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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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 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.
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

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 $\frac{4!}{0!}\left\{ {4 \atop 4} \right\}=24$ $\frac{4!}{1!}\left\{ {4 \atop 3} \right\}=144$ $\frac{4!}{2!}\left\{ {4 \atop 2} \right\}=84$ $\frac{4!}{3!}\left\{ {4 \atop 1} \right\}=4$ Tags integer, number, solutions Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post devenj Number Theory 1 November 30th, 2012 04:29 AM wonderwall Number Theory 7 April 23rd, 2012 12:42 PM proglote Number Theory 4 June 8th, 2011 01:41 PM calligraphy Number Theory 7 April 13th, 2011 12:56 PM SidT Number Theory 5 June 22nd, 2010 10:02 AM

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