Real Numbers - Case Study 3

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CASE STUDY 3: A Mathematics Exhibition is being conducted in your School and one of your friends is making a model of a factor tree. He has some difficulty and asks for your help in completing a quiz for the audience.

L = 5 x 2783

2783 = M x 253

253 = 11 x N

1. What will be the value of L , a) 15005 b) 13915 c) 56920 d) 17429

Soln: 

• L = 5 × 2783


• Multiply: 5 × 2783 = 13915

Ans: b) 13915

2. What will be the value of M ? a) 23 b) 22 c) 11 d) 19 

Soln: 

• 2783 = Y × 253


• Divide: Y = 2783 ÷ 253

Check: 253 × 11 = 2783

Ans: c) 11

3. What will be the value of N ? a) 22 b) 23 c) 17 d) 19 

Soln: 

Value of N

• 253 = 11 × Z


• Divide: Z = 253 ÷ 11 = 23

Ans: b) 23

4. According to Fundamental Theorem of Arithmetic 13915 is a a) Composite number b) Prime number c) Neither prime nor composite d) Even number 

Soln: 

• 13915 = 5 × 11 × 23 → Product of more than one prime → Composite number

Ans: a) Composite number

5. The prime factorisation of 13915 is a) 5x11³x13² b) 5x11³x23² c) 5x11²x23 d) 5x11²x13²

• 13915 = 5x11²x23

Ans: c) 5 × 11² × 23