Math Competition Problems - Problem of the Week - 12

Updated: Feb 14, 2019

An old woman goes to market and a horse steps on her basket and crashes the eggs. The rider offers to pay for the damages and asks her how many eggs she had brought. She does not remember the exact number, but when she had taken them out two at a time, there was one egg left. The same happened when she picked them out three, four, five, and six at a time, but when she took them seven at a time they came out even. What is the smallest number of eggs she could have had?

- posed by Brahmagupta (born 598 AD)

Problem of the Week - 12

To read more about Brahmagupta click here


This week I slightly digress from giving a problem from the math competitions. I just wanted to show how old riddles in mathematics are. Some of them even date back to BC era. I love that most of the problems that have been around for generations are in the form of riddles. It piques the interest of the readers and makes them want to solve it. These are good enrichment math problems for kids in Upper Elementary and Middle School.


Solution:

So lets recap the problem.

When she takes out 2, 3, 4, 5 and 6 eggs at a time there was 1 egg left over.

When she takes out 7 eggs they came out even.

We can also see that the number of eggs minus 1 is divisible by 2, 3, 4, 5 and 6.

Lets find out the least number of eggs that's divisible by 2, 3, 4, 5 and 6

So that's basically the LCM of 2, 3, 4, 5 and 6

LCM of 2, 3, 4, 5 and 6

LCM of 2, 3, 4, 5 and 6 is 60.

The other multiples of 2, 3, 4, 5 and 6 are 120, 180, 240, 300, 360 and so on.

The if she takes out 2, 3, 4, 5 and 6 eggs at a time and has one left over there has to be one more than the multiples of 60.

So there could have been 61, 121, 181, 241, 301, 361 etc.

But we also know that when she takes out 7 eggs they came out even. So the number of eggs should be divisible by 7.

Out of all these numbers only 301 is divisible by 7. So the least amount of eggs she could have had is 301.

This problem can also be solved through Chinese Remainder Theorem, but this way is much easier.


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Now try the next week's problem

Problem of the Week 13


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