NCERT Class 10 Solutions: Real Numbers (Chapter 1) Exercise 1.1 – Part 2

Q-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?

Solution:

For the above problem , the maximum number of columns would be the HCF of 616 and 32

We can find the HCF of 616 and 32 by using Euclid Division algorithm.

Therefore,

616=19×32+8

Since remainder 80 , we apply the division lemma to 32 and 8 to obtain

32=4×8+0

Therefore HCF(616,32)=HCF(32,8)=8

Therefore, they can march in 8 columns each.

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

[Hint: Let x be any positive integer then it is of the form 3q,3q+1or3q+2 . Now square each of these and show that they can be rewritten in the form 3mor3m+1 .]

Solution:

Figure explaining two cases of euclids divsion

Euclids Divsion Cases

Figure explaining two cases of euclids divsion

According to Euclid algorithm

We have a=bq+r (Equation 1)

And substituting b=3 in equation 1, we get

a=3q+r, {r0<3}

r=0,1,2

When r=0,a=3q or a2=9q2 (Equation A)

When r=1,a=3q+1 or a2=9q2+1+6q (Equation B)

When r=2,a=3q+2 or a2=9q2+4+12q (Equation C)

We can be rewrite equation A as a2=3(3q2) say 3m

Where, m=3q2

Also equation B can be written as a2=3(3q2+2q)+1 or a2=3m+1

Where, m=3q2+2q

Also equation C can be written as a2=3(3q2+4q+1)+1 or a2=3m+1

Where, m=3q2+4q+1

Hence, the square of any positive integer is either of the form 3mor3m+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+1or9m+8 .

Solution:

We know that by using Euclid’s Division Algorithm, a=bq+r

Substituting b=9 , we get

a=9q+r Where, {r0<9}

r=0,1,2,3,4,5,6,7,8

When r=0,a=9q

a3=729q3=9(81q3)=9m , where m=81q3

When r=1,a=9q+1

a3=729q3+243q2+27q+1

=9(81q3+27q2+3q)+1=9m+1 , where m=81q3+27q2+3q

When r=4,a=9q+4

a3=729q3+972q2+432q+64

=9(81q3+108q2+48q+7)+1

Continuing the process till r=8anda=9q+8 , we get

a3=729q3+1944q2+1728q+512

=9(81q3+216q2+192q+56)+8

=9m+8

Where, m=81q3+216q2+192q+56

Hence, it is proved that any positive integer is either of the form 9m,9m+1or9m+8

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