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There is a basket of eggs. The remainder is

  • $1$ when we put the eggs in groups of $2$.
  • $2$ when we put the eggs in groups of $3$.
  • $3$ when we put the eggs in groups of $4$.
  • $4$ and $5$, respectively, when we put the eggs in groups of $5$ and $6$.

How many eggs are there in the basket?

Community
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Peaceyou
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    I see you've made at least 5 questions on this MSE, but you accepted only one of the answers given to you. You should accept your favorite answer. – Git Gud Jan 14 '13 at 16:13
  • @GitGud perhaps they were not suitable answers? – user64742 Jul 14 '16 at 01:19
  • @TheGreatDuck If that's the case, my comment was unwarranted. – Git Gud Jul 14 '16 at 22:30
  • @GitGud also, I have some pretty infamous questions with ten plus answers. No particular answer stood out, so I chose to not accept specifically because I wanted it to be clear that all the answers were equally sufficient, and equally engaging Tldr - not all questions can necessarily choose one answer to accept due to their nice diversity. – user64742 Jul 14 '16 at 23:30
  • @TheGreatDuck I didn't want to fully explain the situation, but since you took the time to dissect this, let me do it too. Firstly, I take back what I said about my comment being unwarranted because I said the OP should accept their favorite answer - this is pretty obvious that they should do on the assumption that there is favorite answer, otherwise my comment talks about something which doesn't exist, it's meaningless. (Continues). – Git Gud Jul 15 '16 at 17:59
  • @TheGreatDuck Secondly, naturally one doesn't always do things right the first time around, I've improved my standard comment for this sort of situation, something along the lines of this which is more informative than imperative, then the OP can decide for themselves. TLDR: Old comment, I've since changed my ways of taking this sort of issue on. – Git Gud Jul 15 '16 at 18:01
  • @GitGud that is a good improvement. I just didn't want the op to inadvertently feel pressured into accepting, as there are many instances in which people accept an answer out of obligation and then never receive a proper answer because they get ignored due to acceptance. Anyway, I did not mean to bother you. I received an answer here during review and happened to notice your post. I did not know it was an old post. Oops. Sorry about that. :p – user64742 Jul 15 '16 at 20:59

3 Answers3

4

Hint. If you have $n$ eggs, then $n+1$ is divisible by $2$, $3$, $4$, $5$, and $6$.

Thomas Andrews
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1

59 eggs.

LCM of 2,3,4,5 and 6 is 60 which is divisible by 1,2,3,4,5 and 6. so if 1

is subtracted the required result is obtained which is 59. so at least

59 eggs is the answer.

banke bihari singh
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  • Good answer but please add more detail as to why subtracting 1 provides the appropriate number. Maybe point out that it is 59 + 60K as well, where k is any integer. - from first post review – user64742 Jul 14 '16 at 01:21
  • @TheGreatDuck $k$ cannot be any integer, as $60 k + 59$ must be nonnegative (it's the number of eggs). – Rodrigo de Azevedo Jul 14 '16 at 02:33
  • @RodrigodeAzevedo That kinda goes without saying... We are obviously talking about positive numbers here. – user64742 Jul 14 '16 at 02:38
0

Since this question is over three years old, why not just give the solution? In Haskell:

λ take 10 $ filter (\n->(rem n 2==1) && (rem n 3==2) && (rem n 4==3) && (rem n 5==4) && (rem n 6==5)) [1..]
[59,119,179,239,299,359,419,479,539,599]

Hence, the number of eggs is in the set $\{ 60 m - 1 \mid m \in \mathbb N^+ \}$. What is so special about $60$?

Rodrigo de Azevedo
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