# Ex.1.1 Q2 Real Numbers Solution - NCERT Maths Class 10

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## Question

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

Video Solution
Real Numbers
Ex 1.1 | Question 2

## Text Solution

What is known?

Some integer $$q.$$

What is unknown?

That any positive odd integer is of the form $$6q+1,$$ or $$6q+3,$$ or $$6q+5,$$ where $$q$$ is some integer.

Reasoning:

To solve this question, first think about the Euclid’s division algorithm. Suppose there is any positive integer '$$a$$' and it is of the form $$6q+ r$$. where $$q$$ is some integer. This means that $$0 \le r < 6$$ i.e. $$r=0$$ or $$1$$ or $$2$$ or $$3$$ or $$4$$ or $$5$$ but it can’t be $$6$$ because $$r$$ is smaller than $$6.$$. So, by Euclid’s division lemma possible values for '$$a$$' can be $$6q$$ or $$6q+1$$or $$6q+2$$ or $$6q+3$$ or $$6q+4$$ or $$6q+5.$$

Steps:

Let '$$a$$' be any positive integer. Then according to Euclid’s algorithm,

$$a = 6q + r$$ for some integer $$q \ge 0$$

and $$r= 0,1,2,3,4,5$$ because $$0\le r < 6$$

Therefore,

$$a = 6q + 0$$ or $$6q + 1$$ or $$6q + 2$$ or $$6q + 3$$ or $$6q + 4$$ or $$6q + 5.$$

Now, $$6q + 1 = 2 \times 3q + 1 = 2 {k_1}+1$$

(where $$k_1$$ is a positive integer)

$$6q+3 = 6q+2 +1= 2 (3q + 1)+1 =2{k_2}+1$$

(where $$k_2$$ is a positive integer)

$$6q+5=6q+4+1=2 (3q+2)+1=2{k_3}+1$$

(where $$k_3$$ is a positive integer)

Clearly, $$6q{ }+{ }1,{ }6q{ }+{ }3$$ and $$6q+5$$ are of the form $$2k+1,$$ where $$k$$ is an integer.

Therefore, $$6q+1, 6q+3$$ and $$6q+5$$ are not exactly divisible by $$2.$$

Hence, these expressions of numbers are odd numbers and therefore any odd integers can be expressed in the form $$6q+1$$ or $$6q+3$$ or $$6q+5.$$

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