10th Number system solved important question

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.