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If a,b,c,d are non negative integers and a+b+c+d = 4 then what is the number of...

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christiano-cr7 | Salutatorian

Posted July 10, 2013 at 11:05 AM via web

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If a,b,c,d are non negative integers and a+b+c+d = 4 then what is the number of different ways that a^2+b^2+c^2+d^2 can take?

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted July 10, 2013 at 11:33 AM (Answer #1)

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If a,b,c and d are non negative integers they can only take 0,1,2 and 3. In the question it does not say that a,b,c and d are different integers. So we can take same integers for any one of a,b,c and d which satisfies the conditions.

 

a+b+c+d = 4

`(1) rarr 0+0+0+4^2 = 16`    (a = b = c = 0 and d = 4)

`(2) rarr 0+0+1^2+3^2 = 10`   

`(3) rarr 0+0+2^2+2^2 = 8`

`(4) rarr 0+0+3^2+1^2 = 10`   

(note that (4)=(2).So it can be neglected)

`(5) rarr 0+0+4^2+0 = 16`   

(note that (5)=(1).So it can be neglected)

`(6) rarr 0+1^2+1^2+2^2 = 6`

`(7) rarr 0+1^2+3^2+0 = 10`  

(note that (7)=(4).So it can be neglected)

`(8) rarr 1^2+1^2+1^2+1^2 = 4`

Other options will look like one of the above. So they can be neglected.

So we have 5 ways to find (`a^2+b^2+c^2+d^2` )

 

Note

You have to apply integers to a,b,c abd d systematically as you see above.

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