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?
+118609 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. :)
