Number of integer solutions..

Mar 2019
7
0
bangalore
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:confused:
 
Last edited:
Feb 2010
739
162
How many integer solutions does this expression have:
x1*x2*3*x4 = 770

So, 2*5*7*10 = 770

Where am I going wrong:confused:
Right off the bat ... \(\displaystyle 2 \cdot 5 \cdot 7 \cdot 10 = 700 \neq 770\)
 
Mar 2019
7
0
bangalore
Sorry...read it as 2â‹…5â‹…7â‹…11
 

Denis

Math Team
Oct 2011
14,592
1,026
Ottawa Ontario, Canada
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:
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skipjack

Forum Staff
Dec 2006
21,482
2,472
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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Mar 2019
7
0
bangalore
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!?
 
Jun 2016
25
2
Hong Kong
That should be Stirling numbers of the second kind if you're distributing n elements into k parts

$\frac{4!}{0!}\left\{ {4 \atop 4} \right\}=24$
$\frac{4!}{1!}\left\{ {4 \atop 3} \right\}=144$
$\frac{4!}{2!}\left\{ {4 \atop 2} \right\}=84$
$\frac{4!}{3!}\left\{ {4 \atop 1} \right\}=4$