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primes, multiples, sums

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by kanha81 » Wed Mar 04, 2009 12:36 pm
q: How many positive integers less than 28 are prime numbers, odd multiples of 5, or the sum of positive multiple of 2 and positive multiple of 4?

1) 27
2) 25
3) 24
4) 22
5) 20


What should be the approach to solve it efficiently? For me this is a time consuming problem?
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by DanaJ » Wed Mar 04, 2009 1:22 pm
I'd go with the following line of thought:
a. prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23 - 9 numbers
b. odd multiples of 5: 15, 25 - 2 numbers (5 has already been counted)
c. positive multiples of 4 and 2: start with 4*1 + 2*1. Then every even number can be counted in this series, just by considering every number as 4*1 + 2*n, with n from 1 to 11. This takes us to another 11 numbers.

So the number we're looking for is 9 + 2 + 11 = 22.

My guess is D

What is the OA?
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Re: primes, multiples, sums

by kamu » Wed Mar 04, 2009 2:35 pm
kanha81 wrote:q: How many positive integers less than 28 are prime numbers, odd multiples of 5, or the sum of positive multiple of 2 and positive multiple of 4?

1) 27
2) 25
3) 24
4) 22
5) 20


What should be the approach to solve it efficiently? For me this is a time consuming problem?
primes = 9 numbers.
5 multiples = 15,25 = 2 new numbers.

The next part i'm a lil confused.

sum of multiples of 2 and multiples of 4 less than 28.

Therefore,
4 can be added to all even numbers less than 24. 11 combos.
8 can be added to all even numbers less than 18. 9 combos.
similarly 12,16,20..

I'm missing something! Not sure what..
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by kanha81 » Wed Mar 04, 2009 3:45 pm
DanaJ wrote:I'd go with the following line of thought:
a. prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23 - 9 numbers
b. odd multiples of 5: 15, 25 - 2 numbers (5 has already been counted)
c. positive multiples of 4 and 2: start with 4*1 + 2*1. Then every even number can be counted in this series, just by considering every number as 4*1 + 2*n, with n from 1 to 11. This takes us to another 11 numbers.

So the number we're looking for is 9 + 2 + 11 = 22.

My guess is D

What is the OA?
That's very neat, but would you not consider: 4*n + 2*1, with n from 1 to 5. This takes to another 5 numbers.

So the number is 9 + 2 + 11 + 5 = 27.

Hence [spoiler][A][/spoiler]
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by DanaJ » Wed Mar 04, 2009 9:58 pm
kanha81 wrote: That's very neat, but would you not consider: 4*n + 2*1, with n from 1 to 5. This takes to another 5 numbers.

So the number is 9 + 2 + 11 + 5 = 27.

Hence [spoiler][A][/spoiler]
Well, 4*1 + 2*1 = 4*1 + 2*1
4*2 + 2*1 = 8 + 2 = 10 = 4*1 + 2*3
My point is that if you consider 4*n + 2*1, with n from 1 to 5, EVERY answer will be repeated, since any combination of 4*n + 2*1 can be re-written as 4*1 + 2*k. This is why I believe you counted the 5 numbers twice. Please let me know if you think I'm wrong.
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by kanha81 » Thu Mar 05, 2009 9:10 am
DanaJ wrote:
kanha81 wrote: That's very neat, but would you not consider: 4*n + 2*1, with n from 1 to 5. This takes to another 5 numbers.

So the number is 9 + 2 + 11 + 5 = 27.

Hence [spoiler][A][/spoiler]
Well, 4*1 + 2*1 = 4*1 + 2*1
4*2 + 2*1 = 8 + 2 = 10 = 4*1 + 2*3
My point is that if you consider 4*n + 2*1, with n from 1 to 5, EVERY answer will be repeated, since any combination of 4*n + 2*1 can be re-written as 4*1 + 2*k. This is why I believe you counted the 5 numbers twice. Please let me know if you think I'm wrong.
Nope. I think you're absolutely right. Thank you.
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