Mathematics Questions and Answers – Euclid’s Division Lemma

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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

Explanation: Let a be any arbitrary number.
Then, by Euclid’s division lemma,
a = 3q + r where 0 ≤ r ≤ 3
a2 = (3q + r)2 = 9q2 + r2 + 6qr
When r = 0,
Then, a2 = 9q2 + 02 + 6q(0)
= 9q2 = 3(3q2) = 3m where m = 3q2
Now, r = 1
a2 = (3q+r)2
= 9q2 + (1)2 + 6q(1)
= 9q2 + 1 + 6q
= 3(3q2 + 2q) + 1 = 3m + 1 where m = 3q2 + 2q
When r = 2,
a2 = (3q+r)2
= 9q2 + (2)2 + 6q(2)
= 9q2 + 12q + 4
= 3(3q2) + 3(4)q + 3 × 1 + 1
= 3[(3q2) + (4)q + 1] + 1
= 3m + 1 where m = (3q2) + (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

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

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

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

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 6n, where n belongs to natural numbers, can never end with the digit
a) 0
b) 1
c) 2
d) 3

Explanation: If 6n 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, 6n 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

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