Farmer jill has broken all her eggs and doesn’t remember how many she had but she remembers how she tried to pack the eggs she tried to put the eggs in groups of 2,3,4,5,6 and they all had one egg left she can only put he eggs in groups of 7 how many eggs did she have

This kind of problem is pretty tricky if you've never seen anything

like it before.  

What you're looking for is a number that leaves a remainder of 1 when

divided by 2, 3, 4, 5, or 6, but that is evenly divisible by 7. So as

you already noticed, you only need to look at multiples of 7. Let's

try 7. We get off to a good start. 7 leaves a remainder of 1 when

divided by 2, 3, or 6. Unfortunately it doesn't work for 4. So let's

see what we would have to do to get to a number which leaves a

remainder of 1 when you divide it by 4, without losing what we've

already got.

Well, if you take a number that leaves a remainder of 1 when you

divide by 2 and add 2 to that number, you get another number that

leaves a remainder of 1. So if you add any multiple of 2, you still

get a number that leaves a remainder of 1 when you divide by 2.

Simliarly, if you take a number that leaves a remainder of 1 when you

divide it by 3 and add any multiple of 3 to it, you get another number

that leaves a remainder of 1 when you divide it by 3.

Anything that leaves a remainder of 1 when divided by 2 or 3

automatically leaves a remainder of 1 when you divide by 6, so we get

that for free.

So if you take 7 and add anything which is a multiple of 2 and 3 and 7

to it, you get another multiple of 7 which leaves a remainder of 1

when you divide it by 2 or 3.  So we don't lose anything if we add a

multiple of 42 (that's 2 x 3 x 7) to 7.  So lets' see if we can

satisfy another of our conditions.

7 + 42 = 49.  Hey, that's good!  We've got a number that leaves a

remainder of 1 when you divide by 4.  It still doesn't work for 5,

though.  Can we use the same trick again?

If we add a multiple of 4 we still get a number that leaves a

remainder of 1 when you divide by 4.  So if we add a multiple of

3 x 4 x 7, we'll get another number that satsifies all the conditions

so far.  There's only one to go!

Okay,  49 + 84 = 133.  No good.

     133 + 84 = 217.  No good.

     217 + 84 = 301.  Bingo!!

So 301 is a possible solution.  Are there any others?  Well, yes,

because you can add a multiple of 3 x 4 x 5 x 7 to get another one.  

But usually what you're looking for is the smallest solution, so 301

is probably the answer you want.

thanx

hey

answer is d-29

step-by-step explanation:

since you didn't include the picture, i can give you the description so you know how to pick the correct answer. the third quartile of a data is always shown in a box plot. in the middle of your graph should be a box. the third quartile is the value that goes with the far right edge of the box.