Learn › Class 10 CBSE · Mathematics ›
Real Numbers › Fundamental Theorem of Arithmetic
Fundamental Theorem of Arithmetic — Class 10 CBSE Mathematics MCQs with Solutions
Free Class 10 CBSE Mathematics Fundamental Theorem of Arithmetic MCQs with step-by-step solutions (8 questions). Part of Real Numbers. Practise online on Prepizo — no login needed.
▶ Practise Fundamental Theorem of Arithmetic online (free)
Questions with solutions
Q1 — Fundamental Theorem of Arithmetic · easy · theory
The prime factorisation of $140$ is:
A. $2 \times 5 \times 7$
B. $2^2 \times 5 \times 7$ ✓ Correct
C. $2^2 \times 5^2 \times 7$
D. $2 \times 5 \times 7^2$
Solution: Divide by primes: $140 = 2 \times 70 = 2 \times 2 \times 35 = 2 \times 2 \times 5 \times 7$. Hence $140 = 2^2 \times 5 \times 7$. Multiplying back, $4 \times 5 \times 7 = 140$, which confirms the factorisation.
Q2 — Fundamental Theorem of Arithmetic · easy · theory
If the HCF of two numbers is $12$ and their product is $1800$, then their LCM is:
A. $120$
B. $1800$
C. $12$
D. $150$ ✓ Correct
Solution: For any two positive integers, HCF $\times$ LCM $=$ product of the two numbers. Therefore LCM $= \frac{1800}{12} = 150$.
Q3 — Fundamental Theorem of Arithmetic · easy · theory
The LCM of $12$ and $18$ is:
A. $36$ ✓ Correct
B. $6$
C. $72$
D. $216$
Solution: Write $12 = 2^2 \times 3$ and $18 = 2 \times 3^2$. The LCM is the product of the highest powers of all primes involved: $2^2 \times 3^2 = 36$. Note that $6$ is the HCF, a common mix-up.
Q4 — Fundamental Theorem of Arithmetic · medium · theory
The HCF of $96$ and $404$ is:
A. $8$
B. $101$
C. $12$
D. $4$ ✓ Correct
Solution: By prime factorisation, $96 = 2^5 \times 3$ and $404 = 2^2 \times 101$. The only common prime is $2$, taken with its lowest power $2^2$. Hence HCF $= 4$, and then LCM $= \frac{96 \times 404}{4} = 9696$.
Q5 — Fundamental Theorem of Arithmetic · medium · theory
Given that $\text{LCM}(306, 657) = 22338$, the value of $\text{HCF}(306, 657)$ is:
A. $3$
B. $6$
C. $9$ ✓ Correct
D. $11$
Solution: HCF $\times$ LCM $= 306 \times 657 = 201042$, so HCF $= \frac{201042}{22338} = 9$. As a check, $306 = 2 \times 3^2 \times 17$ and $657 = 3^2 \times 73$, whose common part is $3^2 = 9$.
Q6 — Fundamental Theorem of Arithmetic · medium · theory
The number $6^n$, where $n$ is a natural number, ends with the digit $0$ for:
A. $n = 5$
B. all multiples of $5$
C. $n = 10$ only
D. no value of $n$ ✓ Correct
Solution: A number ends in $0$ only if it is divisible by $10 = 2 \times 5$. Now $6^n = (2 \times 3)^n = 2^n \times 3^n$, and by the Fundamental Theorem of Arithmetic its prime factorisation is unique, so it can never contain the prime $5$. Hence $6^n$ never ends with the digit $0$.
Q7 — Fundamental Theorem of Arithmetic · hard · theory
If $\text{HCF}(x, 18) = 2$ and $\text{LCM}(x, 18) = 36$, then $x$ equals:
A. $2$
B. $4$ ✓ Correct
C. $6$
D. $9$
Solution: Since HCF $\times$ LCM $= x \times 18$, we get $x = \frac{2 \times 36}{18} = 4$. Check: HCF$(4, 18) = 2$ and LCM$(4, 18) = 36$, so $x = 4$ satisfies both conditions; no other option does.
Q8 — Fundamental Theorem of Arithmetic · hard · theory
The smallest number which when divided by $12$, $16$ and $24$ leaves a remainder of $7$ in each case is:
A. $43$
B. $48$
C. $55$ ✓ Correct
D. $103$
Solution: The required number is $7$ more than the smallest common multiple of $12$, $16$ and $24$. Since $12 = 2^2 \times 3$, $16 = 2^4$ and $24 = 2^3 \times 3$, the LCM is $2^4 \times 3 = 48$, so the answer is $48 + 7 = 55$. ($103 = 2 \times 48 + 7$ also leaves remainder $7$, but it is not the smallest.)