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Prime Factor Problem
[#permalink]
26 Dec 2010, 09:42
Hi
Could someone please help with this, I know that we can use the FTA to multiply the number of factors, but I'm unable to reckon clearly beyond that.
The number 105840 can be expressed in prime factors as 2^4 * 3^3 * 5 * 7^2 (a) How many of these factors in (excluding 1 and 105840) are divisible by 9? (b) How many of these factors in (excluding 1 and 105840) are divisible by 4 but not by 8?
Answer (a) - 59
Answer (b) - 24
Regards, Subhash
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Re: Prime Factor Problem
[#permalink]
26 Dec 2010, 22:07
Expert Reply
subhashghosh wrote:
Hi
Could someone please help with this, I know that we can use the FTA to multiply the number of factors, but I'm unable to reckon clearly beyond that.
The number 105840 can be expressed in prime factors as 2^4 * 3^3 * 5 * 7^2 (a) How many of these factors in (excluding 1 and 105840) are divisible by 9? (b) How many of these factors in (excluding 1 and 105840) are divisible by 4 but not by 8?
Answer (a) - 59
Answer (b) - 24
Regards, Subhash
\(N = 2^4 * 3^3 * 5 * 7^2\)
The total number of factors of N = 5*4*2*3 = 120
(a) How many of these factors in (excluding 1 and 105840) are divisible by 9?
For a factor to be divisible by 9, it should have at least two 3s. So you can take 3s in two ways (either take two 3s or three 3s) You can take 2s in five ways, 5s in two ways and 7s in three ways. Total such factors = 2*5*2*3 = 60 factors. But this includes the number 105840 as a factor which we need to exclude. So total such factors = 60 - 1 = 59
(b) How many of these factors in (excluding 1 and 105840) are divisible by 4 but not by 8 For a factor to be divisible by 4 but not by 8, it should have exactly two 2s. So you can take 2s in only one way. You can take 3s in four ways, 5s in two ways and 7s in three ways. Total such factors = 1*4*2*3 = 24 factors. (This does not include 105840 as a factor because 105840 has four 2s, not two 2s.)
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