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Can anyone solve this folk puzzle?

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Saina1712
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Saina1712 | 06-01-2014 17:11
I can solve this puzzle with the help of trial and error method but I want to know if there's another method to solve it.
[Note- Panchayat means a village council.]
•••
A trader was moving along a road selling eggs. An idler who didn't have much work to do, started to get the trader into a wordy duel. This grew into a fight, he pulled the basket with eggs and dashed it on the floor. The eggs broke. The trader requested the Panchayat to ask the idler to pay for the broken eggs. The Panchayat asked the trader how many eggs were broken. He gave the following response:

If counted in pairs, one will remain;
If counted in threes, two will remain;
If counted in fours, three will remain;
If counted in fives, four will remain;
If counted in sixes, five will remain;
If counted in sevens, nothing will remain;
My basket cannot accommodate more than 150 eggs.


So, how many eggs were there?
8 comments
Krunegard
0
Krunegard | 06-01-2014 18:09
119 eggs were in the basket.

I just thought through the highest numbers divisible with 7 and below 150, and then tried my way out. Not much of a method.
Saina1712
0
Saina1712 | 06-01-2014 18:54
Right answer and that's what I did, trial and error method.
scardy_cat
0
scardy_cat | 07-01-2014 00:36
why not 147 eggs?
Captain_Keeta
0
Captain_Keeta | 07-01-2014 01:11
I somehow managed to get 121. lol
Saina1712
0
Saina1712 | 07-01-2014 13:34
Guys, from the given constraints/conditions, we know that the number of eggs in basket is divisible by 7 but not 2,3,4,5 and 6.

And it's not 147 because it's divisible by 3; it's not 121 because when we divide it by 3, the remainder is not 2.
FireWaterBurn6
0
FireWaterBurn6 | 07-01-2014 13:46
This guy shouldn't be selling eggs roadside if he's this good at math.
Saina1712
0
Saina1712 | 07-01-2014 14:50
Don't you know about the unemployment problem? Poor guy couldn't get a better job.
Mathisboring
1
Mathisboring | 30-07-2017 08:24
Let eggs=x
When divided by 2,3,4,5 and 6, remainder is always one less than divisor
Thus, x is one less than LCM(2,3,4,5,6) OR one less than a multiple of the same.

LCM=60
now,
x= (60-1) OR (120-1) OR (180-1)
60-1= 59
NOT divisible by 7

120-1=119
Maybe x= 119

180-1= 179
Cannot be over 150.

We can safely conclude that x= 119

The trader was carrying 119 eggs. Hope the Panchayat appreciates the help. (Though it came 3 years late)
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