In an old town, there lived a pot maker who wanted to sell the last stock of his craft to a merchant. When the merchant asked him how many pots he had, he replied that he was unable to count after 100. But he gave him certain clues to figure out for himself:
If the number of pots are divided by two, there will be one left.
If they are divided by three, there will be one left.
If they are divided by four, there will be one left.
If they are divided by five, there will be one left.
If they are divided by six, there will be one left.
If they are divided by seven, there will be one left.
If they are divided by eight, there will be one left.
If they are divided by nine, there will be one left.
If they are divided by ten, there will be one left.
But if they are divided by eleven, there will be no pot left.
Can you find the number of pots, the pot maker possess?
Solution:
You may solve this question with different methods but we are only giving away one.
What you have to do in that method is look for a number x to which all the numbers from 2 to 10 divide evenly. Multiplying the numbers together (2*3*4*5*6*7*8*9*10) will give you a big number but we have to find a smaller possibility. What we can do is find the prime factors, a subset of which we can use to form a number from 2 to 10.
2*2*2*3*3*5*7 will certainly give us what we need. The product is 2520 and it is the lowest possible number which can be evenly divided by 2-10.
Now if we add 1 to the number, the result will be 2521 which will satisfy all the conditions from 2 to 10. But will not hold true in case of 11.
If you calculate, you will see that you can multiply 2520 by any integer and add 1 to it to satisfy all the conditions except the divisibility by 11. What we need is an integer Y that can be multiplied by 2520 and if we add 1 to the product, it becomes divisible by 11.
2520/11 leaves a remainder 1. Now let us keep increasing the number of 2520. Suppose if we divide two 2520 by 11, the remainder then will be 1 + 1 = 2. If we keep going on like that. Ten 2520 when divided by 11 will give 10 as remainder and if we add 1 to them, it will be completely divisible. Also it will satisfy all the conditions from 2-10.
Thus our answer is 25201.