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john !!
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#1
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if the lowest common denominator of (10!)(18!) and (12!)(17!) is expressed as (a!b!)/c!, hwat is the value of abc?

a set is defined as :
{n|n is a multiple of 3 less than 100} ∩ {n|n is a multiple of 4 less than 100}
how many subsets does this set have?

for all values of x,
(x-a)(x-10) + 1 = (x-b)(x-c)
give all possible integer combinations of a,b and c
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RichE
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(Original post by mik1w)
a set is defined as :
{n|n is a multiple of 3 less than 100} ∩ {n|n is a multiple of 4 less than 100}
how many subsets does this set have?
The set is then multiples of 12 less than 100 of which there are 8 positive ones.

A set with n elements has 2^n subsets.

So the answer is 2^8=256.
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RichE
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(Original post by mik1w)
for all values of x,
(x-a)(x-10) + 1 = (x-b)(x-c)
give all possible integer combinations of a,b and c
Expanding gives

x^2 - (a+10)x + (10a+1) = x^2 - (b+c)x + bc

So, comparing coefficients,

a+10 = b+c
10a+1 = bc

Then

10b+10c-100+1 = bc

bc - 10b -10c +99 = 0

(b-10)(c-10) = 1

As b and c are integers then

b-10 = c-10 = 1 or -1

So

b = c = 11 and a = 12

or

b = c = 9 and a = 8
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RichE
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(Original post by mik1w)
if the lowest common denominator of (10!)(18!) and (12!)(17!) is expressed as (a!b!)/c!, hwat is the value of abc?
Do you mean highest common factor (or greatest common divisor)?
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john !!
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I think I meant lowest common multiple. not sure where denominator came from.
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john !!
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(Original post by RichE)
Expanding gives
x^2 - (a+10)x + (10a+1) = x^2 - (b+c)x + bc
So, comparing coefficients,
a+10 = b+c
10a+1 = bc
Then
10b+10c-100+1 = bc
bc - 10b -10c +99 = 0
(b-10)(c-10) = 1
As b and c are integers then
b-10 = c-10 = 1 or -1
So
b = c = 11 and a = 12
or
b = c = 9 and a = 8
yup good one
there is a quicker solution though :cool:
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RichE
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(Original post by mik1w)
yup good one
there is a quicker solution though :cool:
Do share :cool:
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RichE
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(Original post by mik1w)
I think I meant lowest common multiple. not sure where denominator came from.
10! 18! and 12! 17! both divide 12! 18! clearly.

They go in 11.12 and 18 times which have a highest common factor of 6.

So

lcm (10!18!,12!17!) = 12!18!/6 = 12!18!/3!.

So abc = 12.18.3 = 648
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