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Does the integer K have a factor P such that 1<P<K? 1)

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Does the integer K have a factor P such that 1<P<K? 1) [#permalink]

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26 May 2004, 15:54
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Does the integer K have a factor P such that 1<P<K?
1) K>4!
2)13!+2<=K<=13!+13

please show & explain workings
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26 May 2004, 19:26
A) K>4! K = { 25,26,27,28,29...} 29 is a prime and has no factor p such that 1<p<k
But 25 and 26 do So Insufficient

2)
13!+2 is divisible by 2 hence 1<2<(13!+2)
13!+3 is divisible by 3 hence 1<3<(13!+3)
13!+4 is divisible by 4 hence 1<4<(13!+4)
13!+5 is divisible by 5 hence 1<5<(13!+5)
13!+6 is divisible by 6 hence 1<6<(13!+6)
13!+7 is divisible by 7 hence 1<7<(13!+7)
13!+8 is divisible by 8 hence 1<8<(13!+8)
13!+9 is divisible by 9 hence 1<9<(13!+9)
13!+10 is divisible by 10 hence 1<10<(13!+10)
13!+11 is divisible by 11 hence 1<11<(13!+11)
13!+12 is divisible by 12 hence 1<12<(13!+12)
13!+13 is divisible by 13 hence 1<12<(13!+13)

All the integers between 13!+2 and 13!+13 have one factor p
such that 1<p<k
Sufficient.
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27 May 2004, 06:29
anandnk wrote:
A) K>4! K = { 25,26,27,28,29...} 29 is a prime and has no factor p such that 1<p<k
But 25 and 26 do So Insufficient

2)
13!+2 is divisible by 2 hence 1<2<(13!+2)
13!+3 is divisible by 3 hence 1<3<(13!+3)
13!+4 is divisible by 4 hence 1<4<(13!+4)
13!+5 is divisible by 5 hence 1<5<(13!+5)
13!+6 is divisible by 6 hence 1<6<(13!+6)
13!+7 is divisible by 7 hence 1<7<(13!+7)
13!+8 is divisible by 8 hence 1<8<(13!+8)
13!+9 is divisible by 9 hence 1<9<(13!+9)
13!+10 is divisible by 10 hence 1<10<(13!+10)
13!+11 is divisible by 11 hence 1<11<(13!+11)
13!+12 is divisible by 12 hence 1<12<(13!+12)
13!+13 is divisible by 13 hence 1<12<(13!+13)

All the integers between 13!+2 and 13!+13 have one factor p
such that 1<p<k
Sufficient.

wow! I admire your patience. But anyone with a shortcut method?
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hi [#permalink]

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27 May 2004, 08:08
Hi Anand,

Is there a general number property like

n!+k is divisible by k or did you do actually figure it out for each one of these numbers?

Thanks
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27 May 2004, 09:21
Hi,

No formula for this. I just did it in my brain but I wanted to explain in detail so I wrote so much. Actually kpadma asked me to solve the same problem once during yahoo chat. I remembered the explanation I gave and just reproduced it here.

any number x!+y is divisible by y as long as y <= x because y is a factor of x
if y > x then cannot say for sure.

Anand.
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Thanks [#permalink]

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27 May 2004, 11:11
Hi Anand,

Thank you for your explanation, it makes sense now.
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Re: DS:inequalities :challanging [#permalink]

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28 May 2004, 08:44
mirhaque wrote:
Does the integer K have a factor P such that 1<P<K?
1) K>4!
2)13!+2<=K<=13!+13

please show & explain workings

The question can be rephrased as is K prime, because if K being prime is the only time that 1<P<K will not hold true.

(1) is very insufficient. It just says that K>24. Plenty of primes and non-primes out there.

(2) You gotta find out if there are any primes btwn 13!+2 and 13!+13

If you add a number to a multiple of that number, the sum will also be a multiple of the number (i.e. 4 is a multiple of 2, so 4+2 will also be a multiple of 2). Which means that the sum cannot be prime.

13! is a multiple of every # btwn 1 and 13, inclusive. Therefore, if you add any number btwn 1 and 13 to 13!, the sum will be a multple of the addend and thus will not be prime. Hence (2) is sufficient.

B is the answer.
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Re: DS:inequalities :challanging [#permalink]

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29 May 2004, 02:32
mirhaque wrote:
Does the integer K have a factor P such that 1<P<K?
1) K>4!
2)13!+2<=K<=13!+13

please show & explain workings

1 is not sufficient: 25 and 29 are the examples: 25 = 5*5, 29 is prime number.

2 is sufficient to answer the question: for every 2 <= m <= 13, 13! + m :: m.
Re: DS:inequalities :challanging   [#permalink] 29 May 2004, 02:32
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