# Chinese Remainder Theorem 3

• Mar 26th 2007, 08:05 PM
jmc1979
Chinese Remainder Theorem 3
Problem:

Ancien Indian Problem: If eggs are removed form a basket 2, 3, 4, 5, and 6 at a time, there remain, respectively, 1, 2, 3, 4, and 5 eggs. But if eggs are removed 7 at a time, no eggs remain. What is the least number of eggs that could have been in the basket?

So far this is what I believe this exercise implies but I am not sure:

x ≡ 1 (mod 2)
x ≡ 1 (mod 3)
x ≡ 1 (mod 4)
x ≡ 1 (mod 5)
x ≡ 1 (mod 6)
x ≡ 0 (mod 7)

Can anyone help? Thank you!

• Mar 27th 2007, 10:35 AM
ThePerfectHacker
Quote:

Originally Posted by jmc1979

x ≡ 1 (mod 2)
x ≡ 1 (mod 3)
x ≡ 1 (mod 4)
x ≡ 1 (mod 5)
x ≡ 1 (mod 6)
x ≡ 0 (mod 7)

x=1(mod 2)
x=1(mod 3)
x=1(mod 4) implies x=1(mod 2) (you can drop this one).
x=1(mod 5)
x=1(mod 6) implies x=1(mod 2 and 3) (you can drop this one).
x=0(mod 7)

Thus, you get,
x=1(mod 3)
x=1(mod 4)
x=1(mod 5)
x=0(mod 7)

Solve that.

(I deleted you other question on Chinese Remainder Theorem, because it seems to me you just want the answer. You want me to help you, you show the work you done so far. Because otherwise that makes me angry. You insult my intellegence! :mad: )
• Mar 27th 2007, 11:27 AM
jmc1979
Thank you!
ThePerfectHacker thank you for you help. I wrote the complete problem, that did not meant that I wanted you to solve the exercise for me. I just wanted to know that I was on the right path. Sorry if this insulted your intelligence since this was not the intention. Thank you again!