X class ncert solutions chpater -1 exercise1_1

EXERCISE 1.1

1. Use Euclid’s division algorithm to find the HCF of :

(i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255

(i)

Here 225 > 135 we always divide greater number with smaller one.

Divide 225 by 135 we get 1 quotient and 90 as remainder so that

225= 135*1 + 90

Divide 135 by 90 we get 1 quotient and 45 as remainder so that

135= 90*1 + 45

Divide 90 by 45 we get 2 quotient and no remainder so we can write it as

90 = 2*45+ 0

As there are no remainder so deviser 45 is our HCF

(ii)

38220>196 we always divide greater number with smaller one.

Divide 38220 by 196 then we get quotient 195 and no remainder so we can write it as

38220 = 196 * 195 + 0

As there is no remainder so deviser 196 is our HCF

(iii)

           
867>255  we always divide greater number with smaller one.

divide 867 by 255 then we get quotient 3 and remainder is 102

so we can write it as

867 = 255 * 3 + 102

Divide 255 by 102 then we get quotient 2 and remainder is 51

So we can write it as

255 = 102 * 2 + 51

Divide 102 by 51 we get quotient 2 and no remainder

So we can write it as

102 = 51*2+ 0

As there is no remainder so deviser 51 is our answer

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

Let take a as any positive integer and b = 6.

Then using Euclid’s algorithm we get  a = 6q + r  here r is remainder and value of q is more 

than or equal to 0  and r = 0, 1, 2, 3, 4, 5 because 0 ≤r < b  and the value of b is 6 

So total possible forms  will  6q+0 , 6q+1 , 6q+2,6q+3,6q+4,6q+5 

6q+0   6 is divisible by 2 so it is a even number

6q+1   6 is divisible by 2 but 1 is not divisible by 2 so it is a odd number

6q+2  6 is divisible by 2   and 2 is also divisible by 2 so it is a even number

6q+3  6 is divisible by 2 but 3 is not divisible by 2 so it is a odd number

6q+4  6 is divisible by 2 and 4 is also divisible by 2 it is a even number

6q+5  6 is divisible by 2 but 5 is not divisible by 2 so it is a odd number

So odd numbers will in form of 6q + 1, or 6q + 3, or 6q + 5

3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

To get the maximum number column here we always find HCF and for minimum number we find LCM

So can use Euclid’s algorithm to find the HCF.

Here 616> 32 so always divide greater number with smaller one

When we divide 616 by 32 we get quotient 19 and remainder 8

So we can write it as

616 = 32 * 19 + 8

Now divide 32 by 8 we get quotient 4 and no remainder

So we can write it as

32 = 8 * 4 + 0

As there are no remainder so our HCF will 8

So that maximum number of columns in which they can march is 8.

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.

Let take a as any positive integer and b = 3.

Then using Euclid’s algorithm we get  a = 3q + r  here r is remainder and value of q is more 

than or equal to 0  and r = 0, 1, 2 because 0 ≤r < b  and the value of b is 3 So our possible 

values will 3q+0 , 3q+1 and 3q+2

Now find the square of values 

Use the formula (a+b)^2 = a^2 + 2ab +b^2 to open the square bracket 

(3q)^2             = 9q^2   if we divide by 3 we get no remainder  

we can write it as 3*(3q^2)  so it is in form of 3m  here m = 3q^2

(3q+1)^2         = (3q)^2 + 2*3q*1  + 1^2       

                        =9q^2 + 6q +1 now divide by 3 we get 1 remainder

so we can write it as 3(3q^2 + 2q) +1 so we can write it in form of 

3m+1 and value of m is 3q^2 + 2q  here

(3q+2)^2         = (3q)^2 + 2*3q*2  + 2^2 

                        =9q^2 + 12q +4  now divide by 3 we get 1 remainder

so we can write it as 3(3q^2 + 4q +1) +1 so we can write it in form 

of 3m +1 and value of m will 3q^2 + 4q +1

Square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

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.

According to Euclid’s Division Lemma

Let take a as any positive integer and b = 9. 

Then using Euclid’s algorithm we get  a = 9q + r  here r is remainder and value of q is more 

than or equal to 0  and r = 0, 1, 2, 3, 4, 5 , 6 , 7 , 8, because 0 ≤r < b  and the value of b is 9

Sp possible forms will  9q, 9q+1, 9q+2,9q+3,9q+4,9q+5,9q+6,9q+7 and 9q+8

to get the cube of these values use the formula

(a+b)^3           = a^3 + 3a^2b+ 3ab^2 + b^3

      In this formula value of a is always 9q

So plug the value we get

(9q+b)^3          = 729q^3 + 243q^2b + 27qb^2 + b^3

Now divide by 9 we get quotient = 81q^3 + 27q^2b + 3qb^2  and remainder is b^3

So we have to consider the value of b^3

b = 0                     we get 9m+0  = 9m

b = 1                     then  1^3  = 1 so we get  9m +1

b = 2                     then 2^3 = 8 so we get 9m + 8

b = 3                     then 3^3 = 27  and it is divisible by 9  so we get 9m

b = 4                     then 4^3 =64 divide by 9 we get 1 as remainder so we get 9m +1

b=5                       then 5^3=125 divide by 9 we get 8 as remainder so we get 9m+8

b=6                       then 6^3=216 divide by 9 no remainder there so we get 9m

b=7                       then 7^3 = 343 divide by 9 we get 1 as remainder so we get 9m+1

b=8                       then 8^3 = 512 divide by 9 we get 8 as remainder so we get 9m+8

So all values are in form of 9m , 9m+1 or 9m+8