1. Prove that one of every three consecutive integers is divisible by 3.
Ans: n,n+1,n+2 be three consecutive positive integers
We know that n is of the form 3q, 3q +1, 3q + 2
So we have the following cases
Case – I when n = 3q
In the this case, n is divisible by 3 but n + 1 and n + 2 are not divisible by 3
Case - II When n = 3q + 1
Sub n = 2 = 3q +1 +2 = 3(q +1) is divisible by 3. but n and n+1 are not divisible by 3
Case – III When n = 3q +2
Sub n = 2 = 3q +1 +2 = 3(q +1) is divisible by 3. but n and n+1 are not divisible by 3
Hence one of n, n + 1 and n + 2 is divisible by 3
2. If the H C F of 657 and 963 is expressible in the form of 657x + 963x - 15 find x.
Ans: Using Euclid’s Division Lemma
a= bq+r , o £ r < b
963=657×1+306
657=306×2+45
306=45×6+36
45=36×1+9
36=9×4+0
\ HCF (657, 963) = 9
now 9 = 657x + 963× (-15)
657x=9+963×15
=9+14445
657x=14454
x=14454/657
\ x =22
3. If d is the HCF of 30, 72, find the value of x & y satisfying d = 30x + 72y.
Ans: Using Euclid’s algorithm, the HCF (30, 72)
72 = 30 × 2 + 12
30 = 12 × 2 + 6
12 = 6 × 2 + 0
HCF (30,72) = 6
6=30-12×2
6=30-(72-30×2)2
6=30-2×72+30×4
6=30×5+72×-2
\ x = 5, y = -2
Also 6 = 30 ×5 + 72 (-2) + 30 × 72 – 30 × 72
Solve it, to get
x = 77, y = -32
Hence, x and y are not unique
4. Show that for odd positive integer to be a perfect square, it should be of the form
8k +1.
Let a=2m+1
Ans: Squaring both sides we get
a2 = 4m (m +1) + 1
\ product of two consecutive numbers is always even
m(m+1)=2k
a2=4(2k)+1
a2 = 8 k + 1
Hence proved
5. Use Eculid’s division demma, to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Solution. Let x be any positive integer. Then, it is of the form 3q or, 3q + 1 or, 3q + 2.
So, we have the following cases :
Case I : When x = 3q.
then, x3 = (3q)3 = 27q3 = 9 (3q3) = 9m, where m = 3q3
Case II : When x = 3q + 1
then, x3 = (3q + 1)3
= 27q3 + 27q2 + 9q + 1
= 9 q (3q2 + 3q + 1) + 1
= 9m + 1, where m = q (3q2 + 3q + 1)
Case III. When x = 3q + 2
then, x3 = (3q + 2)3
= 27 q3 + 54q2 + 36q + 8
= 9q (3q2 + 6q + 4) + 8
= 9 m + 8, where m = q (3q2 + 6q + 4)
Hence, x3 is either of the form 9 m or 9 m + 1 or, 9 m + 8.
6. Check whether 6n can end with the digit 0 for any natural number n.
Solution. We have, 6n = (2 × 3)n = 2n × 3n. Therefore, prime factorisation of 6n does not contain 5 as a factor. Hence, 6n can never end with the digit 0 for any natural number n.
7. If p is a prime number, prove that √p is irrational.
Solution : Let p be a prime number and if possible, let √p is rational.
Let its simplest form be √p = m/n
where m and n are integers having no common factor other than 1, and n ¹ 00.
Now, Sq. both side
p = m2/n2
p n 2 = m2
p divides m2
p divides m. ...(i)
Let m = p q for some integer q.
putting m = pq in (i), we get
pn2 = p2q2
n2 = pq2
p divides n2
p divides n. ...(ii)
from (i) and (ii), we observe that p is a common factor of m and n, which contradicts our
assumption.
Hence, √p is irrational.
8. (ii) Show that √(n -1) + √(n + 1) is irrational for every natural number n.
