enigma123 wrote:
How many numbers that are not divisible by 6 divide evenly into 264,600?
(A) 9
(B) 36
(C) 51
(D) 63
(E) 72
Any idea how to solve this please?
Finding the Number of Factors of an IntegerFirst make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.
The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\).
NOTE: this will include 1 and n itself.
Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)
Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.
For more on number properties check:
math-number-theory-88376.htmlBACK TO THE ORIGINAL QUESTION:
How many numbers that are not divisible by 6 divide evenly into 264,600?(A) 9
(B) 36
(C) 51
(D) 63
(E) 72
\(264,600=2^3*3^3*5^2*7^2\), thus it has total of \((3+1)(3+1)(2+1)(2+1)=144\) differernt positive factors, including 1 and the number itself.
# of factors that ARE divisible by 6 will be \(3*3*(2+1)(2+1)=81\): we are not adding 1 to the powers of 2 and 3, this time, to exclude all the cases with 2^0*... and 3^0*... (thus to exclude all the factors which are not divisible by 2 or 3), hence ensuring that at least one 2 and at least one 3 are present to get at least one 6 in the factors we are counting.
So, # of factors that ARE NOT divisible by 6 is 144-81=63.
Answer: D.
Another solution here:
new-set-of-good-ps-85440.html#p642384Hope now it's clear.
why did you multiply by (2+1) and (2+1) again, shouldn't it have been (2+1)(2+1)(2+1) since there's a 2^3 in the prime factorization?