**Class 10, CBSE Board Mathematics**

**Chapter 1, Real Numbers**

__NCERT Mathematics Solutions __

**Maths Textbook Exercise 1.1 Solutions**

(Page 7)

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

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

Solution:

(i) We have,

a = bq + r

Applying division lemma to 225 and 135 we obtain,

225 = 135 x 1 + 90

and 135 = 90 x 1 + 45

and 90 = 45 x 2 + 0

Therefore, HCF of 225, 135 = 45

(ii) We have,

a = bq + r

Applying division lemma to 196 and 38220 we obtain,

38220 = 196 x 195 + 0

Therefore, HCF of 196 and 38220 = 196

(iii) We have,

a = bq + r

Applying division lemma to 867 and 255 we obtain,

867 = 255 x 3 + 102

255 = 102 x 2 + 51

102 = 51 x 2 + 0

Therefore, HCF of 867 and 255 is 51.

Q.2: Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or6q + 5, where *q* is some integer.

Solution:

Let *a* be any positive number and *a* = 6. Then, by Euclid’s algorithm,

*a* = 6*q* + *r *(0 ≤** ***r* < 6)

say, *r* = 0, 1, 2, 3, 4, 5

or, *a* = 6*q* or 6*q* + 1 or 6*q* + 2 or 6*q + *3 or 6*q* + 4 or 6*q* + 5

Also, 6*q* + 1 = 2 × 3*q* + 1 = 2k_{1} + 1, where* *k_{1} is a positive integer,

Similarly, 6*q* + 3 = (6*q* + 2) + 1 = 2 (3*q* + 1) + 1 = 2k_{2} + 1, where* *k_{2} is a positive integer,

and, 6*q* + 5 = (6*q* + 4) + 1 = 2 (3*q* + 2) + 1 = 2k_{3} + 1, where* *k_{3} is a positive integer.

From these we observe that 6*q* + 1, 6*q* + 3, 6*q* + 5 are of the form 2*k* + 1. So, these numbers are not divisible by 2 and hence, are odd positive integers.

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:

To find the maximum number of columns, we have to find HCF of 616 and 32.

Applying Euclid’s algorithm to find the HCF we get,

616 = 32 × 19 + 8

32 = 8 × 4 + 0

Or, the HCF (616, 32) = 8.

Therefore, maximum number of column is 8.

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

Solution:

a = bq + r;

Let *a* be any positive integer, *b* = 3 and *r* = 0, 1, 2 because 0 ≤ *r* < 3

Then *a* = 3*q* + *r* for some integer *q* ≥ 0

Therefore, *a* = 3*q* + 0 or 3*q* + 1 or 3*q* + 2

From the above we can say that the square of any positive integer is either of the form 3*m* or 3*m* + 1.

Q.5: Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9*m*, 9*m *+ 1 or 9*m + *8.

[Taking hint from the above do it yourselves]

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