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?

Guest Feb 10, 2022

#2**+1 **

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.

Melody Feb 10, 2022