Determine the total number of positive divisors of 1890*1930*1970 such that none of the said divisors is divisible by 45.
(In reply to
solution by Jer)
In the product 1890*1930*1970 = 7185969000, factors are:
2^3
3^3
5^3
7
193
197
We want positive divisors such that if 5 is present as a factor, 3 can be a factor only in the first power if at all.
When 5 is not a factor there are 4*4*2*2*2 = 128 choices for inclusion of the other prime factors. This includes 1, but not the number itself, of course.
When 5 is a factor at least once, there are 4*2*3*2*2*2 = 192 choices, as total absence is not a choice for 5 and 3 is either included once or excluded. This set includes the number itself, but not 1, as the smallest such divisor is 5.
That makes the total possibilities 128 + 192 = 320.
|
|
Posted by Charlie
on 2015-03-06 10:39:41 |
re: solution -- same answer, a different way
