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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?

 Feb 10, 2022
 #1
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Sum( 423 , 843 , 1263 ==2,529)

 Feb 10, 2022
 #2
avatar+117175 
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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.

 Feb 10, 2022

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