Skip to main content

Least common multiple

SA I write this post because I failed yesterday to solve the topcoder TCO 250 problem , it was about find the LCM -Least Common multiple -simply it is smallest number which accept the division on -non zero- numbers .

how we calculate it? there exist two ways lets discuss both:
for example 3,4,6
first way find the multiple of all numbers 2,3,4...etc.
Multiples of 2: 4,6,8,10,12,14,16.
Multiples of 3: 6,9,12,15.
Multiples of 4: 8,12,16,20.

as you see 12 is the smallest common number which appear on all numbers .

second way factor each number into it prime component and do the following steps :
1-Count the number of times each prime number appears in each of the factorizations.

2-For each prime number, take the largest of these counts.

3-Write down that prime number as many times as you counted for it in step 2.

4-The least common multiple is the product of all the prime numbers written down.

example 5,6,15.

Factor into primes

Prime factorization of 5 is 5

Prime factorization of 6 is 2 x 3

Prime factorization of 15 is 3 x 5

Notice that the different primes are 2, 3 and 5.

Now, we do Step #1 - Count the number of times each prime number appears in each of the factorizations...

The count of primes in 5 is one 5

The count of primes in 6 is one 2 and one 3

The count of primes in 15 is one 3 and one 5.

Step #2 - For each prime number, take the largest of these counts. So we have...

The largest count of 2s is one

The largest count of 3s is one

The largest count of 5s is one

Step #3 - Since we now know the count of each prime number, you simply - write down that prime number as many times as you counted for it in step 2.

Here they are...

2, 3, 5

Step #4 - The least common multiple is the product of all the prime numbers written down.

2 x 3 x 5 = 30

Therefore, the least common multiple of 5, 6 and 15 is 30.

So there you have it. A quick and easy method for finding least common multiples.

reference see the Link

Comments

Popular posts from this blog

DevExpress C# Chart Tutorial

Hello , this blog post about Charting in DevExpress , I faced many problems I visited hundreds of pages to solve this problems , actually it was the first time I use DevExpress Charts, Work is not like school or university , in school if you don't know thing you can learn it any time you want , in work you have to do today task yesterday, I like to share my knowledge and skills with you , I hope this Post help you . // clear series on chart  chart.Series.Clear(); // clear Titles of chart chart.Titles.Clear(); // legend alignment  chart.Legend.AlignmentHorizontal = LegendAlignmentHorizontal.Center; chart.Legend.AlignmentVertical = LegendAlignmentVertical.Bottom; chart.Legend.Direction = LegendDirection.LeftToRight; // creating series // series view Type example Line DevExpress.XtraCharts.Series series = new DevExpress.XtraCharts.Series("series name ", ViewType.Line); // add title  DevExpress.XtraCharts.ChartTitle chtTitle = new DevExpress.XtraCharts.Cha...

Recursion

Recursion is a programming technique which function calls itself , this seems strange , however it is one of the most interested and effective technique in programming. it helps you write code in less statements and provides unique conceptual frame work to solve problems. lets discus some problems to know how it works : first example Triangular Numbers : it is series like : 1,3,6,10,15,21........ etc our program should get the nth term in the series, the numbers in this series called triangular numbers due to they can be visualized as a triangle. triangular numbers fig 1 finding the nth term using a loop see this figure : loop adding 1+2+3+4 and so on code is easy  , a loop will add numbers from n to 0 like this int triangle(int n) { int total = 0; while(n > 0) // until n is 1 { total = total + n; // add n (column height) to total --n; // decrement column height } return total; } it easy enough and for sure it is fast . before we start the recu...

LeetCode Problem #19 Remove Nth Node From End of List

Solution for LeetCode Problem #19 Remove Nth Node From End of List it is linked list problem can be solved with multiple solutions, but I'm gonna solve it with fastest solution in java. Problem link we should assign dummy_head it is common while solving linked list problems  Assign 2 iterators slow and fast both equals to dummy_head we don't know the length of the list but we know it is always valid, so we will move with a window of size n+1 when we reach the end by the fast iterator the slow iterator will be in the node n+1 from the last , right? we will assign slow.next to be equal slow.next.next we are dropping the node n from last, got it? we will return dummy_head.next Thank you, AhmedG