A woman was carrying a basket of eggs to market when a
passer-by bumped her. She dropped the basket and all
the eggs broke. The passer-by, wishing to pay for
her loss, asked, 'How many eggs were in your basket?
I don't remember exactly, the woman replied, but I do
recall that whether I divided the eggs by 2,3,4,5 or 6
there was always one egg left over. When I took the
eggs out in groups of seven, I emptied the basket.
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19 comments:
721 is the answer
How u got it...
Multiply 2*3*4*5*6 and add 1 with that
There must be infinite answers - perhaps the problem should be formulated as 'atleast' how many. You dont need to multiply all - 3*4 cover 2*6 as well. I come up with (minimum amount) 301. Then the series, summing (60 * 7) to each, i.e. 721, 1141,...
91 eggs
you should multiply only the prime numbers
2*3*5 + 1 =31 but does not divide by 7
61 does not divide by 7
91 / 7 = 13
61 is the least possible no. of eggs with her
301 is least poss.
91 div by 4 leaves a remainder of 3.
61 is not div by 7.
You need to READ the prob then do ALL the math.
ya anonymous i ver intelligent great u have really give me a gud idea
301 is the only least possible number
301 is the least possible number satisfying the conditions ( multiple of (2,3,4,5,6) +1 and divisible by 7)
yes 301 is the least value
91 is the least possible answer..
To satisfy the condition..the number should be 3,13,23,33,43...rd multiples of 7...
considering first condition i.e 21...the first set of conditions r not satisfied..
13th multiple satisfies all the conditions..so 91 is the least possible condition
"God" is wrong again. 91 div by 4 leaves 3 not 1. READ the problem as stated, then when you think you have a sol'n, CHECK it against ALL the criteria of the original problem.
We can ignore division by 2 and 3 as they are subsets of multiples of 4 and 6 respectively. The (answer-1) must be divisible by 5, so ending in 5 or 0, and also by 4 and 6, therefore even.
So the answer must end in a 1. To achieve division by 7, the multiplier must end in a 3.
7x3=21, 20 is not a common multiple of 4, 5 and 6
7x13=91, ditto 90
7x23=161, ditto 160
7x33=231, ditto 230
7x43=301, and 4, 5 and 6 all divide 300. Here is our solution
I tried to solve this algebraically, but there aren't enough simultaneous equations, even assuming infinite solutions. I'd be interested if anyone has any insights on a written solution.
Ben
Jeya says 301 is the right answer.
Good for people to know.
I don't believe for a second that the woman would know all those mathematical facts about the number of eggs but not know how many there were.
Also if she is smart enough to know all of those details about what numbers divide evenly or not, then she can probably answer the riddle herself pretty quickly. I'm guessing that the passer-by is probably not able to get the answer as quickly as she is so why even tell him?
301 is least figure
first statement
Conditoin 1) Divisible by 2 at reminder 1 states that its ODD number
Conditoin 2) Divisible by 3 at reminder states that sum of all digits is 4. (all numbers devisible by 3 has sum of its number=3)
Conditoin 4) divisible by 5 at reminder 1 states that its last digit is 1 or 6 but condition 1 states its ODD so its only 1 at unit position.
this made it simple to solve this puzzle
jumboninja@gmail.com
301 is the least ans i got.
LCM(2,3,4,5,6)=60.
so the ans = 60*x+1, which should be divisible by 7.
I found 301 satisfying this.
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