**Frequently Asked Questions on Chapter 1 Real Numbers**

**Euclid’s division algorithm to find the HCF of 135 and 225?**

**135 and 225**

As you can see, from question 225 is greater than 135. Therefore, by Euclid’s division algorithm, we have, 225 = 135 × 1 + 90 Now, remainder 90 ≠ 0, thus again using division lemma for 90, we get, 135 = 90 × 1 + 45 Again, 45 ≠ 0, repeating the above step for 45, we get, 90 = 45 × 2 + 0 The remainder is now zero, so our method stops here. Since, in the last step, the divisor is 45, therefore, HCF (225,135) = HCF (135, 90) = HCF (90, 45) = 45. Hence, the HCF of 225 and 135 is 45.

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

Let a be any positive integer and b = 6. Then, by Euclid’s algorithm, a = 6q + r, for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5, because 0≤r<6. Now substituting the value of r, we get, If r = 0, then a = 6q Similarly, for r= 1, 2, 3, 4 and 5, the value of a is 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5, respectively. If a = 6q, 6q+2, 6q+4, then a is an even number and divisible by 2. A positive integer can be either even or odd Therefore, any positive odd integer is of the form of 6q+1, 6q+3 and 6q+5, where q is some integer.

**Express each number as a product of its prime factors 5005?**

5005 By Taking the LCM of 5005, we will get the product of its prime factors. Hence, 5005 = 5 × 7 × 11 × 13 × 1 = 5 × 7 × 11 × 13

**Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers 26 and 91.**

6 and 91 Expressing 26 and 91 as product of its prime factors, we get, 26 = 2 × 13 × 1 91 = 7 × 13 × 1 Therefore, LCM (26, 91) = 2 × 7 × 13 × 1 = 182 And HCF (26, 91) = 13**Verification**

Now, product of 26 and 91 = 26 × 91 = 2366 And Product of LCM and HCF = 182 × 13 = 2366 Hence, LCM × HCF = product of the 26 and 91.

**The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form, p q what can you say about the prime factors of q 43.123456789?**

43.123456789 Since it has a terminating decimal expansion, it is a rational number in the form of p/q and q has factors of 2 and 5 only.

**Find the HCF of 135 and 225 using Euclid’s division**

By using Euclid’s division algorithm, the HCF of 135 and 225 can be calculated as:

We know that 225 > 135

So, 225 = 135 × 1 + 90

Since, remainder 90 ≠ 0

135 = 90 × 1 + 45

Remainder 45 ≠ 0

So, 90 = 45 × 2 + 0

Hence the remainder is = 0, so the method stops here.

Therefore HCF of (135, 225) is 45.

**Find the maximum number of columns in which they can march? When an army contingent of 616 members is to march behind an army band of 32 members in a parade. Consider the two groups marching in the same number of columns.**

Given:

Number of army contingent members = 616

Number of army band members = 32

If the two groups have to march in the same column, we have to find out the highest common factor between the two groups.

HCF(616, 32), gives the maximum number of columns in which they can march.

By Using Euclid’s algorithm, we get

We know that 616 > 32

616 = 32 × 19 + 8

Since, 8 ≠ 0

32 = 8 × 4 + 0

Now we have got remainder as 0.

Therefore, HCF (616, 32) = 8.

Hence, the maximum number of columns in which they can march is 8.

**Express 5005 as a product of its prime factors:**

**To find the product of its prime factors, firstly we need to take LCM of 5005**

So, 5005 = 5 × 7 × 11 × 13 × 1

Hence, 5005 = 5 × 7 × 11 × 13