can you figure out this brain teaser??

Question

a farmer is taking her eggs to the market, and she drops her basket and breaks all of her eggs. she needs to know how many eggs she had to begin with. when she put them in groups of 2,3,4,5 or 6 there was always one left over. but when she put them in groups on seven, they fit evenly.

how many eggs did she have?

(p.s the answer is not 49)

Answer 1

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.

Simlarly, 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 satisfies all the conditions so far. One more to go!

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.

Answer 2

Just check multiples of 7.

Note that the number multiplied by 7 can't be divisible by 2, 3, 4, 5 or 6. Since it can't be divisible by 2 or 5 it must end in 1, 3, 7 or 9; these will give products ending in 7, 1, 9 and 3 respectively. To have a remainder of 1 when dividing by 5 the product must end in 1 or 6, so the number multiplied by 7 must end in 3.

13*7 = 91 - doesn't work for 4
23*7 = 161 - doesn't work for 3 or 6
33*7 is divisible by 3
43*7 = 301 - works for 2, 3, 4, 5 and 6. So this is the smallest number that will work.

We can also work this using modular arithmetic, as others have done; then you can show that the possible answers are of the form 301 + 420k for any nonnegative integer k.

Answer 3

301

Date: 12/12/96 at 19:41:38
From: Doctor Wilkinson
Subject: Re: Broken eggs

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.

-Doctor Wilkinson, The Math Forum
Check out our web site! http://mathforum.org/dr.math/

Answer 4

hrm...

you're looking for a number that is of the form:
k mod 2 = 1
k mod 3 = 1
k mod 4 = 1
k mod 5 = 1
k mod 6 = 1
k mod 7 = 0

(k mod 2 means the remainder that you get when you divide k by 2...so 5 mod 2 = 1 because 5 = 2*2 + 1)

a number that does fit this is actually 301

How do I get that? well, from properties of abstract algebra, it can be shown to be something of the form k = n*LCM(2, 3, 4, 5, 6) + 1, where n is a natural number (positive integer)

LCM(2, 3, 4, 5, 6) = 60

so start checking values of n...

n = 1, k = 61...k mod 7 is not 0
n = 2, k = 121...same
n = 3, k = 181...same
n = 4, k = 241...same
n = 5, k = 301...k mod 7 is 0, as k divides 301 exactly 43 times...

and 301-1 = 300 is clearly divisible by 2, 3, 4, 5, 6, as it was generated by the LCM of those integers...

so the number of eggs she had originally was 301...

that number seems impossibly enormous...but it does fit the thing...hehe

that was kinda fun =P

Answer 5

7

Answer 6

721 Multiply 2,3,4,5,and6 =720 they all have that common multiple, but then add 1 and none do. Seven does 721/7 = 103

Answer 7

7

That is the only option without going above 49. 7 is not divisible by 2, 3, 4 ,5 or 6 evenly.

Answer 8

the different dollar went nowhere, each and all the money remains accounted for. the guy on the table have been given his $25, he provides the $5 back to the adult adult males, the bellboy takes his tip of two$ and that they've one each and each making $30. table ($25) + adult adult males ($3) + Bellboy($2) = $30 this is all in how this is phrased, giving the phantasm of the lacking dollar.

Answer 9

7 eggs. Nice one!!!

Answer 10

I think 7.