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I'm working on a problem: A woman has a basket of eggs and she drops them all. All she knows is that when she puts them in groups of 2, 3, 4, 5, and 6, there is one left over. When she puts them into groups of 7, there are none left over. What is the minimum number of eggs she could have in her basket?

Here's where I've gotten.

Since all 2, 3, 4, 5, and 6 have a remainder of one, the number must be a multiple of their lcm + 1. So, we know that $$ x = 60t + 1. $$ So, I checked integer values of t and then found their remainder when divided by 7. The solution was when $t = 5, x = 301$. What I want to know is, is there a 'better' way to do this? And if so, how? I ran into the Chinese Remainder Theorem, but everything I saw didn't make much sense to me.

Thanks!

Patrick Feltes
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2 Answers2

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The first half is a constant-case optimization of CRT: $\ 2,3,4,5,6\mid x\!-\!1\!\iff\! 60\mid x\!-\!1,\,$ since $\,{\rm lcm}(2,3,4,5,6)=60.\,$ So $\, x = 1\!+\!60t.\,$ Further $\, x\equiv 0\pmod 7\ $ so, by Easy CRT

$\qquad\qquad\qquad\! {\rm mod}\ 7\!:\ 0 \equiv x \equiv 1\!+\!60\color{}t\equiv 1\!-\!3t\iff 3t\equiv1\equiv -6\iff \color{#c00}{t\equiv -2\equiv 5}$.

Therefore $\ \color{#c00}{t = 5\!+\!7n}\,$ thus $\, x = 1\!+\!60\color{#c00}{\,t} = 1\!+\!60(\color{#c00}{5\!+\!7n}) = 301+420n$

Bill Dubuque
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  • Something's funny with your link; it appears to be interpreting user:me as, well, me, as opposed to what I think you intended, which is to say you. possibly user:BillDubuque instead? – Dan Uznanski Apr 27 '15 at 02:30
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You have $$x\equiv 1\pmod{60}\\x\equiv 0\pmod{7}$$

Applying Chinese Remainder Theorem is just finding $u,v$ such that $60u+7v=1$. Then the solution is $$x\equiv 1\cdot (7v) + 0\cdot (60 u)=7v\pmod{7\cdot 60}$$

Thomas Andrews
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