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 4. What is the smallest possible value of N?
n mod 9==5,
n mod 7==2,
n mod 5==4, solve for n
LCM[9, 7, 5] ==315
n ==315m + 149, where m==0, 1, 2, 3........etc.
When m ==3, then:
n ==[315 * 3] + 149 ==1,094 - the smallest 4-digit integer.
+118608 Thanks guest answerer :)
I did this by hand and got the same answer.
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 4. What is the smallest possible value of N?
N=9A+5 N=7B+2 N=5C-1
Solve diophantine equations.
9A+5=7B+2
A whole stack of work got me down to A=-12+7T B=-15+9T
Which both gave me
N=-103+63T
So now I have N=-103+63T and N=5C-1
Solve diophantine equations.
-103+63T = 5C-1
A whole stack more work got me to
T=204+5X and C=2550+63X
Which both gave me the general solution N=12749+315X
I'm only interested in 4 digit solutions so
\(1000 \le N <10000\\ 1000\le12749+315X<100000\\ -11749\le315X<-2449\\ -37.3
When X=-37 N=1094 When X=-8 N=9914
-8--37+1=30
There are 30 four digit solutions, the smallest is 1094 and the largest is 9914
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looking at what I have done, I should have simplified the general term first.
N=12749+315X
12749/315=approx 40.5
12749-40*315=149
Nicer general term
N=149+315Y
It is easeier to check that this is correct too.
If anyone wants a more detailed solution then just ask.
