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# For all positive integers f, f◎ equals the distinct pairs of

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For all positive integers f, f◎ equals the distinct pairs of [#permalink]

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15 Nov 2010, 13:34
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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
[Reveal] Spoiler: OA

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15 Nov 2010, 14:17
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Great question - and my strategy for something like this is to first look at what they're asking:

The greatest number...

Which means that I want to maximize the number of factors, meaning that I also want to minimize the value of each factor (so that I can use it more often):

For example, if I use 99 (9 * 11), that 11 takes up too much space...I can't use it very often. Whereas if I use smaller factors (2 and 3), I can use them more often. 2 can become 4 and then 8 before it takes up as much space as does 11.

So if my goal is to find a number less than 100 that has as many small factors as possible, I'm looking at 96, because:

99 = 3*3*11 (and 11 is too big a prime factor)
98 = 2*7*7 (and 7 is too big a prime factor)
97 = not divisible by 2 or 3, so you're not using your smallest available factors
96 = 2*2*2*2*2*3 (or 32*3) ---> this is perfect because you're getting maximum value out of your factors. The only other that would work is 64 (drop the 3 and add another 2)

So for 96, break out the factors into pairs:

1, 96
2, 48
4, 24
8, 12 (these are easy to do - just double one number and halve the other for a fresh pair)
16, 6
32, 3

And that's as far as you can go. My fear is that we may not have enough unique factors (there's a lot of repetitiveness with the 2s), so I'll check that by trying the next smallest prime factor, 5. The biggest I can go with that is 90, or 2*3*3*5, and that gets us:

1, 90
2, 45
3, 30
5, 18
6, 15
9, 10

We're still at 6 factors, and it's going to get harder and harder to incorporate larger prime factors and have room for multiple factors in between (we've seen that 7 and 11 limit us a lot), so the answer must be 6.
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16 Nov 2010, 09:19
Used the same logic, but worked with 80.
What's the OA?

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05 Feb 2011, 21:17
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Is it 96? With 6 pairs?
Method:
Maximum power of 2 < 100 is 2^6=64

But we need to create distinct pairs right? so I removed a two and put in a 3 instead :D

2^5x3=96
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05 Feb 2011, 23:09
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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05 Jan 2013, 00:59
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.
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Re: For all positive integers f, f◎ equals the distinct pairs of [#permalink]

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05 Feb 2014, 12:55
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]

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18 Jul 2015, 22:02
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Re: For all positive integers f, f◎ equals the distinct pairs of [#permalink]

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19 Nov 2015, 05:38
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
Re: For all positive integers f, f◎ equals the distinct pairs of   [#permalink] 19 Nov 2015, 05:38
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