Fundamental Theorem of Arithmetic 10th

1.Fundamental Theorem of Arithmetic 10th

Fundamental Theorem of Arithmetic 10th

We will discuss on real number in this article “Fundamental Theorem of Arithmetic 10th”.We begin with this important properties of positive integers in sections 2 namely Fundamental Theorem of Arithmetic.

Also Read This Article: – Arithmetic Mean in Class 10th

2.Important Points of Fundamental Theorem of Arithmetic 10th

Fundamental Theorem of Arithmetic 10th

The Fundamental Theorem of Arithmetic (FTA) is a very important part of Class 10 Math (Chapter 1).Its key points are given below:
(1.)Core Statement
Statement:Every composite number can be expressed (factorized) as a product of primes,and this factorization is unique,apart from the order in which the prime factors occur.
(That is,every composite number can be expressed as a product of prime numbers,and this factorization is absolutely unique, except that their order can be forward or backward).
(2.)Example
If we consider the number 30:
◦You can write it as 3×2×5 or 5×3 ×2,but the prime factors will always be 2,3 and 5. Nothing else can make 30 a prime number.
(3.)Key Concepts Asked in Exams:
•Composite Number:A number with more than 2 factors (e.g., 4,6,8,9…).
•Prime Number:A number that is divisible only by 1 and itself (e.g., 2,3,5,7…). Remember:1 is neither prime nor composite. 2 is the smallest and only even prime number.
•Uniqueness Property:The most important feature of this theorem is that the “Prime Product Profile” of every composite number is unique.
(4.)Formula to Find HCF and LCM (Most Important)
This theorem is used to find the LCM and HCF for two positive integers a and b:
•HCF (Highest Common Factor):Product of the smallest power of each common prime factor involved in the numbers.
•LCM (Lowest Common Multiple):Product of the greatest power of each prime factor involved in the numbers.
(5.)Important Tips for Board Exam:
This theorem is used to prove two types of questions:
(i).To prove a number is irrational (e.g., Prove that \sqrt{2} or \sqrt{3} is irrational).
2.To check whether a number ends with zero (0) or not (e.g.,Check whether 6^n can end with the digit 0 for any natural number n). [Logic: 2×5 must be among the prime factors to end in zero.]

Also Read This Article:- Fundamental theorem of arithmetic

3.Examples Based on Fundamental Theorem of Arithmetic 10th

Example:1.Express each number as a product of its prime factors:
Example:1(i).140
Solution: 140 = 2^2 \times 5 \times 7 \\ \begin{array}{r|l} 2 & 140\\ \hline 2 & 70\\ \hline 5 & 35\\ \hline 7 & 7\\ \hline & 1 \end{array}
Example:1(ii).156
Solution:156
156 = 2^2 \times 3 \times 13 \\ \begin{array}{r|l} 2 & 156\\ \hline 2 & 78\\ \hline 3 & 39\\ \hline 13 & 13\\ \hline & 1 \end{array}
Example:1(iii).3825
Solution:3825
3825= 3^2 \times 5^2 \times 17\\ \begin{array}{r|l} 3 & 3825\\ \hline 3 & 1275\\ \hline 5 & 425\\ \hline 5 & 85\\ \hline17 & 17\\ \hline & 1 \end{array}
Example:1(iv).5005
Solution:5005
\begin{array}{r|l} 5 & 5005\\ \hline 7 & 1001\\ \hline 11 & 143\\ \hline 13 & 13\\ \hline & 1 \end{array} \\ \\ 5005 = 5 \times 7 \times 11 \times 13
Example:1(i).7429
Solution:7429
\begin{array}{r|l} 17 & 7429\\ \hline 19 & 437\\ \hline 23 & 23\\ \hline & 1 \end{array} \\ 7429 = 17 \times 19 \times 23
Example:2.Find the LCM and HCF of the following pairs of integers and verify that LCM×HCF=product of the two numbers:
Example:2(i).26 and 91
Solution:26 and 91
\begin{array}{r|l} 2 & 26\\ \hline 13 & 13\\ \hline & 1 \end{array} \qquad \begin{array}{r|l} 7 & 91\\ \hline 13 & 13\\ \hline & 1 \end{array} \\ 26 = 2 \times 13\\ 91 = 7 \times 13
HCF=Product of smallest power of each common prime factor in numbers
HCF=13
LCM=Product of greatest power of each common prime factor in numbers
\Rightarrow LCM=182
LCM×HCF=13×182=2366
Product of Numbers=26×91=2366
Hence,LCM×HCF=product of two numbers
Example:2(ii).510 and 92
Solution:510 and 92
\begin{array}{r|l} 2 & 510\\ \hline 3 & 255\\ \hline 5 & 85\\ \hline 17 & 17\\ \hline & 1 \end{array} \qquad \begin{array}{r|l} 2 & 92\\ \hline 2 & 46\\ \hline 23 & 23\\ \hline & 1 \end{array} \\ 510 = 2 \times 3 \times 5 \times 17\\ 92 = 2^2 \times 23
HCF(510,92)=Product of smallest power of each common prime factor in numbers
\Rightarrow HCF=2
LCM(510,92)=Product of greatest power of each common prime factor in numbers
\Rightarrow LCM=23460
LCM×HCF=23460×2=46920
Product of Numbers=510×92=46920
Hence,LCM×HCF=product of two numbers
Example:2(i).336 and 54
Solution:336 and 54
\begin{array}{r|l} 2 & 336\\ \hline 2 & 168\\ \hline 2 & 84\\ \hline 2 & 42\\ \hline 3 & 21\\ \hline7 & 7\\ \hline & 1 \end{array} \qquad \begin{array}{r|l} 2 & 54\\ \hline 3 & 27\\ \hline 3 & 9\\ \hline 3 & 3\\ \hline & 1 \end{array} \\ 336 = 2^4 \times 3 \times 7,\\ 54 = 2 \times 3^3
HCF(336,54)=Product of smallest power of each common prime factor in numbers
\Rightarrow HCF=2×3=6
LCM(336,54)=Product of greatest power of each common prime factor in numbers
=2^4 \times 3^3 \times 7
\Rightarrow LCM=3024
LCM×HCF=6×3024=18144
Product of Numbers=336×53=18144
Hence,LCM×HCF=product of two numbers

Fundamental Theorem of Arithmetic 10th

Example:3.Find the LCM and HCF of the following integers by applying the prime factorisation method.
Example:3(i).12,15 and 21
Solution:12,15 and 21
\begin{array}{r|l} 2 & 12\\ \hline 2 & 6\\ \hline 3 & 3\\ \hline & 1 \end{array} \qquad \begin{array}{r|l} 3 & 15\\ \hline 5 & 5\\ \hline & 1 \end{array} \qquad \begin{array}{r|l} 3 & 21\\ \hline 7 & 7\\ \hline & 1 \end{array} \\ 12 = 2^2 \times 3 \\ 15 = 3 \times 5 \\ 21 = 3 \times 7
HCF(12,15,21)=Product of smallest power of each common prime factor in numbers
\Rightarrow HCF=3
LCM(12,15,21)=Product of greatest power of each common prime factor in numbers
= 2^2 \times 3 \times 5 \times 7 \\ =420
\Rightarrow LCM=420
Example:3(ii).17,23 and 29
Solution:17,23 and 29
17=17×1,23=23×1,29=29×1
HCF(17,23,29)=Product of smallest power of each common prime factor in numbers
\Rightarrow HCF=1
LCM(17,23,29)=Product of greatest power of each common prime factor in numbers
LCM=17×23×29
\Rightarrow LCM=11339
Example:3(iii).8,9 and 25
Solution:8,9 and 25
\begin{array}{r|l} 2 & 8\\ \hline 2 & 4\\ \hline 2 & 2\\ \hline & 1 \end{array} \qquad \begin{array}{r|l} 3 & 9\\ \hline 3 & 3\\ \hline & 1 \end{array} \qquad \begin{array}{r|l} 5 & 25\\ \hline 5 & 5\\ \hline & 1 \end{array} \\ 8 = 2^3 \times 1 \\ 9 = 3^2 \times 1 \\ 25 = 5^2 \times 1
HCF(8,9,25)=Product of smallest power of each common prime factor in numbers
\Rightarrow HCF=1
LCM(8,9,25)=Product of greatest power of each common prime factor in numbers
\Rightarrow \mathrm{LCM}=2^{3} \times3^{2} \times 5^{2}=1800
Example:4.Given that HCF(306,657)=9,find LCM(306,657)
Solution: \mathrm{[LCM]}(a,b)=\frac{a\times b}{\mathrm{HCF}(a,b)}
Put a=306 and b=657
=\frac{306\times657}{9}=22338
\Rightarrow LCM=22338
Example:5.Chack whether 6^n can end with the digit 0 for any natural number n.
Solution:We know that any positive integer ending with the digit zero is divisible by 2 and 5 so its prime factorization must contain the prime 2 and 5.
The only prime factors of 6^n are 2 and 3.
We have 6^n=(2\times3)^n=2^n\times3^n
There is no other prime factors except 2 and 3
(By uniqueness of fundamental theorem of arithmetic)
5 does not occur in the prime factorization of 6^n for any.Thus, 6^n does not end with the digit zero for any natural number n.
Example:6.Explain why 7×11×13+13 and 7×6×5×4×3×2×1+5 are composite numbers.
Solution:7×11×13+13=13(7×11+1)
=13×78=13×2×3×13
Which is not prime number because it has greater than two factors viz 13,2,3
\therefore It is a composite number
Now 7×6×5×4×3×2×1+5=5(7×6×4×3×2+1)
=5×(1008+1)
=5×1009
Which is not also prime number because it has greater than two factors viz 5,1009
\therefore It is a composite number
Example:7.There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field,while Ravi takes 12 minutes for the same.Suppose they both start at the same point and at the same line,and go in the same direction.After how many minutes will they meet again at the starting point?
Solution:They will meet again at the starting point=LCM(18,12)
18=2\times3^{2} and 12=2^{2} \times 3
\operatorname{LCM} (18,12)=2^{2}\times3^{2}=36
=36 minutes
With the above Examples,one can understand the Fundamental Theorem of Arithmetic 10th.

Also Read This Article:-  Arithmetic and Algebra

4.Questions of Fundamental Theorem of Arithmetic 10th for Students

Fundamental Theorem of Arithmetic 10th

(1.)Find the HCF and LCM of 144,180 and 192 by prime factorisation method.
(2.)The HCF of two numbers 145 and their LCM is 2175.If one number is 725,find the other number.
Answers:(1.)HCF=12,LCM=2880 (2.)435
By solving the above questions,you can understand the Fundamental Theorem of Arithmetic 10th well because the concept is well understood when you solve it practically.

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5.Frequently Asked Questions Related to Fundamental Theorem of Arithmetic 10th

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*”8 बजकर कितने मिनट समय हुआ है यदि 74 मिनट पहले यह सात बजने के बाद के कुल मिनटों का आधा था।”*
**Today’s Question to Students”*
*”How many minutes is it past 8 if 74 minutes ago it was half of the total minutes after 7 o’clock?”*

दिनांक 11.07.2026 के प्रश्न का उत्तर
माना चौड़ाई=x,लम्बाई=\frac{140x}{100} \\ \frac{140x}{100}-x=20 \\ \Rightarrow \frac{40x}{100}=20 \\ \Rightarrow x=\frac{20\times100}{40}=50
लम्बाई=\frac{140}{100}\times50 =70
आयत का क्षेत्रफल=50×70=3500 वर्ग मीटर
Answer to Question Dated 11.07.2026
L let Width=x,Length=\frac{140x}{100} \\ \frac{140x}{100}-x=20 \\ \Rightarrow \frac{40x}{100}=20 \\ \Rightarrow x=\frac{20\times100}{40}=50
Length=\frac{140}{100}\times50 =70
Area of Rectangle=50×70=3500 Square Meter
This article has been prepared by **Satyam Coaching Centre** on the **Satyam Mathematics** blog.”*