Once upon a time, and old lady went to sell her vast quantity of eggs at the local market.

When asked how many she had, she replied:

Son, I can't count past 100 but I know that.

If you divide the number of eggs by 2 there will be one egg left.
If you divide the number of eggs by 3 there will be one egg left.
If you divide the number of eggs by 4 there will be one egg left.
If you divide the number of eggs by 5 there will be one egg left.
If you divide the number of eggs by 6 there will be one egg left.
If you divide the number of eggs by 7 there will be one egg left.
If you divide the number of eggs by 8 there will be one egg left.
If you divide the number of eggs by 9 there will be one egg left.
If you divide the number of eggs by 10 there will be one egg left.

Finally. If you divide the Number of eggs by 11 there will be NO EGGS left!

How many eggs did the old lady have?

very cheap

vry cheap,

oya tell us

Show pikin share the answer with us if you know na.

Haba Gil.
U wan blow we head? tel we ans now.

Make you exercise your intelligence joh!

nothing go remain
shebi i go chop yam and egg

please post the answer before someones goes insane here,abeg.

very very very simple

how simple

please talk jare

@poster

is there any answer to this at all i'm loosing my temper

dy/dx = k + 100 - (x * 5) where k = poster has mental problem

lightest:

dy/dx = k + 100 - (x * 5) where k = poster has mental problem

rofl. k is for kolomental

Me I no wan chop cane.

The old lady had 11 eggs

no thats no correct Gilgee say the answr abi u sef no know

11 eggs = very correct!

which of the women are u even refering to?

After several days of silence, Gilgee has finally decided to post the answer to the puzzle.

All of una no sabi maths.

5,201 eggs. This puzzle has a few different methods for finding the solution, one of which is:

Find a number X into which all of the numbers from 2 to 10 divide evenly. You can do this by simply using 2*3*4*5*6*7*8*9*10, but you can find a smaller number by finding the prime factors, a subset of which can be used to form any number from 2 to 10. 2*2*2*3*3*5*7 will do. This comes out to be 2520, and is the lowest number into which all the numbers 2-10 divide evenly.

We can add 1 to this number to satisfy the first 9 constraints of the puzzle (the remainder of 2521/2, 2521/3 , 2521/10 is one), but this does not satisfy the last constraint, divisibility by 11.

Fortunately, we can multiply X (=2520) by any integer and add 1 and we will still satisfy constraints 1-9. So what Y do we multiply X by so that (X*Y) + 1 is divisible by 11. 2520/11 has a remainder of 1. Thus two 2520s divided by eleven would have a remainder of 1+1 = 2, and so forth, so ten 2520s divided by 11 would have a remainder of 10. This number plus one would divide eleven evenly, as well as also satisfy the first 9 constraints - thus 25201 is the answer.

^^so this na beans?

lysaa, where did u dig this post from? must be from the bin, lol

Well, u and i know who bin is, don't we? Lol.

5201 and 25201 which is the correct answer? Come non go dey Form rubbish oooo, u better answer fast before i released my acidic missile now.

Issue! Lol

non mind ham, e non no wetin e wan right again, just there posting rubbish u cnt even justify,

Shoooooz, d wahala plenty. Una dey check date sef?

Lysa, hahahah now i no, so na dead person i dey talk to sinceeeeeeee, since 2008!!!! naim be say him go don kpia now.

Abeg o, i nor talk say e don die o. My hand na aka nchawa lol

lysaa:

Well, u and i know who bin is, don't we? Lol.

bin is not a person but a place where worn-out stuffs are kept

Thank u bibs, u should check around more often cos there're so many bins here.