An Old Egg Problem dating to...

Number Theory Level 3

An old woman went to the market and a horse stepped on her basket and smashed her eggs. The rider offered to pay for the eggs and asked her how many there were. She did not remember the exact number, but when she had taken them two at a time there was one egg left, and the same happened when she took three, four, five, and six at a time. But when she took them seven at a time, they came out even. What is the smallest number of eggs she could have had?

The answer is 301.

2 solutions

X X
May 27, 2018

$\text{lcm}(2,3,4,5,6)=60$ ,when she took 60 out there willl be one egg left,so consider the sequence $61,121,181,241,\boxed{301}$

Donglin Loo
May 27, 2018

Let the number of eggs be $n$

From the problem,

$n\equiv1(mod2)$

$n\equiv1(mod3)$

$n\equiv1(mod4)$

$n\equiv1(mod5)$

$n\equiv1(mod6)$

and $n\equiv0(mod7)$

$\Rightarrow n-1\equiv0(mod2, 3,4,5,6)$

$L. C. M.$ of $2,3,4,5,6=60$

$n-1\equiv0(mod60)$

Let $n=60p+1$ for positive integer $p$

$60p+1=7k$ for positive integer $k$

When $p=7a+5$ , $7k=420a+301$

$\therefore n=420a+301$

Minimum value of $n=301$

An Old Egg Problem dating to...

The answer is 301.

2 solutions

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