Question

yup. another question. help please?

Answer 1

Find the smallest positive integer that solves the systems of congruences
\[N \equiv 1\pmod{7},\]
\[N \equiv 7\pmod{13}\]
\[N \equiv 13\pmod{20}.\]

Answer 2

Oh man. I used to love this stuff but it's been sooooo long since I had it.

Answer 3

lol :)

Answer 4

calculate the first 2
N \(\equiv\)7 (mod 13)
N \(\equiv\) 1(mod 7)
therefore N = 7(49) +1(13) [mod91] = 356 [mod 91] = 83[mod 91]

Answer 5

now work with the last one
N = 83 [mod 91]
N = 17 [mod 20]
so N = 83(20*20) +17(20) [mod91*20]= 33540[mod 1820] = 32760[mod1820]=780[mos1820]
therefore, the smallest number is

Answer 6

Oh, woman!! I don't know how to do the last step,

Answer 7

I got 1170 but it's wrong.

Answer 8

how do you get it? you should check before submit, right?

Answer 9

I got 2633, but my solution was idiotic...

Answer 10

I got another glass of wine.

Answer 11

@Loser66 I didn't do it your way... My solution was rather adhoc.

Could you explain this step please?
N = 7(49) +1(13) [mod91] = 356 [mod 91] = 83[mod 91]
My brain is tired... ;)

Answer 12

erin, it says its wrong... augh! I'm going to give up and copy the solution

Answer 13

Try 813... I think I missed one.

Answer 14

I give up!!

Answer 15

ok. this is interesting but here it is

The first two congruences tell us that \[N = 7a + 1 = 13b + 7 \] for some integers a and b, which simplifies as \[7a = 13b + 6 \].

One solution is \[a = 12\] and \[b=6\], for which \[N = 7a + 1 = 13b + 7 = 85 \]. Therefore, a solution to \[N \equiv 1 \pmod{7} \] and \[N \equiv 7 \pmod{13}\] is \[N \equiv 85 \pmod{91}\].

Then \[N = 91c + 85 = 20d + 13 \] for some integers c and d, which simplifies as \[91c + 72 = 20d. \]

One solution is\[c=8\] and \[d=40], for which \[N = 91c + 85 = 20d + 13 = 813 \]. Therefore, a solution to the system is \[N \equiv 813 \pmod{1820} \].

Since 7, 13, and 20 are relatively prime, all of the integer solutions to this system differ by some multiple of 1820, so the smallest positive solution is 813.

Whew!

Answer 16

aww...erin. I gave up before you said to try 813 :(

Answer 17

Therefore, a solution to the system is
N≡813(mod1820)
.

Answer 18

Here's how I did it: We know N is congruent to 13 mod 20, so
\[N=20m+13\] for some integer m. Then
\[N=(13+7)m+13=7m+13(m+1)\equiv 7m \mod 13,\] so N is congruent to 7 mod 13 iff
\[7m \equiv 7 \mod 13.\] Now 7 and 13 are coprime, so we can divide 7, i.e.
\[m\equiv 1 \mod 13.\] And at this point my solution became stupid and I decided to just try values of m. So there's 1,14,27,40... and lo and behold, m=40 works!

Answer 19

this problem was... crazy. good thing i have another one like it! lol

Answer 20

thank you so much everyone!!

Answer 21

It was more interesting than the previous problem! ;)