Real Numbers

$\displaystyle \textbf{Question 1. }\text{Express each of the following integers as a product of its prime factors:}$
$\displaystyle (i)\ 420\qquad (ii)\ 468\qquad (iii)\ 945\qquad (iv)\ 7325\qquad (v)\ \frac{1}{\sqrt{5}}\qquad [\mathrm{CBSE}\ 2025]$
$\displaystyle \text{Answer:}$
$\displaystyle (i)\ 420=2\times210$
$\displaystyle =2\times2\times105$
$\displaystyle =2\times2\times3\times35$
$\displaystyle =2^{2}\times3\times5\times7$
$\displaystyle (ii)\ 468=2\times234$
$\displaystyle =2\times2\times117$
$\displaystyle =2^{2}\times3\times39$
$\displaystyle =2^{2}\times3^{2}\times13$
$\displaystyle (iii)\ 945=3\times315$
$\displaystyle =3^{2}\times105$
$\displaystyle =3^{3}\times35$
$\displaystyle =3^{3}\times5\times7$
$\displaystyle (iv)\ 7325=5\times1465$
$\displaystyle =5^{2}\times293$
$\displaystyle =5^{2}\times293$
$\displaystyle (v)\ \frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}$
$\displaystyle =\frac{\sqrt{5}}{5^{1}}$
$\displaystyle =5^{-\frac{1}{2}}$
$\\$

$\displaystyle \textbf{Question 2. }\text{Determine the prime factorisation of each of the following positive integers:}$
$\displaystyle (i)\ 20570\qquad (ii)\ 58500\qquad (iii)\ 45470971$
$\displaystyle \text{Answer:}$
$\displaystyle (i)\ 20570=2\times10285$
$\displaystyle =2\times5\times2057$
$\displaystyle =2\times5\times11\times187$
$\displaystyle =2\times5\times11^{2}\times17$
$\displaystyle \therefore 20570=2\times5\times11^{2}\times17$
$\displaystyle (ii)\ 58500=2\times29250$
$\displaystyle =2^{2}\times14625$
$\displaystyle =2^{2}\times3\times4875$
$\displaystyle =2^{2}\times3^{2}\times1625$
$\displaystyle =2^{2}\times3^{2}\times5^{3}\times13$
$\displaystyle \therefore 58500=2^{2}\times3^{2}\times5^{3}\times13$
$\displaystyle (iii)\ 45470971=7\times6495853$
$\displaystyle =7\times17\times382109$
$\displaystyle =7\times17\times19\times20111$
$\displaystyle =7\times17\times19\times23\times874$
$\displaystyle =2\times7\times17\times19\times23\times19$
$\displaystyle =2\times7\times17\times19^{2}\times23$
$\displaystyle \therefore 45470971=13\times17\times19\times23\times47$
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$\displaystyle \textbf{Question 3. }\text{Explain why }7\times11\times13+13\text{ and }7\times6\times5\times4\times3\times2\times1+5$
$\displaystyle \text{are composite numbers.}$
$\displaystyle \text{Answer:}$
$\displaystyle 7\times11\times13+13$
$\displaystyle =13(7\times11+1)$
$\displaystyle =13\times78$
$\displaystyle \therefore \text{It has factors other than }1\text{ and itself. Hence it is composite.}$
$\displaystyle 7\times6\times5\times4\times3\times2\times1+5$
$\displaystyle =5(7\times6\times4\times3\times2\times1+1)$
$\displaystyle =5\times1009$
$\displaystyle \therefore \text{It has factors other than }1\text{ and itself. Hence it is composite.}$
$\\$

$\displaystyle \textbf{Question 4. }\text{Explain why }3\times5\times7+7\text{ is a composite number.}$
$\displaystyle \text{Answer:}$
$\displaystyle 3\times5\times7+7$
$\displaystyle =7(3\times5+1)$
$\displaystyle =7(15+1)$
$\displaystyle =7\times16$
$\displaystyle \text{Since it has factors other than }1\text{ and itself, it is a composite number.}$
$\\$

$\displaystyle \textbf{Question 5. }\text{Check whether }6^{n}\text{ can end with the digit }0\text{ for any natural number }n.\ [\mathrm{CBSE}\ 2023]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{If a number ends with }0,\text{ then it is divisible by }10.$
$\displaystyle \therefore \text{It must have both }2\text{ and }5\text{ as prime factors.}$
$\displaystyle 6^{n}=(2\times3)^{n}$
$\displaystyle =2^{n}\times3^{n}$
$\displaystyle \text{Prime factorisation of }6^{n}\text{ does not contain the factor }5.$
$\displaystyle \therefore 6^{n}\text{ is not divisible by }10.$
$\displaystyle \therefore 6^{n}\text{ cannot end with the digit }0.$
$\\$

$\displaystyle \textbf{Question 6. }\text{Can the number }(15)^{n},\ n\text{ being a rational number, end with the digit }0?$
$\displaystyle \text{Give reasons.}\qquad [\mathrm{CBSE}\ 2024]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{If a number ends with }0,\text{ then it must be divisible by }10.$
$\displaystyle \therefore \text{It must have both }2\text{ and }5\text{ as prime factors.}$
$\displaystyle 15=3\times5$
$\displaystyle \therefore (15)^{n}=(3\times5)^{n}$
$\displaystyle \text{The prime factorisation of }(15)^{n}\text{ does not contain the factor }2.$
$\displaystyle \therefore (15)^{n}\text{ is not divisible by }10.$
$\displaystyle \therefore (15)^{n}\text{ cannot end with the digit }0.$
$\\$

$\displaystyle \textbf{Question 7. }\text{Find the values of }x,y,z\text{ and }w\text{ in the following factor tree. Also write}$
$\displaystyle \text{the prime factorisation of }x.\qquad [\mathrm{CBSE}\ 2024]$ $\displaystyle \text{Answer:}$
$\displaystyle w=13\times7=91$
$\displaystyle z=3\times w$
$\displaystyle =3\times91=273$
$\displaystyle 819=3\times z$
$\displaystyle =3\times273$
$\displaystyle y=2\times819=1638$
$\displaystyle x=2\times y$
$\displaystyle =2\times1638=3276$
$\displaystyle \therefore w=91,\ z=273,\ y=1638,\ x=3276$
$\displaystyle \text{Prime factorisation of }x=2\times2\times3\times3\times13\times7$
$\displaystyle =2^{2}\times3^{2}\times7\times13$
$\\$

$\displaystyle \textbf{Question 8. }\text{State true or false for each of the following statements and justify in each case:}$
$\displaystyle (i)\ 2\times3\times5\times7+7\text{ is a composite number.}$
$\displaystyle (ii)\ 2\times3\times5\times7+1\text{ is a composite number.}\qquad [\mathrm{CBSE}\ 2025]$
$\displaystyle \text{Answer:}$
$\displaystyle (i)\ \text{True}$
$\displaystyle 2\times3\times5\times7+7$
$\displaystyle =7(2\times3\times5+1)$
$\displaystyle =7\times31$
$\displaystyle \therefore \text{It has factors other than }1\text{ and itself. Hence it is a composite number.}$
$\displaystyle (ii)\ \text{False}$
$\displaystyle 2\times3\times5\times7+1=211$
$\displaystyle \sqrt{211}<15$
$\displaystyle \text{211 is not divisible by }2,3,5,7,11\text{ or }13.$
$\displaystyle \therefore 211\text{ is a prime number.}$
$\\$

$\displaystyle \textbf{Question 9. }\text{Let }p,\ q\text{ and }r\text{ be three distinct prime numbers. Check whether }$
$\displaystyle pqr+q\text{ is a composite number or not. Further, give an example of }3\text{ distinct prime numbers }$
$\displaystyle p,q,r \text{ such that }(i)\ pqr+1\text{ is a composite number. }(ii)\ pqr+1\text{ is a prime number.}\ [\mathrm{CBSE}\ 2025]$
$\displaystyle \text{Answer:}$
$\displaystyle pqr+q=q(pr+1)$
$\displaystyle \text{Since }q\text{ is a prime factor of }pqr+q\text{ and }pr+1>1,$
$\displaystyle \therefore pqr+q\text{ is a composite number.}$
$\displaystyle (i)\ \text{Let }p=2,\ q=3,\ r=5.$
$\displaystyle pqr+1=2\times3\times5+1=31$
$\displaystyle \text{Since }31\text{ is prime, this example is not suitable.}$
$\displaystyle \text{Take }p=3,\ q=5,\ r=7.$
$\displaystyle pqr+1=3\times5\times7+1=106$
$\displaystyle =2\times53$
$\displaystyle \therefore pqr+1\text{ is a composite number.}$
$\displaystyle (ii)\ \text{Let }p=2,\ q=3,\ r=5.$
$\displaystyle pqr+1=2\times3\times5+1=31$
$\displaystyle \therefore 31\text{ is a prime number.}$
$\\$