+0

+1
598
4
+109

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
edited by 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
+109
+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
#4
+109
+1

THANK YOU SO MUCH YOU JUST SAVED MY BUTT FROM FAILING MY CLASS GOD BLESS YOU YOU ARE AN AMAZING HUMAN BEING DON'T LET ANYONE CHANGE YOU THANK YOU SO MUCH!!!!!

Jeff123  Sep 17, 2017