A famous problem goes as follows (source: NCERT class X Mathematics book):

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.

The problem, of course, is to find the number of eggs in the basket. This is a very good example of the concept of division of two integers, to obtain a quotient and a remainder. Let us solve this problem step by step.

Let the number of eggs be x. Note that x must be a multiple of 7, that is:

\[x = 7a,\;a \in \mathbb{Z}\]

Note that if counted in pairs, one egg remains, that is:

\[x = 2b + 1,\;b \in \mathbb{Z}\]

In other words, x is odd. Let us write out all odd multiples of 7 which are less than 150:

\[\begin{align}&7,\;21,\;35,\;49,\;63,\;77,\\&91,\;105,\;119,\;133,\;147\end{align}\]

Note also that x is not a multiple of 3 or 5. Thus, we cross out all multiples of 3 or 5 on our list:

\[\begin{align}&7,\;\rlap{--} 2\rlap{--} 1,\;\rlap{--} 3\rlap{--} 5,\;49,\;\rlap{--} 6\rlap{--} 3,\;77,\\&91,\;\rlap{--} 1\rlap{--} 0\rlap{--} 5,\;119,\;133,\;\rlap{--} 1\rlap{--} 4\rlap{--} 7\\ &\Rightarrow \;\;\;\left\{ {7,\;49,\;77,\;91,\;119,\;133} \right.\end{align}\]

We now have 6 possible numbers on our list. If counted in fours, three will remain, which means that x must be one less than a multiple of 4. Thus, we cross out all those numbers which are not one less than a multiple of 4:

\[\begin{align}&7,\;\rlap{--} 4\rlap{--} 9,\;77,\;91,\;119,\;\rlap{--} 1\rlap{--} 3\rlap{--} 4\\ &\Rightarrow \;\;\;\left\{ {7,} \right.\;91,\;119\end{align}\]

Also, x should be one less than a multiple of 3, and one less than a multiple of 5, and so:

\[\begin{align}&\rlap{--} 7,\;\rlap{--} 9\rlap{--} 1,\;119\\ &\Rightarrow \;\;\;119\end{align}\]

You can verify that the number 119 satisfies all the conditions of the problem:

\[119 = \left\{ \begin{array}{i}2\left( {59} \right) + 1\\3\left( {39} \right) + 2\\4\left( {29} \right) + 3\\5\left( {23} \right) + 4\\6\left( {19} \right) + 5\\7\left( {17} \right) + 0\end{array} \right.\]

What we have seen here is an example of division of integers. When you divide one integer by another non-zero integer, you are left with a quotient and a remainder. For example, when 119 is divided by 4, the quotient is 29 and the remainder is 3.

Let us formalize this discussion.

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