Q.1. What do you mean by Euclid's division lemma?
Solution: For any two given positive integers a and b there exist unique whole numbers q and r such that
a = bq + r, when 0 ≤ r < b.
Here, we call 'a' as a dividend, b as a divisor, q as the quotient and r as the remainder.
Dividend = (Divisor × Quotient) + Remainder
Q.2. A number when divided by 61 gives 27 as quotient and 32 as remainder. Find the number.
Solution: Using Euclid’s division Lemma
Dividend = (Divisor × Quotient) + Remainder
= (61 × 27) + 32
= 1647 + 32
= 1679
Required number = 1679
Q.3. By what number should 1365 be divided to get 31 as the quotient and 32 as the remainder?
Solution: By Euclid's Division Algorithm, we have:
Dividend = (Divisor X Quotient) + Remainder
⇒ 1365 = (Divisor × 31) + 32
⇒ (1365 - 32) / 31 = Divisor
⇒ 1333 / 31 = Divisor
Therefore, Divisor = 43
Q.4. Using Euclid's algorithm, find the HCF of:
(i) 405 and 2520
(ii) 504 and 1188
(iii) 960 and 1575
Solution:
(i) On dividing 2520 by 405, we get
(ii) Dividing 1188 by 504, we get
(iii) Dividing 1575 by 960, we get
When a = 2q + 1 for some integer q, then dearly a is odd.
Thus, every positive integer is either even or odd.
Q.6. Show that any positive odd integer is of the form (6m + 1), (6m + 3) or (6m + 5), where m is some integer.
Solution: Let a be a given positive integer.
On dividing a by 6, let q be the quotient and r be the remainder.
Then, by Eudid's algorithm, we have
a = 6q + r, where 0 < r < 6
⇒ a = 6q + r, where r = 0,1,2,3,4,5,6
⇒ a = 6q or a = 6q + 1 or a = 6q + 2 or a = 6q + 3 or a = 6q + 4 or a = 6q + 5
But, a = 6q, a = 6q + 2, a = 6q + 4 gives even values of a.
Thus, when a is odd, it is of the form
6q + 1, 6q + 3 or 6q + 5 for some integer q.
Q.7. Show that any positive odd integer is of the form (4m + 1) or (4m + 3), wherein is some integer.
Solution:
Let a be a given positive odd integer.
On dividing by 4, let q be the quotient and r be the remainder.
Then, by Eudid's algorithm, we have
a = 4m + r, where 0 < r < 4
⇒ a = 4m + r, where r = 0,1,2,3
⇒ a = 4m or a = 4m + 1 or a = 4m + 2 or a = 4m + 3
But, a = 4m and a = 4m + 2 = 2(2m + 1) are clearly even.
Thus, when a is odd, it is of the form a = (4m + 1)or (4m + 3) for some integer m.
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