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 March 20th, 2019, 05:18 AM #1 Newbie   Joined: Mar 2019 From: bangalore Posts: 7 Thanks: 0 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.
March 20th, 2019, 05:25 AM   #2
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Quote:
 Originally Posted by mohish 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 ... $\displaystyle 2 \cdot 5 \cdot 7 \cdot 10 = 700 \neq 770$

 March 20th, 2019, 05:36 AM #3 Newbie   Joined: Mar 2019 From: bangalore Posts: 7 Thanks: 0 Sorry...read it as 2⋅5⋅7⋅11
March 20th, 2019, 07:15 AM   #4
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 Originally Posted by mohish 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.

Your i) and iv) are correct

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

Last edited by Denis; March 20th, 2019 at 07:33 AM.

 March 20th, 2019, 08:07 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,921 Thanks: 2203 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 from mohish
March 20th, 2019, 08:24 AM   #6
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 Originally Posted by skipjack 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!?

March 21st, 2019, 06:51 AM   #7
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 Originally Posted by mohish 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.

 July 6th, 2019, 04:07 AM #8 Newbie     Joined: Jun 2016 From: Hong Kong Posts: 25 Thanks: 2 That should be Stirling numbers of the second kind if you're distributing n elements into k parts $\frac{4!}{0!}\left\{ {4 \atop 4} \right\}=24$ $\frac{4!}{1!}\left\{ {4 \atop 3} \right\}=144$ $\frac{4!}{2!}\left\{ {4 \atop 2} \right\}=84$ $\frac{4!}{3!}\left\{ {4 \atop 1} \right\}=4$

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