+330 You have a total supply of $1000$ pieces of candy, and an empty vat. You also have a machine that can add exactly $52$ pieces of candy per scoop to the vat, and another machine that can remove exactly $39$ pieces of candy with a different scoop from the vat. When these two machines are done, there is only one piece of candy left in the vat. What is the smallest possible number of times the the first machine added candy to the vat?
+1188
My answer is that it cannot be done.
First, I intuited this because both 52 and 39 are multiples of 13.
Any subtaction of (52 • n1) – (39 • n2) will leave a multiple of 13.
Second, I graphed 52x – 39y = 1 on Desmos. I expanded the gridlines so that the
distance between each large division was 1. This was just to make it easier to read.
I checked the path of the curve to see if it passes through any intersection of x and y
perfectly, up to x = 20, because any x > 19 exceeds the 1,000 candies in the supply.
The curve does not pass perfectly through any intersection. This means that there
is no solution that consists of two integers.
.
