\displaystyle \textbf{Question 1. }\text{If }x=ab^{3}\text{ and }y=a^{3}b,\text{ where }a\text{ and }b\text{ are prime numbers, then }
\displaystyle [\mathrm{HCF}(x,y)-\mathrm{LCM}(x,y)]\text{ is equal} \text{to:} \hspace{0.2cm}\text{[CBSE 2025]}
\displaystyle \text{(a) }1-a^{3}b^{3}
\displaystyle \text{(b) }ab(1-ab)
\displaystyle \text{(c) }ab-a^{4}b^{4}
\displaystyle \text{(d) }ab(1-ab)(1+ab)
\displaystyle \text{Answer:}
\displaystyle  (d)\ x=ab^{3}\text{ and }y=a^{3}b
\displaystyle \text{HCF of }(x,y)=ab
\displaystyle \text{LCM of }(x,y)=a^{3}b^{3}
\displaystyle \text{Now, }[\text{HCF }(x,y)-\text{LCM }(x,y)]=ab-a^{3}b^{3}
\displaystyle =ab(1-a^{2}b^{2})
\displaystyle =ab(1-ab)(1+ab)
\\

\displaystyle \textbf{Question 2. }\left(1+\sqrt{3}\right)^{2}-\left(1-\sqrt{3}\right)^{2}\text{ is:} \hspace{0.2cm}\text{[CBSE 2025]}
\displaystyle \text{(a) a positive rational number}
\displaystyle \text{(b) a negative integer}
\displaystyle \text{(c) a positive irrational number}
\displaystyle \text{(d) a negative irrational number}
\displaystyle \text{Answer:}
\displaystyle  (c)\ (1+\sqrt{3})^{2}-(1-\sqrt{3})^{2}
\displaystyle =1+3+2\sqrt{3}-1-3+2\sqrt{3}
\displaystyle =4\sqrt{3}
\displaystyle \therefore \text{It is a positive irrational number.}
\\

\displaystyle \textbf{Question 3. }\text{Prove that }\sqrt{2}\text{ is an irrational number.} \hspace{0.2cm}\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle  \text{To prove: }\sqrt{2}\text{ is an irrational number.}
\displaystyle \text{Proof: Let us suppose that }\sqrt{2}\text{ is a rational number.}
\displaystyle \text{Then by definition, }\sqrt{2}=\frac{a}{b},\ \text{where }a\text{ and }b\text{ are coprime and }b\neq0
\displaystyle \text{On squaring both sides, we get}
\displaystyle 2=\frac{a^{2}}{b^{2}}
\displaystyle \Rightarrow 2b^{2}=a^{2} \qquad (1)
\displaystyle \Rightarrow 2\text{ divides }a^{2}
\displaystyle \Rightarrow 2\text{ divides }a\ (\because\ 2\text{ is a prime})
\displaystyle \Rightarrow a=2c,\ \text{where }c\text{ is a positive integer}
\displaystyle \text{Substituting in }(1),\ 2b^{2}=4c^{2}
\displaystyle \Rightarrow b^{2}=2c^{2}
\displaystyle \Rightarrow 2\text{ divides }b^{2}
\displaystyle \Rightarrow 2\text{ divides }b
\displaystyle \Rightarrow 2\text{ is a common factor of }a\text{ and }b
\displaystyle \text{This is a contradiction since }a\text{ and }b\text{ are coprime.}
\displaystyle \text{Hence our supposition is wrong.}
\displaystyle \therefore \sqrt{2}\text{ is an irrational number.}
\\

\displaystyle \textbf{Question 4. }\text{Let }x\text{ and }y\text{ be two distinct prime numbers and }p=x^{2}y^{3},
\displaystyle q=xy^{4},\ r=x^{5}y^{2}. \text{Find the HCF and LCM of }p,q\text{ and }r.\text{ Further} \\ \text{check if }\mathrm{HCF}(p,q,r)\times \mathrm{LCM}(p,q,r)=(p,q,r)=p\times q\times r\text{ or not.} \hspace{0.2cm}\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle  \text{Given : }p=x^{2}y^{3},\ q=xy^{4},\ r=x^{5}y^{2}
\displaystyle \text{HCF }(p,q,r)=xy^{2}
\displaystyle \text{LCM }(p,q,r)=x^{5}y^{4}
\displaystyle \text{Now, HCF }(p,q,r)\times\text{LCM }(p,q,r)=xy^{2}\times x^{5}y^{4}=x^{6}y^{6}
\displaystyle p\times q\times r=x^{2}y^{3}\times xy^{4}\times x^{5}y^{2}=x^{8}y^{9}
\displaystyle \text{So, the two numbers are not equal.}
\displaystyle \therefore \text{LCM }(p,q,r)\times\text{HCF }(p,q,r)\neq p\times q\times r
\\

\displaystyle \textbf{Question 5. }\text{Directions: In the following question, a statement of assertion }(A) \\ \text{is followed by a statement of reason }(R).
\displaystyle \text{Choose the correct answer out of the following choices.}
\displaystyle \text{Assertion }(A):4^{n}\text{ ends with digit }0\text{ for some natural number }n.
\displaystyle \text{Reason }(R):\text{ For a number }x^{n}\text{ having }2\text{ and }5\text{ as its prime factors, } x^{n}\text{ always ends} \\ \text{with digit }0\text{ for every natural number }n. \hspace{0.2cm}\text{[CBSE 2025]}
\displaystyle \text{(a) Both Assertion }(A)\text{ and Reason }(R)\text{ are true and Reason }(R)\text{ is the correct} \\ \text{explanation of Assertion }(A).
\displaystyle \text{(b) Both Assertion }(A)\text{ and Reason }(R)\text{ are true and Reason }(R)\text{ is not the correct} \\ \text{explanation of Assertion }(A).
\displaystyle \text{(c) Assertion }(A)\text{ is true but Reason }(R)\text{ is false.}
\displaystyle \text{(d) Assertion }(A)\text{ is false but Reason }(R)\text{ is true.}
\displaystyle \text{Answer:}
\displaystyle 4^n=(2^2)^n=2^{2n}
\displaystyle \text{Hence }4^n\text{ has only the prime factor }2\text{ and never has the factor }5.
\displaystyle \text{A number ends with digit }0\text{ only if it is divisible by }10=2\times5.
\displaystyle \therefore 4^n\text{ can never end with digit }0.
\displaystyle \therefore \text{Assertion (A) is false.}
\displaystyle \text{The Reason (R) is true because any number having both }2\text{ and }5\text{ as prime factors}
\displaystyle \text{contains the factor }10\text{ and hence ends with digit }0.
\displaystyle \therefore \text{Assertion (A) is false but Reason (R) is true.}
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 6. }\text{If two positive integers }p\text{ and }q\text{ can be expressed as }p=18a^{2}b^{4}
\displaystyle \text{and }q=20a^{3}b^{2},\text{ where }a\text{ and }b\text{ are prime numbers, then }\mathrm{LCM}(p,q)\text{ is: } \hspace{0.2cm}\text{[CBSE 2024]}
\displaystyle \text{(a) }2a^{2}b^{2}
\displaystyle \text{(b) }180a^{2}b^{2}
\displaystyle \text{(c) }120a^{2}b^{2}
\displaystyle \text{(d) }180a^{3}b^{4}
\displaystyle \text{Answer:}
\displaystyle  (d)\ \text{As, }p=18a^{2}b^{4}=2\times3\times3\times a\times a\times b\times b\times b\times b
\displaystyle q=20a^{3}b^{2}=2\times2\times5\times a\times a\times a\times b\times b
\displaystyle \text{LCM }(p,q)=2\times2\times3\times3\times5\times a\times a\times a\times b\times b\times b\times b
\displaystyle =180a^{3}b^{4}
\\

\displaystyle \textbf{Question 7. }\text{Prove that }5-2\sqrt{3}\text{ is an irrational number. It is given that } \\ \sqrt{3}\text{ is an irrational number.} \hspace{0.2cm}\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let us assume that }5-2\sqrt{3}\text{ is a rational number, such that}
\displaystyle 5-2\sqrt{3}=\frac{p}{q},\ \text{where }p\text{ and }q\text{ are coprime and }q\neq0
\displaystyle \Rightarrow 5-\frac{p}{q}=2\sqrt{3}
\displaystyle \Rightarrow \frac{5q-p}{2q}=\sqrt{3}
\displaystyle \Rightarrow \sqrt{3}\text{ is a rational number as }\left(\frac{5q-p}{2q}\right)\text{ is a rational number.}
\displaystyle \text{But it is given that }\sqrt{3}\text{ is an irrational number.}
\displaystyle \text{So, this is a contradiction.}
\displaystyle \text{Hence, our assumption is wrong.}
\displaystyle \therefore (5-2\sqrt{3})\text{ is an irrational number.}
\\

\displaystyle \textbf{Question 8. }\text{Show that the number }5\times 11\times 17+3\times 11\text{ is a composite number.} \hspace{0.2cm}\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle  \ 5\times11\times17+3\times11=11\times(5\times17+3)
\displaystyle =11\times88=11\times2\times2\times2\times11=2^{3}\times11^{2}
\displaystyle \text{Because factorisation of the number contains more than one prime.}
\displaystyle \text{So the given number is a composite number.}
\\

\displaystyle \textbf{Question 9. }\text{In a teachers' workshop, the number of teachers teaching French, Hindi} \\ \text{and English are }48,80\text{ and }   144\text{ respectively. Find the minimum number of rooms} \\ \text{required if in each room the same number of teachers are seated and all of them are of} \\ \text{the same subject.} \hspace{0.2cm}\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle  \text{Number of French teachers}=48,\ \text{Number of Hindi teachers}=80,
\displaystyle \text{Number of English teachers}=144
\displaystyle \text{The prime factorisation of }48,\ 80\text{ and }144:
\displaystyle 48=2\times2\times2\times2\times3
\displaystyle 80=2\times2\times2\times2\times5
\displaystyle 144=2\times2\times2\times2\times3\times3
\displaystyle \text{HCF }(48,80,144)=16
\displaystyle \therefore \text{Minimum number of rooms required}=\frac{48}{16}+\frac{80}{16}+\frac{144}{16}
\displaystyle =3+5+9=17
\\

\displaystyle \textbf{Question 10. }\text{The ratio of HCF to LCM of the least composite number and the }
\displaystyle \text{least prime number is:} \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }1:2
\displaystyle \text{(b) }2:1
\displaystyle \text{(c) }1:1
\displaystyle \text{(d) }1:3
\displaystyle \text{Answer:}
\displaystyle  (a)\ \text{Least composite number is }4\text{ and least prime number is }2.
\displaystyle \text{HCF of }4\text{ and }2=2
\displaystyle \text{LCM of }4\text{ and }2=4
\displaystyle \therefore \text{Ratio}=2:4=1:2
\\

\displaystyle \textbf{Question 11. }\text{Two numbers are in the ratio }2:3\text{ and their LCM is }180. \\ \text{What is the HCF of these numbers?} \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let numbers be }2x\text{ and }3x.
\displaystyle \text{LCM of two numbers}=2\times3\times x=6x
\displaystyle \Rightarrow 6x=180
\displaystyle \Rightarrow x=30
\displaystyle \therefore \text{Numbers are }2\times30=60\text{ and }3\times30=90
\displaystyle \text{Now, HCF}\times\text{LCM}=a\times b
\displaystyle \Rightarrow \text{HCF}\times180=60\times90
\displaystyle \Rightarrow \text{HCF}=\frac{60\times90}{180}=30
\\

\displaystyle \textbf{Question 12. }\text{Check whether }6^{n}\text{ can end with the digit }0\text{ for any} \\ \text{natural number }n. \hspace{0.2cm}\text{[CBSE 2023(C)]}
\displaystyle \text{Answer:}
\displaystyle  \text{Prime factorisation of }6^{n}\text{ is given by,}
\displaystyle 6^{n}=(2\times3)^{n}
\displaystyle \text{So, prime factorisation of }6^{n}\text{ contains only prime numbers }2\text{ and }3.
\displaystyle \text{Now, }6^{n}\text{ may end with digit }0\text{ for some natural number }n
\displaystyle \text{if }5\text{ and }2\text{ are present in its prime factorisation, which is not present.}
\displaystyle \text{So, there is no natural number }n\text{ for which }6^{n}\text{ ends with digit }0.
\\

\displaystyle \textbf{Question 13. }\text{February }14\text{ is celebrated as International Book Giving Day and} \\ \text{many countries in this world celebrate this day.}
\displaystyle \text{Some people in India also started celebrating this day and donated the following} \\ \text{number of books of various subjects to a public library:}
\displaystyle \text{History }=96,\ \text{Science }=240,\ \text{Mathematics }=336
\displaystyle \text{These books have to be arranged in minimum number of stacks such that each stack } \\ \text{contains books of only one subject and the number of books on each stack is the same.}
\displaystyle \text{Based on the above information, answer the following questions:} \hspace{0.2cm}\text{[CBSE 2023(C)]}
\displaystyle \text{(i) How many books are arranged in each stack?}
\displaystyle \text{(ii) How many stacks are used to arrange all the Mathematics books?}
\displaystyle \text{(iii) Determine the total number of stacks that will be used for arranging all the books.}
\displaystyle \text{OR}
\displaystyle \text{If the thickness of each book of History, Science and Mathematics is }1.8\text{ cm, }2.2\text{ cm and } \\ 2.5\text{ cm respectively, then find the height of each stack of History, Science and Mathematics} \\ \text{books.}
\displaystyle \text{Answer:}
\displaystyle  (i)\ \text{History}=96,\ \text{Science}=240,\ \text{Mathematics}=336
\displaystyle \text{Prime factorisation of }96,\ 240\text{ and }336:
\displaystyle 96=2\times2\times2\times2\times2\times3
\displaystyle 240=2\times2\times2\times2\times3\times5
\displaystyle 336=2\times2\times2\times2\times3\times7
\displaystyle \text{HCF}(96,240,336)=2\times2\times2\times2\times3=48
\displaystyle \text{So, }48\text{ books are arranged in each stack.}
\displaystyle \text{(ii) Number of stacks needed for Mathematics books}=\frac{336}{48}=7
\displaystyle \text{(iii) Total books}=96+240+336=672
\displaystyle \text{Number of stacks to arrange all books}=\frac{672}{48}=14
\displaystyle \text{OR}
\displaystyle \text{Thickness of each History book}=1.8\text{ cm}
\displaystyle \text{Thickness of }48\text{ History books}=1.8\times48=86.4\text{ cm}
\displaystyle \therefore \text{Height of each stack of History books}=86.4\text{ cm}
\displaystyle \text{Thickness of each Science book}=2.2\text{ cm}
\displaystyle \therefore \text{Height of each stack of Science books}=2.2\times48=105.6\text{ cm}
\displaystyle \text{Thickness of each Mathematics book}=2.5\text{ cm}
\displaystyle \therefore \text{Height of each stack of Mathematics books}=2.5\times48=120\text{ cm}

\displaystyle \textbf{Question 14. }\text{The exponent of }5\text{ in the prime factorisation of }3750\text{ is:} \hspace{0.2cm}\text{[CBSE 2021]}
\displaystyle \text{(a) }3 \qquad \text{(b) }4
\displaystyle \text{(c) }5 \qquad \text{(d) }6
\displaystyle \text{Answer:}
\displaystyle  (b)\ \text{The prime factorisation of }3750\text{ is,}
\displaystyle 3750=2\times3\times5^{4}
\displaystyle \text{The exponent of }5\text{ is }4.
\\

\displaystyle \textbf{Question 15. }\text{What is the greatest possible speed at which a girl can walk }95\text{ m and } \\ 171\text{ m in an exact number of minutes?} \hspace{0.2cm}\text{[CBSE 2021]}
\displaystyle \text{(a) }17\text{ m/min} \qquad \text{(b) }19\text{ m/min}
\displaystyle \text{(c) }23\text{ m/min} \qquad \text{(d) }13\text{ m/min}
\displaystyle \text{Answer:}
\displaystyle  (b)\ \text{The prime factorisation of }95\text{ and }171:
\displaystyle 95=5\times19;\ 171=3\times3\times19
\displaystyle \text{HCF }(95,171)=19
\displaystyle \therefore \text{Greatest possible speed is }19\ \text{m/min}.
\\

\displaystyle \textbf{Question 16. }\text{Three alarm clocks ring their alarms at regular intervals of }20\text{ min, } \\ 25\text{ min and }30\text{ min respectively. }   \text{If they first beep together at }12\text{ noon, at what} \\ \text{time will they beep again for the first time?} \hspace{0.2cm}\text{[CBSE 2021]}
\displaystyle \text{(a) }4:00\text{ pm} \qquad \text{(b) }4:30\text{ pm}
\displaystyle \text{(c) }5:00\text{ pm} \qquad \text{(d) }5:30\text{ pm}
\displaystyle \text{Answer:}
\displaystyle  (c)\ \text{The prime factorisation of }20,\ 25\text{ and }30:
\displaystyle 20=2\times2\times5;\ 25=5\times5;\ 30=2\times3\times5
\displaystyle \text{LCM }(20,25,30)=2^{2}\times3\times5^{2}=300
\displaystyle \text{So, alarm clocks will beep again after }300\text{ min or }5\text{ hours.}
\displaystyle \text{So they will beep at }5{:}00\text{ pm together.}
\\

\displaystyle \textbf{Question 17. }\text{If }a\text{ and }b\text{ are two coprime numbers, then }a^{3}\text{ and }b^{3}\text{ are} \hspace{0.2cm}\text{[CBSE 2021]}
\displaystyle \text{(a) Coprime} \qquad \text{(b) Not coprime}
\displaystyle \text{(c) Even} \qquad \text{(d) Odd}
\displaystyle \text{Answer:}
\displaystyle  (a)\ \text{Since }a\text{ and }b\text{ are coprime, so HCF }(a,b)=1.
\displaystyle \text{Also, HCF }(a^{3},b^{3})=1.
\displaystyle \text{So }a^{3}\text{ and }b^{3}\text{ are also coprime.}
\\

\displaystyle \textbf{Question 18. }\text{If }n\text{ is a natural number, then }2(5^{n}+6^{n})\text{ always ends with} \hspace{0.2cm}\text{[CBSE 2021]}
\displaystyle \text{(a) }1 \qquad \text{(b) }4
\displaystyle \text{(c) }3 \qquad \text{(d) }2
\displaystyle \text{Answer:}
\displaystyle  (d)\ \text{If }n\text{ is a natural number, then }5^{n}\text{ always ends with }5
\displaystyle \text{and }6^{n}\text{ always ends with }6.
\displaystyle \text{Also, }(5^{n}+6^{n})\text{ always ends with }1
\displaystyle \text{and hence }2(5^{n}+6^{n})\text{ always ends with }2\text{ when }n\text{ is a natural number.}
\\

\displaystyle \textbf{Question 19. }\text{The LCM of two numbers is }2400.\text{ Which of the following} \\ \text{CANNOT be their HCF?} \hspace{0.2cm}\text{[CBSE 2021]}
\displaystyle \text{(a) }300 \qquad \text{(b) }400
\displaystyle \text{(c) }500 \qquad \text{(d) }600
\displaystyle \text{Answer:}
\displaystyle  (c)\ \text{As, HCF of }2\text{ numbers is always a factor of the LCM of }2\text{ numbers.}
\displaystyle 500\text{ is not a factor of }2400.
\displaystyle \text{So it can't be the HCF.}
\\

\displaystyle \textbf{Question 20. }\text{The total number of factors of a prime number is} \hspace{0.2cm}\text{[CBSE 2020]}
\displaystyle \text{(a) }1 \qquad \text{(b) }0
\displaystyle \text{(c) }2 \qquad \text{(d) }3
\displaystyle \text{Answer:}
\displaystyle  (c)\ \text{Total number of factors of a prime number is }2.
\\

\displaystyle \textbf{Question 21. }\text{The HCF and the LCM of }12,\,21,\,15\text{ respectively are} \hspace{0.2cm}\text{[CBSE 2020]}
\displaystyle \text{(a) }3,\,140 \qquad \text{(b) }12,\,420
\displaystyle \text{(c) }3,\,420 \qquad \text{(d) }420,\,3
\displaystyle \text{Answer:}
\displaystyle  (c)\ \text{As }12=2^{2}\times3;\ 21=3\times7;\ 15=3\times5
\displaystyle \text{HCF}=3
\displaystyle \text{LCM}=2^{2}\times3\times7\times5=420
\\

\displaystyle \textbf{Question 22. }\text{Two positive integers }a\text{ and }b\text{ can be written as }a=x^{3}y^{2}\text{ and }
\displaystyle b=xy^{3}; \ x,y\text{ are prime numbers, then find the }\mathrm{LCM}(a,b). \hspace{0.2cm}\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle  \ a=x^{3}y^{2}\text{ and }b=xy^{3}
\displaystyle \text{HCF}=xy^{2}
\displaystyle \text{LCM}=\frac{a\times b}{\text{HCF}}=\frac{x^{3}y^{2}\times xy^{3}}{xy^{2}}=x^{3}y^{3}
\\

\displaystyle \textbf{Question 23. }\text{Prove that }\frac{2+\sqrt{3}}{5}\text{ is an irrational number, given that }\sqrt{3} \\ \text{is an irrational number.} \hspace{0.2cm}\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle  \text{It is given that }\sqrt{3}\text{ is an irrational number.}
\displaystyle \text{Let }\frac{2+\sqrt{3}}{5}\text{ be a rational number.}
\displaystyle \therefore \frac{2+\sqrt{3}}{5}\text{ can be written as }\frac{a}{b},\ \text{where }a\text{ and }b\text{ are coprime}
\displaystyle \text{and }b\neq0.
\displaystyle \frac{2+\sqrt{3}}{5}=\frac{a}{b}
\displaystyle \Rightarrow \sqrt{3}=\frac{5a-2b}{b}
\displaystyle \text{Here, RHS represents a rational number but LHS represents }\sqrt{3},
\displaystyle \text{which is an irrational number (given).}
\displaystyle \text{This gives a contradiction.}
\displaystyle \therefore \text{Our assumption is wrong.}
\displaystyle \therefore \frac{2+\sqrt{3}}{5}\text{ is an irrational number.}
\\

\displaystyle \textbf{Question 24. }\text{Find HCF and LCM of }404\text{ and }96\text{ and verify that} \\ \text{HCF }\times\text{ LCM }=\text{ Product of the two given numbers.} \hspace{0.2cm}\text{[CBSE 2018]}
\displaystyle \text{Answer:}
\displaystyle  \text{The prime factorisation of }404\text{ and }96:
\displaystyle 404=2\times2\times101
\displaystyle 96=2\times2\times2\times2\times2\times3
\displaystyle \text{HCF}=2\times2=4
\displaystyle \text{LCM of }404\text{ and }96=2\times2\times101\times2\times2\times2\times3=9696
\displaystyle \text{Verification:}
\displaystyle \text{HCF}\times\text{LCM}=4\times9696=38784
\displaystyle \text{Product of two numbers}=404\times96=38784
\displaystyle \text{Clearly, HCF}\times\text{LCM}=\text{Product of two numbers.}
\displaystyle \therefore \text{Hence verified.}
\\

\displaystyle \textbf{Question 25. }\text{Two tankers contain }850\text{ litres and }680\text{ litres of petrol respectively.}
\displaystyle \text{Find the maximum capacity of a container which can measure the petrol of} \\ \text{either tanker in exact number of times.} \hspace{0.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle  \text{Maximum capacity of a container, which can measure the petrol in exact number of times}
\displaystyle =\text{HCF of }850\text{ and }680
\displaystyle 850=2\times5\times5\times17
\displaystyle 680=2\times2\times2\times5\times17
\displaystyle \text{HCF of }850\text{ and }680=2\times5\times17=170\text{ litres}
\\

\displaystyle \textbf{Question 26. }\text{Explain whether the number }3\times 5\times 13\times 46+23\text{ is a prime number} \\ \text{or a composite number.} \hspace{0.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle  \text{We have}
\displaystyle 3\times5\times13\times46+23
\displaystyle =23\times(3\times5\times13\times2)+23
\displaystyle =23\times390+23
\displaystyle =23\times391
\displaystyle \text{It is clear that the above number is a multiple of }23.
\displaystyle \text{Therefore, the number }(3\times5\times13\times46+23)\text{ is not a prime number.}
\\

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