I'm terrible at modular arithmetic, so I might be asking a lot of questions on it:

1. Brandon spends his afternoon picking apples from an orchard. He notices that if he groups the apples he picked by 5's or 6's, then he ends up with 3 left over. If he groups by 7's or 8's, he has 5 left over. If Brandon knows that he picked between 1000 and 2000 apples, how many did he pick?

2. When \(N\) is divided by 10, the remainder is \(a\) . When \(N\) is divided by 13, the remainder is \(b\) . What is \(N\) modulo 130, in terms of \(a\) and \(b\) ?

(Your answer should be in the form \(ra+sb\) , where and are replaced by nonnegative integers less than 130.)

3. Find all numbers \(r\) for which the system of congruences

\(\begin{align*} x &\equiv r \pmod{6}, \\ x &\equiv 9 \pmod{20}, \\ x &\equiv 4 \pmod{45} \end{align*}\)

has a solution.

Thank you so much!!!

Jeff123 Sep 16, 2017

#1**+1 **

1)

Since the LCM of 5and 6 =30, and the LCM of 7 and 8 =56, then the LCM of 30 and 56 =840.

N mod 30 =3

N mod 56 =5, and N=?

N = 840n +453, where n=0, 1, 2, 3.......etc.

Since the number apples picked by Brandon is between 1,000 and 2,000, it, therefore, follows that:

**N =840 x 1 + 453 =1,293 apples.**

3)

The only value of r that satisfies the congruences is r = 1

solve N mod 6 = 1 N mod 20 = 9 N mod 45 = 4 for N

N =180n + 49, where n=0, 1, 2, 3.......etc. So that:

**N=49, 229, 409......etc.**

Guest Sep 16, 2017

#2**+1 **

Can you please help with Problem 2? And I don't really understand problem 3. But thanks anyways!!!!!

Jeff123 Sep 17, 2017

#3**+1 **

3)

This one is relatively simple. You have to take each modulus x some constant + remainder = each other as follows:

(6 * a) + r =(20 * b) + 9 =(45 * c ) + 4=x. You simply try, by iteration, to find the values of a, b, and c.

If you look at (45*c) + 4 and try c= 1, then it becomes: 45*1+4=49. and if you look at 20b+9, you will see that b= 2, and it becomes 20*2 + 9 =49. Check and see if it balances 6*a +r. If you take 49 and divide by 6, you will get 8, because 6*8 =48. And so 49 mod 6 =1. Therefore, r can only =1.

Since the LCM of 6, 20, 45 =180, therefore x will be:

x =180C + 49, and C =0, 1, 2, 3, 4, 5.......etc. So:

x =180*0 + 49 =49

x =180*1 + 49 =229

x =180*2 +49 =409

x =180*3+49 =589..........and so on. All these values of x satisfy the congruences.

2)

I don't actually get this question because there are so many unknown variables.

You may have to try trial and error to see if you get anything meaningful: For example, try some number such as 775:

775 mod 10 = 5

775 mod 13 = 8

775 mod 130 =120. Now, 5 and 8 are factors of 120, because:

120/5 =24 and 120/8 =15. Therefore:

(24*5) + (15*8) =120 + 120 = 240 !!!!. Does it make sense???. I haven't done many of these for very long time!.

Guest Sep 17, 2017

edited by
Guest
Sep 17, 2017