Farmer's Eggs POW

Question

A farmer is taking her eggs to market in her cart. Along the way, she hits a pothole, which jars her cart and spills the eggs. Though the farmer is unhurt, every single egg is broken. She goes to her insurance agent, who asks her how many eggs she had. She doesn’t remember, but she does remember some things from the ways she tried packing the eggs.

Here is what she remembers:

1. When she put the eggs in groups of two, there was one left over.
2. When she put the eggs in groups of three, there was also one egg left over.
3. When she put the eggs in groups of four, groups of five, and groups of six, there was also just one egg left over.
4. When she put the eggs in groups of seven, she ended up with complete groups of seven, with no eggs left over.

What can you figure out from this information about how many eggs the farmer had? Is there more than one possibility?

Answer 1

1 there's an odd number of eggs

2 not a multiple of 3

3 not a multiple of 4, 5, or 6

4 the number of eggs is divisible by 7. it's a multiple of 7, an odd multiple of 7, and odd prime multiple of 7

E could = 7, 21, 35, 49, 63, 77, 91, 15(7), 17(7), 19(7) 21(7), 23(7) (25(7) 27(7) 29(7) 31(7) 33(7), 35(7) 37(7) 39(7), 41(7) 43(7)

that rules out all up to 43(7) = 301

301/3 = 100 with remainder of 1 egg

301/4 - 75 with 1 egg remainder

301/5 = 60 with 1 egg remainder

301/6 = 50 with 1 egg remainder

301/7 = 43 with no remainder

answer is 301 eggs for the smallest answer

Yes there's more than one answer

there are probably an infinite number of possibilities

163(7)=1141 for the next smallest answer

1141/3= 380 with remainder = 1

1141/4 = 285 with remainder 1

1141/5 = 248 with remainder 1

1141/6 = 190 with remainder 1

other numbers will also be a prime number times 7