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The number of positive integral solutions to the equation 4x+5y+2z = 111 is? Please provide explanation too.
We'll start by assuming suitable values for y (its coefficient is the largest). You will realize that since 4x + 2z is an even number, y has to be odd so as to produce the integer solutions.
Now, the possible values of y are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19 and 21 (for y = 23 the value, 23*5 = 115, exceeds 111), so we will consider only odd values of y up to 21.
When y = 1, 4x + 2z = 106 or 2x + z = 53. The number of positive integral solutions for this equation would be 26.
When y = 3, 4x + 2z = 96 or 2x + z = 48. The number of positive integral solutions for this equation would be 23.
When y = 5, 4x + 2z = 86 or 2x + z = 43. The number of positive integral solutions for this equation would be 21.
In similar fashion find the number of integral solutions for every value of y.
You should get the following number of results -
26 + 23 + 21 + 18 + 16 + 13 + 11 + 8 + 6 + 3 + 1 = 146, which is the answer.
Notice there is a pattern in the above series. The difference between the odd number terms and even number terms is 5, so after finding a first few solutions you can generalize the sequence.
4*1 + 5*21 + 2*1 = 111;
4*(2) - 5*(2) + 2*(1) = 0
Now see, 4(x+2) + 5(y-2) + 2*(z+1) = 111
Initial +ve integral solutions are (1,21,1).
Others are (3,19,2) , (5,17,3),...(21,1,11), which can be counted as 10 solutions.
So, total 11 positive Integral solutions.
Yes, I made an error while writing down the series. Thanks for pointing it out, I've edited it now.
