When the positive integer x is divided by each of 4, 5, 6, and 7, it has a remainder of 3. What is the sum of the three smallest possible values of x?
+118608
When the positive integer x is divided by each of 4, 5, 6, and 7, it has a remainder of 3. What is the sum of the three smallest possible values of x?
We need a number that all of those go into. So we look at the prime factors of each of them
2*2, 5, 2*3, and 7
So
2*2*5*3*7 = 420 will be the smallest number that is divisable by all of them.
We want a remainder of 3 so the smallest will be 423
The smallest 3 will be
420+3 2*420+3 3*420+3
The sum of these is 6*420 +3*3 = 2529
Just as guest already found.
