# SSC Real Numbers Exercise – 1.2 of 10th class

**SSC Real Numbers :**

- Positive or negative, Large or small, whole numbers or decimal numbers are all
**Real Numbers.**

They are called “**Real Numbers**” because they are not Imaginary Numbers.

**EXERCISE – 1.2**

**1Q. Express each of the following number as a product of its prime factors.**

**i) 140 ii) 156 iii) 3825 iv) 5005 v) 7429**

**Sol** : ** 140**

** 140 = 2 X 2 X 5 X 7 = 2 ^{2} X 5 X 7**

**Sol** : ** ii) 156**

**156 = 2 X 2 X 3 X 13 = 2 ^{2} X 3 X 13**

**Sol** :** iii) 3825**

**3825 = 3 X 3 X 5 X 5 X 17**

**= 3 ^{2} X 5^{2} X 17**

**Sol** : **iv) 5005**

**5005 = 5 X 7 X 11 X 13**

**Sol** : ** v) 7429**

**7429 = 17 X 19 X 23**

### SSC Real Numbers

**2Q. Find the L.C.M and H.C.F of the following integers by the prime factorization method.**

**Sol**. **i) 12, 15 and 21 ii) 17, 23 and 29**

**iii) 8, 9 and 25 iv) 72 and 108**

**Sol : ** **i) 12, 15 and 21**

** 12 = 2 X 2 X 3 = 2 ^{2} X 3**

** 15 = 3 X 5**

** 21 = 3 X 7**

** L.C.M = 2 ^{2} X 3 X 5 X 7 = 420**

** H.C.F = 3**

**Sol :**** ii) 17, 23 and 29**

**The given numbers 17, 23 and 29 are all primes.**

**L.C.M = their product**

**= 17 X 23 X 29 = 11339**

**H.C.F = 1**

**iii) 8, 9 and 25**

** 8 = 2 X 2 X 2 = 2 ^{3}**

** 9 = 3 X 3 = 3 ^{2}**

** 25 = 5 X 5 = 5 ^{2}**

** L.C.M = 2 ^{3} X 3^{2} X 5^{2} = 1800**

** ( OR )**

**8, 9 and 25 are relatively prime, there fore L.C.M is equal to their product. (i.e.) L.C.M = 8 X 9 X 25 = 1800**

**H.C.F = 1**

**iv) 72 and 108**

**72 = 2 ^{3} X 3^{2}**

**108 = 2 ^{2} X 3^{3}**

**L.C.M = 2 ^{3} X 3^{3} = 8 X 27 = 216**

**H.C.F = 2 ^{2} X 3^{2} = 4 X 9 = 316**

**v) 306 and 657**

**306 = 2 X 3 ^{2} X 17**

**657 = 3 ^{2} X 73**

**L.C.M = 2 X 3 ^{2} X 17 X 73 = 22338**

**H.C.F = 3 ^{2} = 9.**

### SSC Real Numbers

### 3Q. Check whether 6^{n} can end with the digit ‘0’ for any natural number n.

**Sol :** **Given number = 6 ^{n} = ( 2 X 3 )^{n}**

**The prime factors here are 2 and 3 only.**

**To be end with 0; 6 ^{n} should have a prime factor 5 and also 2. So,**

**6 ^{n} can’t end with zero.**

### SSC Real Numbers

**4Q. Explain why 7 X 11 X 13 + 13 and 7 X 6 X 5 X 4 X 3 X 2 X 1 + 5 are composite numbers.**

**Sol : Given numbers are**

**7 X 11 X 13 + 13 and**

**7 X 6 X 5 X 4 X 3 X 2 X 1 + 5**

**13 ( 7 X 11 + 1 ) and 5 ( 7 X 6 X 4 X 3 X 2 X 1 + 1 )**

**13 K and 5 L, where K = 78 and**

**L = 7 X 6 X 4 X 3 X 2 X 1 + 1 = 1009**

**As the given numbers can be written as product of two numbers, they are composite.**

### SSC Real Numbers

**5Q. How will you show that ( 17 X 11 X 2 ) + ( 17 X 11 X 5 ) is a composite number? Explain.**

**Sol :**** ( 17 X 11 X 2 ) + ( 17 X 11 X 5 )**

**= ( 17 X 11 ) ( 2 + 5 )**

**= ( 17 X 11 ) ( 7 ) = 187 X 7**

**Now the given expression is written as a product of two integers and hence it is a composite number.**

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