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# help math

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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 4. What is the smallest possible value of N?

Nov 11, 2022

#1
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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.

Nov 11, 2022
#2
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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.

Nov 13, 2022