Real Numbers Exercise-1.1
Maths - Exercise
1. Express each number as a product of its prime factors:
- 140 $$ \begin{array}{r|r} 2&140\\\hline 2&70\\\hline 5&35\\\hline 7&7\\\hline &1 \end{array}\\\\ \begin{aligned} 140&=1\times 2\times2\times5\times7\\ &=2^2\times5\times7 \end{aligned} $$
- 156 $$ \begin{array}{r|r} 2&156\\\hline 2&78\\\hline 3&39\\\hline 13&13\\\hline &1 \end{array}\\\\ \begin{aligned} 156&=1\times 2\times2\times3\times13\\ &=2^2\times3\times13 \end{aligned} $$
- 3825 $$ \begin{array}{r|r} 3&3825\\\hline 3&1275\\\hline 5&425\\\hline 5&85\\\hline 17&17\\\hline &1 \end{array}\\\\ \begin{aligned} 3825&=1\times 3\times3\times5\times5\times17\\ &=3^2\times5^2\times17 \end{aligned} $$
- 5005 $$ \begin{array}{r|r} 5&5005\\\hline 7&1001\\\hline 11&143\\\hline 13&13\\\hline &1 \end{array}\\\\ \begin{aligned} 5005&=1\times 5\times7\times11\times13\\ &=5\times7\times11\times13 \end{aligned} $$
- 7429 $$ \begin{array}{r|r} 17&7429\\\hline 19&437\\\hline 23&23\\\hline &1 \end{array}\\\\ \begin{aligned} 7429&=1\times 17\times19\times23\\ &=17\times19\times23 \end{aligned} $$
2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.
- 26 and 91 $$\begin{aligned}26&=2\times 13\\ 91&=7\times 13\end{aligned}$$ \(LCM\left( 26,91\right)=\) Product of the greatest power of each prime factor, involved in the numbers. \begin{aligned} &=2^{1}\times 13^{1}\times 7^{1}\\&=182\end{aligned} \(HCF\left( 26,91\right) =\) Product of the smallest power of each common prime factor in the numbers. \begin{aligned}&=13\\\\ LCM\times HCF&=\text{Product of two numbers}\\\\ LHS&=182\times 13\\ &=2366\\\\ RHS&=26\times 91\\ &=2366\\\\ LHS&=RHS\\\\&\text{Hence, verified}\end{aligned}
- 510 and 92 \begin{aligned}510&=2\times5\times51\\ 92&=2\times 46\end{aligned} \(LCM\left( 510,92\right)=\) Product of the greatest power of each prime factor, involved in the numbers. \begin{aligned} &=2^{1}\times 5^{1}\times 41^{1}\times 46^{1}\\&=23460\end{aligned} \(HCF\left( 510,92\right) =\) Product of the smallest power of each common prime factor in the numbers. \begin{aligned}&=2^1\\ &=2\\\\ LCM\times HCF&=\text{Product of two numbers}\\\\ LHS&=23460\times 2\\ &=46920\\\\ RHS&=510\times 92\\ &=46920\\\\ LHS&=RHS\\\\&\text{Hence, verified}\end{aligned}
- 336 and 54 \begin{aligned}336&=2\times2\times2\times2\times3\times7\\ 54&=2\times3\times3\times3\end{aligned} \(LCM\left( 336,54\right)=\) Product of the greatest power of each prime factor, involved in the numbers. \begin{aligned} &=2^{4}\times 3^{3}\times 7^{1}\\&=3024\end{aligned} \(HCF\left( 336,54\right) =\) Product of the smallest power of each common prime factor in the numbers. \begin{aligned}&=2^1\times3^1\\ &=6\\\\ LCM\times HCF&=\text{Product of two numbers}\\ LHS&=3024\times 6\\ &=18144\\\\ RHS&=336\times 54\\ &=18144\\\\ LHS&=RHS\\\\&\text{Hence, verified}\end{aligned}
3. Find the LCM and HCF of the following integers by applying the prime factorisation method.
- 12, 15 and 21 $$\begin{aligned} 12&=2\times2\times3\\ &=2^2\times3\\ 15&=3\times5\\ 21&=3\times7 \end{aligned}$$ HCF = Product of the smallest power of each common prime factor in the numbers. $$\begin{aligned} HCF\left( 12,15,21\right) &= 3^1\\&=3\\ \end{aligned}$$ LCM = Product of the greatest power of each prime factor, involved in the numbers. $$\begin{aligned} LCM\left( 12,15,21\right)&=2^2\times3^1\times5^1\times7^1\\&=420 \end{aligned}$$
- 17, 23 and 29 $$\begin{aligned} 17&=1\times17\\ 23&=1\times23\\ 29&=1\times29 \end{aligned}$$ HCF = Product of the smallest power of each common prime factor in the numbers. $$\begin{aligned} HCF\left( 17,23,29\right) &= 1^1\\&=1\\ \end{aligned}$$ LCM = Product of the greatest power of each prime factor, involved in the numbers. $$\begin{aligned} LCM\left( 17,23,29\right)&=17^1\times23^1\times29^1\\&=11339 \end{aligned}$$
- 8, 9 and 25 $$\begin{aligned} 8&=1\times2\times2\times2\\ &=1\times2^3\\ 9&=1\times3\times3\\ &=1\times3^2\\ 25&=1\times5\times5\\ &=1\times5^2 \end{aligned}$$ HCF = Product of the smallest power of each common prime factor in the numbers. $$\begin{aligned} HCF\left(8,9,25\right) &= 1^1\times1^1\times1^1\\&=1\\ \end{aligned}$$ LCM = Product of the greatest power of each prime factor involved in the numbers. $$\begin{aligned} LCM\left(8,9,25\right)&=2^3\times3^2\times5^2\\&=1800 \end{aligned}$$
4. Given that HCF (306, 657) = 9, find LCM (306, 657).
\(HCF (306, 657) = 9\\\\ HCF\times LCM=\text{Product of both numbers}\) $$\require{cancel}\begin{aligned} HCF (306, 657) &= 9\quad\text{(Given)}\\ HCF\times LCM&=306\times 657\\ 9\times LCM&=306\times 657\\\\ LCM&=\frac{306\times 657}{9}\\\\ LCM&=\frac{\cancelto{34}{306}\times 657}{\cancelto{1}{9}}\\\\ LCM&=34\times657\\ &=22338 \end{aligned}$$5. Check whether \(6^n\) can end with the digit 0 for any natural number \(n\).
solution:If the number \(6^n\), for any \(n\), were to end with the digit zero, then it would be divisible by 5.
$$\begin{aligned} 6^n&=(2\times3)^n\\ &=2^n\times3^n\\ \end{aligned}$$2 and 3 are prime numbers and therefore, it is not divisible by 5, hence \(6^n\) can never end with zero digit.
6. 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.
Solution:Both numbers are composite because they have factors other than 1 and themselves, making them divisible by smaller numbers.
\(7 \times 11 \times 13 + 13\) $$\begin{aligned} 7 \times 11 \times 13 + 13 &= 13 \times (7 \times 11 + 1) \\&= 13 \times 78 \\&= 1014 \end{aligned}$$ \(1014\) can be factored as $$ 13 × 78 $$ both \(13\) and \(78\) are greater than \(1\), therefore given number is a composite number.\(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5\) $$\begin{aligned} 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5&=7! + 5\\&=5045 \end{aligned}$$ \(5045\) can be written as product of \(5\) and \(1009\) $$\begin{aligned} 5045=5\times 1009 \end{aligned}$$ Thus, it is a composite number because it is divisible by \(5\) and \(1009\) as well as \(1\) and itself.
7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round, the same time, and goes in the same direction. After how many minutes will they meet again at the starting point?
Solution:
Time taken to complete one round by Sonia = 18 minutes
Time Taken by Ravi to complete one round = 12 minutes
\(\because\) they started at the same point at the same time and in the same direction, they will meet again in
\(LCM(18,12)\) minutes
$$\begin{aligned}
12&=2\times 2\times\times 3\\
18&=2\times 3\times3\\\
LCM(12,18)&=2^2\times3^2\\&=4\times9\\&=36\text{ minutes}
\end{aligned}$$
