(lcm(x,y))=570
Let m be the number of (not necessarily distinct) prime factors in the prime factorization of x, and let n be the number of (not necessarily distinct) prime factors in the prime factorization of y. Find 3m+2n.
Solution:
The two equations are equivalent to x(gcd(x,y))2=1060 and y(lcm(x,y))2= 10570, respectively. Multiplying corresponding sides of the equations leads to xy(gcd(x,y)lcm(x,y))2=(xy)3=10630, so xy=10210. It follows that there are nonnegative integers a,b,c, and d such that (x,y)=(2a5b,2c5d) with a+c=b+d=210. Furthermore,
x(lcm(x,y))2=xyy(lcm(x,y))2=1021010570=10360
Thus max(2a,2c)−a=max(2b,2d)−b=360. Because neither 2a−a nor 2b−b can equal 360 when a+c=b+d=210, it follows that 2c−a=2d−b=360. Hence (a,b,c,d)=(20,20,190,190), so the prime factorization of x has 20+20=40 prime factors, and the prime factorization of y has 190+190=380 prime factors. The requested sum is 3⋅40+2⋅380= 880.
The problems on this page are the property of the MAA's American Mathematics Competitions