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Q1(iv):
Express each number as a product of its prime factors: (iv) 5005
Solution :
Given: The number $5005$.
To Find: The prime factorization of $5005$.
Step 1: Understanding Prime Factorization
Prime factorization is the process of expressing a composite number as a product of prime numbers. We will use the method of successive division by the smallest prime numbers ($2, 3, 5, 7, 11, 13, \dots$) until the quotient becomes $1$.
Step 2: Sequential Division
We test the divisibility of $5005$ by prime numbers in ascending order:
1. Check for 2: The last digit of $5005$ is $5$, which is odd. Therefore, it is not divisible by $2$.
2. Check for 3: The sum of the digits is $5 + 0 + 0 + 5 = 10$. Since $10$ is not divisible by $3$, $5005$ is not divisible by $3$.
3. Check for 5: The last digit is $5$. According to the divisibility rule for $5$, any number ending in $0$ or $5$ is divisible by $5$.
$5005 \div 5 = 1001$
4. Check for 7: We divide $1001$ by $7$.
$1001 \div 7 = 143$
Since the remainder is $0$, $7$ is a prime factor.
5. Check for 11: We divide $143$ by $11$.
$143 \div 11 = 13$
Since the remainder is $0$, $11$ is a prime factor.
6. Check for 13: We divide $13$ by $13$.
$13 \div 13 = 1$
Since the quotient is $1$, the process is complete.
Step 3: Tabular Representation of Factors
| Divisor | Quotient |
|---|---|
| 5 | 1001 |
| 7 | 143 |
| 11 | 13 |
| 13 | 1 |
Step 4: Expressing as a Product
By collecting the divisors used in the successive division, we can write:
$5005 = 5 \times 1001$
$5005 = 5 \times 7 \times 143$
$5005 = 5 \times 7 \times 11 \times 13$
Final Answer: The prime factorization of $5005$ is $5 \times 7 \times 11 \times 13$.
More Questions from Class 10 Mathematics Real numbers EXERCISE 1.1
- Q1(i): Express each number as a product of its prime factors: (i) 140
- Q1(ii): Express each number as a product of its prime factors: (ii) 156
- Q1(iii): Express each number as a product of its prime factors: (iii) 3825
- Q1(v): Express each number as a product of its prime factors: (v) 7429
- Q2(i): Find the LCM and HCF of the following pairs of integers and verify that LCM $\times$ HCF = product of the two numbers. (i) 26 and 91
- Q2(ii): Find the LCM and HCF of the following pairs of integers and verify that LCM $\times$ HCF = product of the two numbers. (ii) 510 and 92
- Q2(iii): Find the LCM and HCF of the following pairs of integers and verify that LCM $\times$ HCF = product of the two numbers. (iii) 336 and 54
- Q3(i): Find the LCM and HCF of the following integers by applying the prime factorisation method. (i) 12, 15 and 21
- Q3(ii): Find the LCM and HCF of the following integers by applying the prime factorisation method. (ii) 17, 23 and 29
- Q3(iii): Find the LCM and HCF of the following integers by applying the prime factorisation method. (iii) 8, 9 and 25
- Q4: Given that HCF (306, 657) = 9, find LCM (306, 657).
- Q5: Check whether $6^n$ can end with the digit 0 for any natural number $n$.
- Q6: Explain why $7 \times 11 \times 13 + 13$ and $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$ are composite numbers.
- Q7: 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 time, and go in the same direction. After how many minutes will they meet again at the starting point?
CBSE Solutions for Class 10 Mathematics Real numbers
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