## How Many Zeros Are There in n factorial

How many trailing zeros are there in 100! (! is read as factorial)? This is one of the most common problems in elementary school and middle school math competitions and for those who have memorized the strategy, this can be solved in less than five seconds. There are (100/5) + (100/25) = 24 trailing zeros in 100!. But why does the trick works?

Small Cases

Example 1: How many zeroes are there in $6!$?

For those who are new to the factorial notation, when we say $6!$, we mean that we multiply $6$ and $5$ and $4$ all the way down to $1$. That is

$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$.

So, where did all the zeros come from? Zero came from 5 multiplied by any even number factor. For example, in $6!$,  if we multiply $6$ and $5$, this will give us 30, a number with one trailing zero. Notice that none of the remaining numbers in the multiplication can add another trailing zero. » Read more

## Introduction to Age Problems in School Mathematics

Age Problems is the second part of the Math Word Problem Solving Series here in Mathematics and Multimedia. In this series, we are going to learn how to solve math word problems involving age.

Age problems are very similar to number word problems. They are easy to solve when you know how to set up the correct equations. Most of the time, this type of problem discusses the age of a certain individual in relation to another in the past, present, or future.

Below, are some of the common phrases used in age problems. In all the phrases, we let x be the age of Hannah now.

• Hannah’s age four years from now (x + 4)
• Hannah’s age three years ago (x – 3)
• Karen is twice as old as Anna (Karen’s age: 2x)
• Karen is thrice as old as Anna four years ago (Karen’s age: 2(x-4))

In solving age problems, creating a table is always helpful. This is one of the strategies that I am going to discuss in this series. As a start, we discuss one sample problem. » Read more

## Solving Word Problems in Numbers using Algebra Part 3

This is the last part of the Solving Number Problems Series. In this post, we are going to solve number problems in disguise, or numbers problems with different contexts. The previous two parts in this series you may want to read are Solving Word Problems in Numbers using Algebra are Part 1 and Part 2.

Example 7

Jack is twice as old as Rose. Two years from now, the sum of their ages is 40. How old are they now?

Solution

This problem is an age problem but it is very similar to number problems.  As stated, there are two points in time: now and 2 years from now.

Now, Jack is twice as old as Rose. That means that if Rose is 15, then Jack is 30. That means that if the age of Jack is $2x$, then Rose’s age is $x$. Therefore, we have the following representation.  » Read more

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