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Re: Try this one! [#permalink]
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What's the idea;

I took numbers in reverse order from 99 and got 6 distinct pairs in 96.

I thought it may be because 96=(2^5*3) has (5+1)*(1+1)=12(perhaps maximum) factors.

1*96,2*48*4*24,8*12,16*6,32*3

However, I was not sure while answering this. Please let us know if there is a better way.

Ans: "A"
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Re: problem solving [#permalink]
tejhpamarthi wrote:
For all positive integers f, f◎ equals the distinct pairs of positive integer factors. For example, 16◎ =3, since there are three positive integer factor pairs in 16: 1 x 16, 2 x 8, and 4 x 4.

What is the greatest possible value for f◎ if f is less than 100?
a)6 b)7 c)8 d)9 e)10


As stated earlier we need to maximize the number of factors. This can be done by using the smallest possible base and the highest possible power.
1: 2^6 = 64 => (1,2,4,8;8,16,32,64) = > this gives us 4 pairs. Though this need not give us the answer it gives us the highest power => 6. So any subsequent answer would have the sum of powers not more than 6.
2: 2^5 * 3 = 96 = > (1,2,3,4,6,8;12,16,24,32,48,96) = > this gives us 6 pairs
Other combinations such as 2^4*3^2, 2^5*5, etc would be more than 100.
Answer: A-6
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Re: For all positive integers f, f◎ equals the distinct pairs of [#permalink]
tejhpamarthi wrote:
For all positive integers f, f◎ equals the distinct pairs of positive integer factors. For example, 16◎ =3, since there are three positive integer factor pairs in 16: 1 x 16, 2 x 8, and 4 x 4.

What is the greatest possible value for f◎ if f is less than 100?

A. 6
B. 7
C. 8
D. 9
E. 10


Brian, can this be thought in the following way?

When one does prime factorization you get one factor pair for every prime number (counting repetitions). So we are basically asked how many primes can we have in this factorization so that x<100.

Well I start with 100, and 2 being the smallest prime I can get and not until 2^6 do I get a number that is smaller than 100. So that's why I chose A

Is this method correct or is it rather flawed?

Thanks
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Re: For all positive integers f, f◎ equals the distinct pairs of [#permalink]
We can also solve this by calculating from the answer option backwards and a simple P&C formula we use to calculate the number of factors, though I was not able to think through the entire thing in the first go

Formula - If a number N can be written as a product like this: P^a * Q^b.... where P,Q etc. are prime numbers and a,b...are the highest powers of these primes in the numbers, then the total number of factors for number N is given by (a+1)(b+1)(c+1) <One more additional fact here is that you'll get odd # factors for perfect squares only>

So now since we need to find pairs of factors in our question --> Let us say we have a total of N factors. Then we need to select 1 number out of N/2 factors i.e. N/2 C 1 to identify the number of pairs (since if we select one from the half, the other from the remaining half will be a fix selection)

So now back calculating from our options:
A. 6 i.e total of 12 factors which is the maximum possible under 100 i.e. for 96
All others simply get eliminated automatically

I hope this makes sense
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Re: For all positive integers f, f◎ equals the distinct pairs of [#permalink]
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Re: For all positive integers f, f◎ equals the distinct pairs of [#permalink]
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