Extra Questions For Class 10 Maths Chapter 1 Real Numbers Based On Euclid’s Division Lemma

Extra Questions On Euclid’s Division Lemma

Given positive integers a and b, there exist unique integers q and r satisfying a = bq + r, where 0 ≤ r < b .
Here we call ‘a’ as dividend, ‘b’ as divisor, ‘q’ as quotient and ‘r’ as remainder.
Dividend = (Divisor x Quotient) + Remainder
If in Euclid’s lemma r = 0 then b would be HCF of ‘a’ and ‘b’.

Question: Show that any positive even integer is of the form 6q, or 6q + 2, or 6q + 4, where q is some
integer.
Solution: Let x be any positive integer such that x > 6. Then, by Euclid’s algorithm,
x = 6q + r for some integer q ≥ 0 and 0 ≤ r < 6.
Therefore, x = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5
Now, 6q is an even integer being a multiple of 2.
We know that the sum of two even integers are always even integers.
Therefore, 6q + 2 and 6q + 4 are even integers
Hence any positive even integer is of the form 6q, or 6q + 2, or 6q + 4, where q is some integer.

Q.1. Show that any positive even integer is of the form 4q, or 4q + 2, where q is some integer.

Q.2. Show that any positive odd integer is of the form 4q + 1, or 4q + 3, where q is some integer.

Q.3. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

Q.4. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Q.5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m+1 or 9m+8.

Q.6. Use Euclid’s division lemma to show that the square of an odd positive integer can be of the form 6q + 1 or 6q + 3 for some integer q.

Q.7. Use Euclid’s division lemma to prove that one and only one out of n, n + 2 and n + 4 is divisible by 3, where n is any positive integer.

Q.8. Use Euclid’s division lemma to show that the square of an odd positive integer is of the form 8m+1, for some whole number m.

Q.9. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 4q or 4q + 1 for some integer q.

Q.10. Use Euclid’s division lemma to show that cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3, for some integer m.

Q.11. Use Euclid’s division lemma to show that the square of any positive integer cannot be of the form 5q + 2 or 5q + 3 for any integer q.

Q.12. Use Euclid’s division lemma to show that the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.

Q.13. Use Euclid’s division lemma to show that the square of any odd integer is of the form 4q + 1, for some integer q.

Q.14. If n is an odd integer, then use Euclid’s division lemma to show that n2 – 1 is divisible by 8.

Q.15. Use Euclid’s division lemma to prove that if x and y are both odd positive integers, then x2 + y2 is even but not divisible by 4.

Q.16. Use Euclid’s division lemma to prove that one of any three consecutive positive integers must be divisible by 3.

Q.17. Use Euclid’s division lemma to show that the product of three consecutive natural numbers is divisible by 6.

Q.18. For any positive integer n, use Euclid’s division lemma to prove that n3 – n is divisible by 6.

Q.19. Use Euclid’s division lemma to show that one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5, where n is any positive integer.

Frequently Asked Questions – FAQs

What is Euclid’s division lemma?

Given positive integers a and b, there exist unique integers q and r satisfying a = bq + r, where 0 ≤ r < b .
Here we call ‘a’ as dividend, ‘b’ as divisor, ‘q’ as quotient and ‘r’ as remainder.

Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m

Let x be any positive integer and y = 3.

By Euclid’s division algorithm;

x =3q + r (for some integer q ≥ 0 and r = 0, 1, 2 as r ≥ 0 and r < 3)

Therefore,

x = 3q, 3q + 1 and 3q + 2

As per the given question, if we take the square on both the sides, we get;

x2 = (3q)2 = 9q2 = 3.3q2

Let 3q2 = m

Therefore,

x2 = 3m ………………….(1)

x2 = (3q + 1)2

= (3q)2 + 12 + 2 × 3q × 1

= 9q2 + 1 + 6q

= 3(3q2 + 2q) + 1

Substitute, 3q2+2q = m, to get,

x2 = 3m + 1 ……………………………. (2)

x2 = (3q + 2)2

= (3q)+ 2+ 2 × 3q × 2

= 9q2 + 4 + 12q

= 3(3q2 + 4q + 1) + 1

Again, substitute, 3q+ 4q + 1 = m, to get,

x2 = 3m + 1…………………………… (3)

Hence, from equation 1, 2 and 3, we conclude that, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

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