Question

Consider the following folk puzzle:
A trader was moving along a road selling eggs. An idler who didn't have much 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?

Answer 1

Hint: Start with assuming a variable for the total number of eggs in the basket. Now use the responses given by the trader to form equations. The statement ‘if counted in pairs one will remain’ will give equation $m = 2n + 1$ . Similarly form six equations. Now use the fact that the unit place of multiples of five is zero or five, to figure out the unit place of $'m'$ . Now write the multiples of seven that are odd, not multiples of five, with the unit place of nine and less than $150$ . Now eliminate the excess possibilities using the relations.

Complete step-by-step answer:
In this problem, we need to find the number of eggs present in the basket according to the statements given by the person. Since all the responses given are in terms of the total eggs, let’s assume the number of eggs to be $'m'$ .
The response ‘If counted in pairs one will remain’ means if we divide all the eggs into groups of two, there will be one left at the end. Also, we can say that if we divide the total number of eggs by $2$ then we will get $1$ remaining unpaired egg.
$ \Rightarrow $ The number is odd and can be represented as: $m = 2n + 1$ (i) Again, the response ‘If counted in threes, two will remain’ means if the total number of eggs is divided into a group of $3$ then $2$ eggs will remain ungrouped.
$ \Rightarrow $ Number of eggs can also be represented as: $m = 3p + 2$ (ii) Similarly, for each of the responses we have:
‘If counted in fours, three will remain’ means: $m = 4q + 3$ (iii)
‘If counted in fives, four will remain’ means: $m = 5r + 4$ (iv)
‘If counted in sixes, five will remain’ means: $m = 6s + 5$ (v)
‘If counted in sevens, nothing will remain’ means that the number of eggs is exactly divisible by the number seven, i.e. $m = 7t$ (vi)
And also the total number of eggs are less than $150$, i.e. $m < 150$
Now, since we know that all the multiples of five always have either $0{\text{ or }}5$ at the units place. So using the relation (iv), we can say that the units place of $'m'$ will be either $0 + 4 = 4{\text{ or }}5 + 4 = 9$ .
Also having four at its units place will make the number $'m'$ an even number. But from relation (i), we know that $'m'$ is an odd number.
Therefore, $9$ will be at the unit place of the number $'m'$ .
As we know that $'m'$ is multiple of seven and hence lie between $7 \times 1 = 7{\text{ and 7}} \times {\text{21 = 147}}$ .
So, from relation (vi) $ \Rightarrow t{\text{ belongs to interval }}\left[ {1,21} \right]$
But from relation (i) and (iv), we know that $t \ne 2,4,6,8......{\text{ or }}t \ne 5,10,15,20,25......$
Therefore, we are left with the values $ \Rightarrow t = 1,3,7,9,11,13,17,19,21$
But also from relation (ii) we know $'m'$ is not a multiple of $3$ . So after eliminating the multiples of $3$ from the set of values of $'t'$ , we will see:
$ \Rightarrow t = 1,7,11,13,17,19$
For the above possible values of $'t'$ , the possible values of $'m'$ are:
$ \Rightarrow {\text{ For }}t = 1,7,11,13,17,19{\text{ ; }}m = 7,49,77,91,119,133$
But since $'m'$ can only have nine at its units place, i.e. possible values of $'m'$ are $m = 49{\text{ or }}m = 119$
For $m = 49$ , we can write it as: $m = 49 = 48 + 1 = 6 \times 8 + 1$ which is not of the form $m = 6s + 5$
Therefore, from relation (v) we can conclude that $m \ne 49$
Thus, we have the final value as: $m = 119$

So, there were a total of $119$ eggs in the basket.

Note: We used the concept of unit place for finding the relation $m = 5r + 4$ since we know the multiples of $5$ always have unit places as $0{\text{ or }}5$ . This was a crucial part of determining the unit place of the required number. We can also verify our answer by using the relations (i), (ii), (iii), (iv) and (v)
$ \Rightarrow 119 = 2 \times 59 + 1 = 3 \times 39 + 2 = 4 \times 29 + 3 = 5 \times 23 + 4 = 6 \times 19 + 5 = 7 \times 17$