Let x1, x2, x3 be non-negative integers such that x1 leaves remainder 0 when divided by 4, x2 leaves remainder 1 when divided by 4, and x3 leaves remainder 2 when divided by 4. Find the number of non-negative integral solutions of x1 + x2 + x3 = 35.

Question

Accepted Answer

45. Substituting x1=4a, x2=4b+1, x3=4c+2 into x1+x2+x3=35: 4a+4b+1+4c+2=35 => 4(a+b+c)=32 => a+b+c=8. Number of non-negative integer solutions = C(8+3-1, 3-1) = C(10,2) = 45.

Let x1, x2, x3 be non-negative integers such that x1 leaves remainder 0 when divided by 4, x2 leaves remainder 1 when divided by 4, and x3 leaves remainder 2 when divided by 4. Find the number of non-negative integral solutions of x1 + x2 + x3 = 35.

Solution

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