CBSE Maths › Class 10 PYQs › Real Numbers

Real Numbers: 2 marks Questions (CBSE Class 10)

48 different 2 marks questions on Real Numbers from CBSE Class 10 Maths board exams 2022–2026, newest first.

1 mark (104)2 marks (48)3 marks (49)4 marks (2)

Q222 marksVery Short AnswerReal Numbers CBSE 2026 · Basic 430/5/1

Find the H.C.F. and L.C.M. of 1530 and 2040.

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Answer: H.C.F. = 510, L.C.M. = 6120

$1530 = 2 \times 3^{2} \times 5 \times 17$
$2040 = 2^{3} \times 3 \times 5 \times 17$
H.C.F. $= 2 \times 3 \times 5 \times 17 = 510$
L.C.M. $= 2^{3} \times 3^{2} \times 5 \times 17 = 6120$

Q242 marksVery Short AnswerReal Numbers CBSE 2026 · Basic 430/5/2

If H.C.F. of $135 x^{2}$ and $189 x^{3}$ is 108, then find the value of $x$ .

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Answer:

x = 2

$135 = 3^{3} \times 5$ and $189 = 3^{3} \times 7$ , so H.C.F. of $135 x^{2}$ and $189 x^{3}$ is $27 x^{2}$ .
$27 x^{2} = 108 \Rightarrow x^{2} = 4 \Rightarrow x = 2$ (taking $x$ positive).
Check: $540 = 2^{2} \times 3^{3} \times 5$ and $1512 = 2^{3} \times 3^{3} \times 7$ have H.C.F. $2^{2} \times 3^{3} = 108$ .

Q232 marksVery Short AnswerReal Numbers CBSE 2026 · Basic 430/5/3

Find the H.C.F. and L.C.M. of 408 and 312.

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Answer: H.C.F. = 24, L.C.M. = 5304

$408 = 2^{3} \times 3 \times 17$
$312 = 2^{3} \times 3 \times 13$
H.C.F. $= 2^{3} \times 3 = 24$
L.C.M. $= 2^{3} \times 3 \times 13 \times 17 = 5304$

Q212 marksVery Short AnswerReal Numbers CBSE 2026 · Standard 30/3/1

Find the length of the plank that can be used to measure the lengths 4 m 20 cm and 5 m 4 cm exactly, in the least time.

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Answer: 84 cm

Lengths: 420 cm and 504 cm; the required length is their HCF
$420 = 2^{2} \times 3 \times 5 \times 7$
$504 = 2^{3} \times 3^{2} \times 7$
HCF $= 2^{2} \times 3 \times 7 = 84$
Length of plank $= 84$ cm

Also asked in: 2026 Standard 30/3/2, 2026 Standard 30/3/3

Q242 marksVery Short AnswerReal Numbers CBSE 2026 · Standard 30/4/1

Prove that $2 + 35$ is an irrational number given that $5$ is irrational number.

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Answer: Proved.

Suppose $2 + 35 = r$ , where r is rational.
Then $5 = \frac{r - 2}{3}$ , which is rational.
This contradicts that $5$ is irrational.
Hence $2 + 35$ is irrational.

Also asked in: 2026 Standard 30/4/2, 2026 Standard 30/4/3

Q24 (OR) (OR)2 marksVery Short AnswerReal Numbers CBSE 2026 · Standard 30/4/1

If the HCF of 210 and 55 is expressed as $210 \times 5 + 55 m$ , then find the value of m.

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Answer:

m = - 19

$210 = 2 \times 3 \times 5 \times 7$ , $55 = 5 \times 11$ , so HCF = 5.
$5 = 210 \times 5 + 55 m = 1050 + 55 m$
$55 m = - 1045 \Rightarrow m = - 19$

Also asked in: 2026 Standard 30/4/2, 2026 Standard 30/4/3

Q212 marksVery Short AnswerReal Numbers CBSE 2026 · Standard 30/5/1

Prove that $2 - 53$ is an irrational number given that $3$ is irrational.

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Answer: Proved.

Suppose $2 - 53$ is rational. Let $2 - 53 = r$ , where r is rational.
Then $53 = 2 - r$ , so $3 = \frac{2 - r}{5}$ .
Since r is rational, $\frac{2 - r}{5}$ is rational, so $3$ would be rational.
This contradicts the fact that $3$ is irrational.
Hence $2 - 53$ is irrational.

Q232 marksVery Short AnswerReal Numbers CBSE 2026 · Standard 30/5/2

Prove that $4 - 25$ is an irrational number given that $5$ is irrational.

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Answer: Proved.

Suppose $4 - 25$ is rational. Let $4 - 25 = r$ , where r is rational.
Then $25 = 4 - r$ , so $5 = \frac{4 - r}{2}$ .
Since r is rational, $\frac{4 - r}{2}$ is rational, so $5$ would be rational.
This contradicts the fact that $5$ is irrational.
Hence $4 - 25$ is irrational.

Q252 marksVery Short AnswerReal Numbers CBSE 2026 · Standard 30/5/3

Prove that $14 - 23$ is an irrational number, given that $3$ is irrational.

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Answer: Proved.

Suppose $14 - 23$ is rational. Let $14 - 23 = r$ , where r is rational.
Then $23 = 14 - r$ , so $3 = \frac{14 - r}{2}$ .
Since r is rational, $\frac{14 - r}{2}$ is rational, so $3$ would be rational.
This contradicts the fact that $3$ is irrational.
Hence $14 - 23$ is irrational.

Q212 marksVery Short AnswerReal Numbers CBSE 2025 · Basic 430/4/1

Show that $4 5^{n}$ can not end with the digit 0, $n$ being a natural number. Write the prime number ‘ $a$ ’ which on multiplying with $4 5^{n}$ makes the product end with the digit 0.

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Answer:

4 5^{n} = 3^{2 n} \times 5^{n}

has no factor 2, so it cannot end with 0;

a = 2

A number ends with 0 only if its prime factorisation contains both 2 and 5.
$4 5^{n} = (3^{2} \times 5)^{n} = 3^{2 n} \times 5^{n}$ , which has no factor 2.
By the uniqueness of prime factorisation, $4 5^{n}$ can never end with the digit 0.
Multiplying by the prime 2 gives $2 \times 3^{2 n} \times 5^{n}$ , which has both 2 and 5 and ends with 0. So $a = 2$ .

Also asked in: 2025 Basic 430/4/2, 2025 Basic 430/4/3

Q212 marksVery Short AnswerReal Numbers CBSE 2025 · Basic 430/5/1

Check whether $1 5^{n} \times 2^{n}$ , n being a natural number, ends with the digit zero.

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Answer: Yes,

1 5^{n} \times 2^{n} = 3 0^{n}

always ends with the digit zero.

$1 5^{n} \times 2^{n} = (3 \times 5)^{n} \times 2^{n} = 2^{n} \times 3^{n} \times 5^{n}$ .
A number ends with 0 if its prime factorisation contains both 2 and 5.
For every natural number n, both 2 and 5 occur, so $1 5^{n} \times 2^{n} = 3 0^{n} = 1 0^{n} \times 3^{n}$ ends with zero.

Also asked in: 2025 Basic 430/5/3

Q232 marksVery Short AnswerReal Numbers CBSE 2025 · Basic 430/5/2

Prove that, for a natural number n, $6^{n}$ can not end with the digit 0. Which prime number must be multiplied with $6^{n}$ so that the resultant ends with the digit zero ?

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Answer: Proved; the prime number is 5.

A number ends with the digit 0 only if it is divisible by 10, i.e. its prime factorisation contains both 2 and 5.
$6^{n} = (2 \times 3)^{n} = 2^{n} \times 3^{n}$ , which has no factor 5.
By the uniqueness of prime factorisation (Fundamental Theorem of Arithmetic), $6^{n}$ has no other prime factors, so $6^{n}$ cannot end with 0.
Multiplying by the prime 5 gives $2^{n} \times 3^{n} \times 5$ , which is divisible by 10, so it ends with 0. The required prime is 5.

Q252 marksVery Short AnswerReal Numbers CBSE 2025 · Basic 430/6/1

Using prime factorisation, find the HCF of 180, 140 and 210.

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Answer: HCF = 10

$180 = 2^{2} \times 3^{2} \times 5$
$140 = 2^{2} \times 5 \times 7$
$210 = 2 \times 3 \times 5 \times 7$
HCF = product of the smallest powers of common primes $= 2 \times 5 = 10$ .

Also asked in: 2025 Basic 430/6/3

Q232 marksVery Short AnswerReal Numbers CBSE 2025 · Basic 430/6/2

Using prime factorisation, find the HCF of 144, 180 and 192.

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Answer: HCF = 12

$144 = 2^{4} \times 3^{2}$
$180 = 2^{2} \times 3^{2} \times 5$
$192 = 2^{6} \times 3$
HCF = product of the smallest powers of common primes $= 2^{2} \times 3 = 12$ .

Q232 marksVery Short AnswerReal Numbers CBSE 2025 · Standard 30/3/1

Find the smallest number which is divisible by both 644 and 462.

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Answer: 21252

$644 = 2^{2} \times 7 \times 23$ .
$462 = 2 \times 3 \times 7 \times 11$ .
Required number $= LCM = 2^{2} \times 3 \times 7 \times 11 \times 23 = 21252$ .

Also asked in: 2025 Standard 30/3/2, 2025 Standard 30/3/3

Q23 (OR) (OR)2 marksVery Short AnswerReal Numbers CBSE 2025 · Standard 30/3/1

Two numbers are in the ratio $4 : 5$ and their HCF is 11. Find the LCM of these numbers.

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Answer: 220

The numbers are $4 \times 11 = 44$ and $5 \times 11 = 55$ (4 and 5 are co-prime).
$44 = 2^{2} \times 11$ , $55 = 5 \times 11$ .
LCM $= 2^{2} \times 5 \times 11 = 220$ .

Also asked in: 2025 Standard 30/3/2, 2025 Standard 30/3/3

Q212 marksVery Short AnswerReal Numbers CBSE 2024 · Basic 430/1/2

Prove that $6 - 45$ is an irrational number, given that $5$ is an irrational number.

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Answer: Proved.

Suppose $6 - 45$ is rational, say $6 - 45 = r$ where $r$ is rational.
Then $5 = \frac{6 - r}{4}$ .
The right side is rational, since rationals are closed under subtraction and division by a non-zero rational.
This contradicts the fact that $5$ is irrational.
Hence $6 - 45$ is irrational.

Also asked in: 2024 Basic 430/1/3

Q21 (OR) (OR)2 marksVery Short AnswerReal Numbers CBSE 2024 · Basic 430/1/2

Show that $11 \times 19 \times 23 + 3 \times 11$ is not a prime number.

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Answer: Proved; the number is

11 \times 440 = 4840

, which is composite.

$11 \times 19 \times 23 + 3 \times 11 = 11 (19 \times 23 + 3)$ .
$= 11 (437 + 3) = 11 \times 440$ .
It has factors other than 1 and itself (for example 11), so it is not a prime number.

Also asked in: 2024 Basic 430/1/3

Q212 marksVery Short AnswerReal Numbers CBSE 2024 · Basic 430/2/1

Find the HCF of 84 and 144 by prime factorisation method.

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Answer: HCF = 12

$84 = 2^{2} \times 3 \times 7$
$144 = 2^{4} \times 3^{2}$
HCF = product of smallest powers of common primes $= 2^{2} \times 3 = 12$

Also asked in: 2024 Basic 430/2/2

Q232 marksVery Short AnswerReal Numbers CBSE 2024 · Basic 430/2/3

Find the LCM of 231 and 396 by prime factorisation method.

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Answer: LCM = 2772

$231 = 3 \times 7 \times 11$
$396 = 2^{2} \times 3^{2} \times 11$
LCM = product of greatest powers of all primes $= 2^{2} \times 3^{2} \times 7 \times 11 = 2772$

Q232 marksVery Short AnswerReal Numbers CBSE 2024 · Basic 430/3/1

Prove that $- 7 - 23$ is an irrational number, given that $3$ is an irrational number.

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Answer: Proved.

Suppose $- 7 - 23$ is rational, say $- 7 - 23 = r$ where $r$ is rational.
Then $3 = \frac{- 7 - r}{2}$ .
The right side is rational because rationals are closed under subtraction and division by a non-zero number.
This makes $3$ rational, which contradicts the given fact.
Hence $- 7 - 23$ is irrational.

Also asked in: 2024 Basic 430/3/2, 2024 Basic 430/3/3

Q23 (OR) (OR)2 marksVery Short AnswerReal Numbers CBSE 2024 · Basic 430/3/1

Explain why $(7 \times 11 \times 13 + 2 \times 11)$ is not a prime number.

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Answer: It equals

11 \times 93 = 1023

, which has factors other than 1 and itself, so it is composite.

$7 \times 11 \times 13 + 2 \times 11 = 11 (7 \times 13 + 2)$
$= 11 \times 93 = 1023$ .
It has factors 11 and 93 besides 1 and itself, so it is a composite number, not prime.

Also asked in: 2024 Basic 430/3/2, 2024 Basic 430/3/3

Q212 marksVery Short AnswerReal Numbers CBSE 2024 · Basic 430/4/1

Given that HCF (306, 1314) = 18, find LCM of (306, 1314).

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Answer: LCM (306, 1314) = 22338

HCF $\times$ LCM $=$ product of the two numbers.
LCM $= \frac{306 \times 1314}{18} = 17 \times 1314 = 22338$ .

Also asked in: 2024 Basic 430/4/3

Q232 marksVery Short AnswerReal Numbers CBSE 2024 · Basic 430/4/2

Find the HCF of 45, 54, 270 using prime factorization method.

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Answer: HCF = 9

$45 = 3^{2} \times 5$
$54 = 2 \times 3^{3}$
$270 = 2 \times 3^{3} \times 5$
HCF = product of the smallest powers of common primes $= 3^{2} = 9$ .

Q252 marksVery Short AnswerReal Numbers CBSE 2024 · Standard 30/1/1

Prove that $5 - 23$ is an irrational number. It is given that $3$ is an irrational number.

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Answer: Proved.

Assume $5 - 23$ is rational, say $5 - 23 = r$ , where $r$ is rational.
Then $3 = \frac{5 - r}{2}$ .
The right side is rational (rational numbers are closed under subtraction and division by a non-zero rational), so $3$ would be rational.
This contradicts the fact that $3$ is irrational.
Hence $5 - 23$ is irrational.

Also asked in: 2024 Standard 30/1/2, 2024 Standard 30/1/3

Q25 (OR) (OR)2 marksVery Short AnswerReal Numbers CBSE 2024 · Standard 30/1/1

Show that the number $5 \times 11 \times 17 + 3 \times 11$ is a composite number.

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Answer: It equals

11 \times 88 = 968

, which has factors other than 1 and itself, so it is composite.

$5 \times 11 \times 17 + 3 \times 11 = 11 (5 \times 17 + 3) = 11 (85 + 3) = 11 \times 88$ .
So the number has the factors 11 and 88 besides 1 and itself.
Hence it is a composite number.

Also asked in: 2024 Standard 30/1/2, 2024 Standard 30/1/3

Q212 marksVery Short AnswerReal Numbers CBSE 2024 · Standard 30/2/1

Can the number $(15)^{n}$ , n being a natural number, end with the digit 0 ? Give reasons.

Show answer & solution

Answer: No,

(15)^{n}

can never end with the digit 0.

A number ending with 0 is divisible by 10 = $2 \times 5$ , so its prime factorisation must contain both 2 and 5.
$(15)^{n} = 3^{n} \times 5^{n}$ , whose only prime factors are 3 and 5.
By the uniqueness of prime factorisation (Fundamental Theorem of Arithmetic), 2 is not a factor of $(15)^{n}$ .
So $(15)^{n}$ cannot end with the digit 0 for any natural number n.

Q232 marksVery Short AnswerReal Numbers CBSE 2024 · Standard 30/2/2

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.

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Answer: Both have factors other than 1 and themselves:

13 \times 78

and

5 \times 1009

$7 \times 11 \times 13 + 13 = 13 (7 \times 11 + 1) = 13 \times 78 = 1014$ .
It has the factor 13 besides 1 and itself, so it is composite.
$7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5 (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1) = 5 \times 1009 = 5045$ .
It has the factor 5 besides 1 and itself, so it is composite.

Q252 marksVery Short AnswerReal Numbers CBSE 2024 · Standard 30/2/3

Can the number $8^{n}$ , n being a natural number, end with the digit 0 ? Give reasons.

Show answer & solution

Answer: No,

8^{n}

can never end with the digit 0.

A number ending with 0 is divisible by 10 = $2 \times 5$ , so its prime factorisation must contain 5.
$8^{n} = 2^{3 n}$ , whose only prime factor is 2.
By the uniqueness of prime factorisation (Fundamental Theorem of Arithmetic), 5 is not a factor of $8^{n}$ .
So $8^{n}$ cannot end with the digit 0 for any natural number n.

Q212 marksVery Short AnswerReal Numbers CBSE 2024 · Standard 30/4/1

Three bells toll at intervals of 9, 12 and 15 minutes respectively. If they start tolling together, after what time will they next toll together ?

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Answer: After 180 minutes, i.e. 3 hours

$9 = 3^{2}$ , $12 = 2^{2} \times 3$ , $15 = 3 \times 5$ .
LCM = $2^{2} \times 3^{2} \times 5 = 180$ .
They toll together again after 180 minutes = 3 hours.

Also asked in: 2024 Standard 30/4/3

Q232 marksVery Short AnswerReal Numbers CBSE 2024 · Standard 30/4/2

In a school, there are two sections of class X. There are 40 students in the first section and 48 students in the second section. Determine the minimum number of books required for their class library so that they can be distributed equally among students of both sections.

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Answer: 240 books

The number of books must be a multiple of both 40 and 48; the minimum is their LCM.
$40 = 2^{3} \times 5$ , $48 = 2^{4} \times 3$ .
LCM = $2^{4} \times 3 \times 5 = 240$ .
Minimum number of books = 240.

Q232 marksVery Short AnswerReal Numbers CBSE 2023 · Basic 430/1/1

Find the LCM and HCF of 92 and 510, using prime factorisation.

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Answer: HCF = 2, LCM = 23460

$92 = 2^{2} \times 23$
$510 = 2 \times 3 \times 5 \times 17$
HCF = product of smallest powers of common primes = 2
LCM = $2^{2} \times 3 \times 5 \times 17 \times 23 = 23460$
Check: HCF $\times$ LCM $= 2 \times 23460 = 46920 = 92 \times 510$ .

Also asked in: 2023 Basic 430/1/2, 2023 Basic 430/1/3

Q222 marksVery Short AnswerReal Numbers CBSE 2023 · Basic 430/2/1

Find the HCF of the numbers 540 and 630, using prime factorization method.

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Answer: 90

$540 = 2^{2} \times 3^{3} \times 5$
$630 = 2 \times 3^{2} \times 5 \times 7$
HCF = product of the smallest powers of common primes $= 2 \times 3^{2} \times 5 = 90$

Also asked in: 2023 Basic 430/2/2, 2023 Basic 430/2/3

Q22 (OR) (OR)2 marksVery Short AnswerReal Numbers CBSE 2023 · Basic 430/2/1

Show that $(15)^{n}$ cannot end with the digit 0 for any natural number ‘n’.

Show answer & solution

Answer: Proved.

A number ending with digit 0 is divisible by 10, so its prime factorisation must contain both 2 and 5.
$(15)^{n} = (3 \times 5)^{n} = 3^{n} \times 5^{n}$
By the Fundamental Theorem of Arithmetic this factorisation is unique, and it has no factor 2.
Hence $(15)^{n}$ cannot end with the digit 0 for any natural number n.

Also asked in: 2023 Basic 430/2/2, 2023 Basic 430/2/3

Q212 marksVery Short AnswerReal Numbers CBSE 2023 · Basic 430/5/1

Find LCM of 576 and 512 by prime factorization.

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Answer: 4608

$576 = 2^{6} \times 3^{2}$ .
$512 = 2^{9}$ .
LCM $= 2^{9} \times 3^{2} = 512 \times 9 = 4608$ .

Q242 marksVery Short AnswerReal Numbers CBSE 2023 · Basic 430/5/2

Find HCF of 660 and 704 by prime factorization.

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Answer: 44

$660 = 2^{2} \times 3 \times 5 \times 11$ .
$704 = 2^{6} \times 11$ .
HCF = product of the smallest powers of common primes $= 2^{2} \times 11 = 44$ .

Q242 marksVery Short AnswerReal Numbers CBSE 2023 · Basic 430/5/3

Find LCM of 480 and 256 using prime factorization.

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Answer: 3840

$480 = 2^{5} \times 3 \times 5$ .
$256 = 2^{8}$ .
LCM $= 2^{8} \times 3 \times 5 = 256 \times 15 = 3840$ .

Q242 marksVery Short AnswerReal Numbers CBSE 2023 · Standard 30/1/1

Find the greatest number which divides 85 and 72 leaving remainders 1 and 2 respectively.

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Answer: 14

The number divides 85 – 1 = 84 and 72 – 2 = 70 exactly.
$84 = 2^{2} \times 3 \times 7$ , $70 = 2 \times 5 \times 7$ .
HCF(84, 70) $= 2 \times 7 = 14$ .
Required number = 14.

Also asked in: 2023 Standard 30/1/3

Q212 marksVery Short AnswerReal Numbers CBSE 2023 · Standard 30/1/2

Find the least number which when divided by 12, 16 and 24 leaves remainder 7 in each case.

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Answer: 55

$12 = 2^{2} \times 3$ , $16 = 2^{4}$ , $24 = 2^{3} \times 3$ .
LCM(12, 16, 24) $= 2^{4} \times 3 = 48$ .
Required number = 48 + 7 = 55.

Q212 marksVery Short AnswerReal Numbers CBSE 2023 · Standard 30/2/1

Prove that $2 + 3$ is an irrational number, given that $3$ is an irrational number.

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Answer: Proved.

Suppose $2 + 3$ is rational, say $2 + 3 = \frac{p}{q}$ with integers $p, q$ , $q \neq = 0$ .
Then $3 = \frac{p}{q} - 2 = \frac{p - 2 q}{q}$ .
The right side is rational, so $3$ would be rational.
This contradicts the fact that $3$ is irrational.
Hence $2 + 3$ is irrational.

Q252 marksVery Short AnswerReal Numbers CBSE 2023 · Standard 30/2/2

Prove that $4^{n}$ can never end with digit 0, where n is a natural number.

Show answer & solution

Answer: Proved.

A number ending with digit 0 is divisible by 10, so its prime factorisation must contain both 2 and 5.
$4^{n} = (2^{2})^{n} = 2^{2 n}$ , whose only prime factor is 2.
By the uniqueness of prime factorisation (Fundamental Theorem of Arithmetic), 5 is not a factor of $4^{n}$ .
Hence $4^{n}$ can never end with digit 0.

Q252 marksVery Short AnswerReal Numbers CBSE 2023 · Standard 30/2/3

Prove that $6 - 7$ is irrational number, given that $7$ is an irrational number.

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Answer: Proved.

Suppose $6 - 7$ is rational, say $6 - 7 = \frac{p}{q}$ with integers $p, q$ , $q \neq = 0$ .
Then $7 = 6 - \frac{p}{q} = \frac{6 q - p}{q}$ .
The right side is rational, so $7$ would be rational.
This contradicts the fact that $7$ is irrational.
Hence $6 - 7$ is irrational.

Q212 marksVery Short AnswerReal Numbers CBSE 2023 · Standard 30/4/1

Two numbers are in the ratio 2 : 3 and their LCM is 180. What is the HCF of these numbers ?

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Answer: 30

Let the numbers be $2 k$ and $3 k$ , where $k$ is their HCF (2 and 3 are co-prime).
LCM $= 6 k = 180 \Rightarrow k = 30$
HCF = 30 (the numbers are 60 and 90).

Also asked in: 2023 Standard 30/4/2, 2023 Standard 30/4/3

Q232 marksVery Short AnswerReal Numbers CBSE 2023 · Standard 30/5/1

Using prime factorisation, find HCF and LCM of 96 and 120.

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Answer: HCF = 24, LCM = 480

$96 = 2^{5} \times 3$ .
$120 = 2^{3} \times 3 \times 5$ .
HCF $= 2^{3} \times 3 = 24$ .
LCM $= 2^{5} \times 3 \times 5 = 480$ .

Also asked in: 2023 Standard 30/5/2

Q212 marksVery Short AnswerReal Numbers CBSE 2023 · Standard 30/5/3

Find the greatest 3-digit number which is divisible by 18, 24 and 36.

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Answer: 936

$18 = 2 \times 3^{2}$ , $24 = 2^{3} \times 3$ , $36 = 2^{2} \times 3^{2}$ .
LCM $= 2^{3} \times 3^{2} = 72$ ; a number divisible by all three is a multiple of 72.
$999 = 72 \times 13 + 63$ .
Greatest 3-digit multiple of 72 $= 72 \times 13 = 936$ .

Q212 marksVery Short AnswerReal Numbers CBSE 2023 · Standard 30/6/1

Show that $6^{n}$ can not end with digit 0 for any natural number 'n'.

Show answer & solution

Answer: Proved.

A number ending in 0 is divisible by 10, so its prime factorisation must contain both 2 and 5.
$6^{n} = 2^{n} \times 3^{n}$ .
By the Fundamental Theorem of Arithmetic this factorisation is unique, and it has no factor 5.
So $6^{n}$ cannot end with the digit 0 for any natural number n.

Also asked in: 2023 Standard 30/6/2, 2023 Standard 30/6/3

Q21 (OR) (OR)2 marksVery Short AnswerReal Numbers CBSE 2023 · Standard 30/6/1

Find the HCF and LCM of 72 and 120.

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Answer: HCF = 24, LCM = 360

$72 = 2^{3} \times 3^{2}$
$120 = 2^{3} \times 3 \times 5$
HCF = $2^{3} \times 3 = 24$
LCM = $2^{3} \times 3^{2} \times 5 = 360$

Q23 (OR) (OR)2 marksVery Short AnswerReal Numbers CBSE 2023 · Standard 30/6/2

Find the LCM and HCF of 72 and 120.

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Answer: LCM = 360, HCF = 24

$72 = 2^{3} \times 3^{2}$
$120 = 2^{3} \times 3 \times 5$
LCM = $2^{3} \times 3^{2} \times 5 = 360$
HCF = $2^{3} \times 3 = 24$

Also asked in: 2023 Standard 30/6/3

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