If 112=q×6+r Find the possible value of r
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If q are positive integers, then
Assuming that r has to be positive, q x 6 will have to equal less than 112. Because of PEMDAS/BIDMAS/BODMAS we do the multiplication first and then add r rather than adding q to 6 x r. Assuming r has to be positive the possible combinatons are:
112 = 1 x 6 + 106(q = 1, r = 106)
112 = 2 x 6 + 100(q = 2, r = 100)
On and on until: 112 = 18 x 6 + 4(q = 18, r = 4)
That makes 18 possible values for r(19 if you include the possibility that q = 0, making r = 112) with each one being 4 higher than the multipe of 6 below it. The 19 possible values of r are: 112, 106, 100, 94, 88, 82, 76, 70, 64, 58, 52, 46, 40, 34, 28, 22, 16, 10 & 4.
If r didn’t have to be positive there would be an infinite number of possible values of r as you just need to add the neccesary negative number to equal 112 because, when you add a negative number, you are actually just subtracting.
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If q are positive integers, then
Assuming that r has to be positive, q x 6 will have to equal less than 112. Because of PEMDAS/BIDMAS/BODMAS we do the multiplication first and then add r rather than adding q to 6 x r. Assuming r has to be positive the possible combinatons are:
112 = 1 x 6 + 106(q = 1, r = 106)
112 = 2 x 6 + 100(q = 2, r = 100)
On and on until: 112 = 18 x 6 + 4(q = 18, r = 4)
That makes 18 possible values for r(19 if you include the possibility that q = 0, making r = 112) with each one being 4 higher than the multipe of 6 below it. The 19 possible values of r are: 112, 106, 100, 94, 88, 82, 76, 70, 64, 58, 52, 46, 40, 34, 28, 22, 16, 10 & 4.
If r didn’t have to be positive there would be an infinite number of possible values of r as you just need to add the neccesary negative number to equal 112 because, when you add a negative number, you are actually just subtracting.