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March 20th, 2019, 05:18 AM   #1
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Number of integer solutions..

How many integer solutions does this expression have:
x1*x2*3*x4 = 770

So, 2*5*7*11 = 770

But the catch is that either one, two, or three of these variables can be 1 as well.

So, 70*11*1*1 = 770
Similarly, 770*1*1*1 = 770.

The way I proceeded with this question, was a follows:

i) No "1": Number of cases = 4! = 24
ii) One "1": Number of cases = C(4,3)*4! = 96
iii) Two "1": Number of cases = C(4,2)*4!/2! =72
iv) Three "1": Number of cases = 4

So, total = 24 + 96 + 72 + 4 = 196

However, I believe (not 100% sure) that the correct answer is 256.

Where am I going wrong

Last edited by mohish; March 20th, 2019 at 05:37 AM.
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March 20th, 2019, 05:25 AM   #2
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Quote:
Originally Posted by mohish View Post
How many integer solutions does this expression have:
x1*x2*3*x4 = 770

So, 2*5*7*10 = 770

Where am I going wrong
Right off the bat ... $\displaystyle 2 \cdot 5 \cdot 7 \cdot 10 = 700 \neq 770$
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March 20th, 2019, 05:36 AM   #3
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Sorry...read it as 2⋅5⋅7⋅11
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March 20th, 2019, 07:15 AM   #4
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Quote:
Originally Posted by mohish View Post
i) No "1": Number of cases = 4! = 24
ii) One "1": Number of cases = C(4,3)*4! = 96
iii) Two "1": Number of cases = C(4,2)*4!/2! =72
iv) Three "1": Number of cases = 4

So, total = 24 + 96 + 72 + 4 = 196

However, I believe (not 100% sure) that the correct answer is 256.
256 is the correct answer.

Your i) and iv) are correct

Your ii) should be 144, your iii) should be 84

24 + 144 + 84 + 4 = 256

Ascending order
1: 1,1,1,770
2: 1,1,2,385
3: 1,1,5,154
4: 1,1,7,110
5: 1,1,10,77
...
252: 154,5,1,1
253: 385,1,1,2
254: 385,1,2,1
255: 385,2,1,1
256: 770,1,1,1
Thanks from mohish

Last edited by Denis; March 20th, 2019 at 07:33 AM.
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March 20th, 2019, 08:07 AM   #5
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For one "1", for example, exactly two of the four primes must be replaced by their product, so the number of cases = C(4,2)*4! = 144.
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March 20th, 2019, 08:24 AM   #6
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For one "1", for example, exactly two of the four primes must be replaced by their product, so the number of cases = C(4,2)*4! = 144.
Thanks. How about for two "1" example? Should it not be C(4,3)*4!/2!?
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March 21st, 2019, 06:51 AM   #7
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Originally Posted by mohish View Post
Thanks. How about for two "1" example? Should it not be C(4,3)*4!/2!?
Can you please respond to this query of mine, whenever you have time.
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