# How Many Eggs ?

A farmer is taking her eggs to the market in a cart, but she hits a pothole, which knocks over all the containers of eggs. Though she is unhurt, every egg is broken.

So she goes to her insurance agent, who asks her how many eggs she had. She says she doesn’t know, but she remembers something from various ways she tried packing the eggs.

When she put the eggs in groups of two, three, four, five, and six there was one egg left over, but when she put them in groups of seven they ended up in complete groups with no eggs left over.

**What can the farmer and agent figure out from this information about the number of eggs she had ? **

**Is there more than one answer ?**

PS:- Though I know the answer but still I want the answer in the simplest understandable manner possible. Thanks for all answers in advance.

The answer is __301__

__EXPLANATION-__

__Condition 1 – __**When she put the eggs in groups of two, three, four, five, and six there was one egg left over**Since there was one egg left when packed in 2,3,4,5,6, we need to find the LCM (lowest common multiple) of 2,3,4,5,6 and add one to get the

*least*possible answer, THat is 60+1=61.

Moreover, we can easily figure that

**60x+1**will always satisfy the first 5 conditions, where

**x**is a positive integer.

__Condition 2- __**when she put them in groups of seven they ended up in complete groups with no eggs left over**

To satisfy this condition, we have to find the *least* value of **X** such that **60x+1** is divisible by **7**.

That is,

(60x+1)%7 = 0 Where % is mod

When,

x=1 —–> 61 % 7 = 5

x=2 —–>121 % 7 = 2

x=3 —–>181 % 7 = 6

x=4 —–>241 % 7 = 3

__x=5 —–>301 % 7 = 0____ —- SATISFIED__

**Hence, 301 eggs is the Minimum Number of eggs.**

** NOTE-**While there are many possible answers satisfies this equation (60x+1)%7 = 0

__Answers__**–**301, 721, 1141, 1561, 1981 …………

Nice Explanation @ SherlockHolmes. But think over no. 21.

2 * 3 + 3 * 1 + 5 * 1 + 6 * 1 = 20, therefore one egg will be left over.

| |

| no. of such group

group of 2

301, 721, etc.

My explanation:

the number is a multiple of LCM of (2,3,4,5,6) + 1 = 60*x + 1

And this number should be divisible by 7.

It’s easy to see the above number will always end up with the number 1. There’s only one multiple of 7 which ends in 1, i.e., 3 (7*3 = 21).

So we only need to check all prime numbers that end in 3. For example, 7 * 13, 7 * 23, 7 * 43, etc. because only these multiples of 7 will result in a number that ends in 1 and also the multiple has to be prime or else it’ll be divisible by either 2, 3 or 5 negating the above condition of leaving one egg.

so we can check all such numbers multiplied by 7 and subtract 1 and check if the number is also a multiple of 60::

7 * 13 = 91 – 1 = 90 (not a multiple of 60)

7 * 23 = 161 – 1 = 160 (not a multiple of 60)

(33 is not a prime number so we don’t need to check that)

**7 * 43** = **301** – 1 = 300 (multiple of 6) (Answer)

and we can find many more numbers like this

7 * 53

7 * 73

7 * 83

**7 * 103 = 721**

### Your Answer

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