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N is a four-digit positive integer. Dividing N by 9, the remainder is 5. Dividing n by 7, the remainder is 2. Dividing n by 5, the remainder is 1. What is the smallest possible value of N?

 Oct 19, 2021
 #1
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All the pronumerals I introduce represent integers

 

N=9t+5    (1)

N=7m+2   (2)

N=5g+1     (3)

 

Looking at 2 and 3

7m+2=5g+1

5g=7m+1

5g=5m+(2m+1)

g=m+(2m+1)/5

 

Let a=(2m+1)/5

5a-1=2m

2m=4a+(a-1)

m=2a+(a-1)/2

 

let b=(a-1)/2

2b+1=a

so   m=2(2b+1)+b = 4b+2+b = 5b+2

and so     g=(5b+2)+a = 5b+2+2b+1 = 7b+3

Equation 3 becomes     N=5(7b+3)+1 = 35b+16

 

So now we have only 2  Diophantine equations to solve simultaneously

N=35b+16   and

N=9t+5

 

You can try solving those 2.

They are actually a little easier than the two I just did.  :)

 Oct 19, 2021

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