anonymous
  • anonymous
Please help!!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
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?
anonymous
  • anonymous
It all boils down to \[N\equiv 3\ (mod 5)\] \[N\equiv 3\ (mod 6)\] \[N\equiv 5\ (mod 7)\] \[N\equiv 3\ (mod 8)\] Solve. I think. But I don't know how to do those! Could someone give the answer and then explain?
anonymous
  • anonymous
This question looks very similar to the ones we were doing last time...

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anonymous
  • anonymous
Yep.
anonymous
  • anonymous
I have a lot on the topic.
anonymous
  • anonymous
I guess your last equation should be \(N≡5 \mod 8 \)?
anonymous
  • anonymous
oops. yeah.
anonymous
  • anonymous
I don't see a clever way of doing it. As far as I can see, you'll just have to solve the first pair, deal with the next pair and finally solve the last two equations. Do you still remember the chinese remainder theorem from last time?
anonymous
  • anonymous
no. I never really understood how to do it.
anonymous
  • anonymous
I do remeber talking about it though
anonymous
  • anonymous
Now this question is practically identical to the other one :) http://openstudy.com/study#/updates/52195471e4b06211a67cafc0 You even have the same number of equations!
anonymous
  • anonymous
ok. I need to leave now. its and...emergency. real life sorry!!!
anonymous
  • anonymous
crt again good luck
anonymous
  • anonymous
got to be odd, right? and a multiple of 3. 6x+3 = n so (2x+1)*3 = n n=3mod5 => 2x+1 =1 mod 5 n=5mod7 => 2x+1 = 4 mod 7 n=5 mod 8 => 2x+1 = 7 mod 8
anonymous
  • anonymous
I don't know...just solve the equations?
anonymous
  • anonymous
31 fits the bill... 31= 1mod 5 31 = 4 mod 7 31 = 7 mod 8 so n should be 3*(c*31), yeah?
anonymous
  • anonymous
wait so 31 works. it works?
anonymous
  • anonymous
wait no. It's 31 times something.
anonymous
  • anonymous
yeah but something is off...
anonymous
  • anonymous
there are four equations?
anonymous
  • anonymous
the number is 1293... but how?
anonymous
  • anonymous
How'd you get that?
anonymous
  • anonymous
It;s right. Wait I've got the solution.
anonymous
  • anonymous
2x+1= 5 mod 6, but why? oh... excel
anonymous
  • anonymous
2x+2 = 6 mod 6 so 2x+1=5 mod 6. because 3(2x+1) + 3 = 0 mod 6
anonymous
  • anonymous
that makes 2x+1 = 431
anonymous
  • anonymous
3*431 = 1293
anonymous
  • anonymous
Let n be the number of apples that Brandon picked. Then \[n \equiv 3 \pmod{5} \]\[n \equiv 3 \pmod{6} \]\[n \equiv 5 \pmod{7} \]\[n \equiv 5 \pmod{8}\]... From the first two congruences, \[n \equiv 3 \pmod{30}. \] From the last two congruences, \[n \equiv 5 \pmod{56}. \] Hence, \[n = 30a + 3 = 56b + 5 \] for some integers a and b, which simplifies as \[30a = 56b + 2, or 15a = 28b + 1 \]. One solution is a = 15 and b = 8. Subtracting 225 from both sides of the equation \[15a = 28b + 1 \], we get \[15a - 225 = 28b - 224 \], which factors as \[15(a - 15) = 28(b - 8) \]. This equation tells us that 15(a - 15) is divisible by 28, but 15 is relatively prime to 28, so a - 15 is divisible by 28. Then a - 15 = 28k for some integer k, so a = 28k + 15, and n = 30a + 3 = 30(28k + 15) + 3 = 840k + 453. The only number of this form between 1000 and 2000 is 1293.
anonymous
  • anonymous
what class are you taking? you have the best questions! i want to take your class.
anonymous
  • anonymous
Just Number Theory.
anonymous
  • anonymous
As in the class is called Number Theory. I sometimes ask questions for my Counting and Probability class too.
anonymous
  • anonymous
nm. I got te answer
anonymous
  • anonymous
what school do you go to?
anonymous
  • anonymous
Klondike Middle School
anonymous
  • anonymous
wow!
anonymous
  • anonymous
heh. It's an online class. :)

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