This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Euclid’s Division Lemma”.

1. For any integer m, square of the number is of the form ______ or ______

a) 3m + 3, 3m – 2

b) 3m – 2, 3m + 2

c) 3m + 2, 3m – 3

d) 3m, 3m + 1

View Answer

Explanation: Let a be any arbitrary number.

Then, by Euclid’s division lemma,

a = 3q + r where 0 ≤ r ≤ 3

a

^{2}= (3q + r)

^{2}= 9q

^{2}+ r

^{2}+ 6qr

When r = 0,

Then, a

^{2}= 9q

^{2}+ 0

^{2}+ 6q(0)

= 9q

^{2}= 3(3q

^{2}) = 3m where m = 3q

^{2}

Now, r = 1

a

^{2}= (3q+r)

^{2}

= 9q

^{2}+ (1)

^{2}+ 6q(1)

= 9q

^{2}+ 1 + 6q

= 3(3q

^{2}+ 2q) + 1 = 3m + 1 where m = 3q

^{2}+ 2q

When r = 2,

a

^{2}= (3q+r)

^{2}

= 9q

^{2}+ (2)

^{2}+ 6q(2)

= 9q

^{2}+ 12q + 4

= 3(3q

^{2}) + 3(4)q + 3 × 1 + 1

= 3[(3q

^{2}) + (4)q + 1] + 1

= 3m + 1 where m = (3q

^{2}) + (4)q + 1

Hence, square of number n is of the form 3m or 3m + 1

2. A number when divided by 60 gives 35 as quotient and leaves 81 as remainder. What is the number?

a) 2121

b) 4151

c) 2181

d) 3171

View Answer

Explanation: According to Euclid’s Division Lemma,

Number = (quotient × divisor) + remainder = 60 × 35 + 81 = 2181

3. For any two given positive integers a and b, there exists unique whole numbers q and r such that a = bq + r, where 0 ≤ r ≤ b.

a) True

b) False

View Answer

Explanation: According to Euclid’s Division lemma, any two numbers can be written in the form of a = bq + r where a and b are integers and q and r whole numbers.

For Example: 28 when divided by 7 give 4 as quotient and 0 as remainder.

So, according to Euclid’s Division Lemma,

28 = 7 × 4 + 0

4. When a number is divided by 3 it leaves remainder as 5. What will be the remainder when 3n + 3 are divided by 3?

a) 0

b) 3

c) 9

d) 6

View Answer

Explanation: Let the number be n.

If the number is divided by 3 it leaves 5 as remainder.

By Euclid’s division lemma,

n = 3q + 5 where q is quotient

3n = 3(3q+5)

3n = 9q + 15

3n + 3 = 9q + 18 = 3 × 3q + 3 × 6 = 3 (3q + 6)

Hence, the remainder is 0.

5. The HCF of 80 and 567 is ___________

a) 5

b) 4

c) 1

d) 6

View Answer

Explanation: 80 = 1×2×2×2×2×5

567 = 1×3×3×3×3×7. The largest common factor between the two numbers is 1.

6. A number in the form of 6^{n}, where n belongs to natural numbers, can never end with the digit

a) 0

b) 1

c) 2

d) 3

View Answer

Explanation: If 6

^{n}ends with 0, then it should have 5 as a factor.

In case of 6 only 3 and 2 are factors of 6.

Also, from the fundamental theorem of arithmetic, prime factorisation of each number is unique.

Hence, 6

^{n}can never end with 0.

7. If the HCF of two numbers is 1 and the LCM is 3395. What is the other number if one of them is 97?

a) 61

b) 57

c) 43

d) 35

View Answer

Explanation: For two numbers a and b, we know that

(a × b) = HCF of (a, b) × LCM of (a, b)

Here a = 97, HCF is 1 and LCM is 3395

97 × b = 1 × 3395

B = \(\frac {3395}{97}\) = 35

**Sanfoundry Global Education & Learning Series – Mathematics – Class 10**.

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