Real Numbers | Exercise 1.1

Question 6

Explain why 7×11×13+137 \times 11 \times 13 + 13 and 7×6×5×4×3×2×1+57 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 are composite numbers.

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Solution
Understand the Question
  • A composite number is a positive integer greater than 11 that has factors other than 11 and itself (i.e., it can be written as a product of two or more integers greater than 11).
  • To show that each given expression is composite, we factor out a common integer to write the entire expression as a product of factors other than 11 and the number itself.

Step 1 · Factorise the First Expression

Taking 1313 as a common factor

7×11×13+13=13×(7×11+1)=13×(77+1)=13×78=13×(13×6)=13×13×6\begin{aligned} 7 \times 11 \times 13 + 13 &= 13 \times (7 \times 11 + 1) \\ &= 13 \times (77 + 1) \\ &= 13 \times 78 \\ &= 13 \times (13 \times 6) \\ &= 13 \times 13 \times 6 \end{aligned}

Since the given number has factors other than 11 and itself (such as 66 and 1313), it is a composite number.

Step 2 · Factorise the Second Expression

Taking 55 as a common factor 7×6×5×4×3×2×1+5=5×(7×6×4×3×2×1+1)7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5 \times (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1)

Evaluating the product inside the bracket 7×6×4×3×2×1=10087 \times 6 \times 4 \times 3 \times 2 \times 1 = 1008

Substituting this back

5×(1008+1)=5×1009\begin{aligned} 5 \times (1008 + 1) &= 5 \times 1009 \end{aligned}

Since the number is expressed as the product of two factors 55 and 10091009 (which are both greater than 11), it has factors other than 11 and itself. Hence, it is a composite number.

Answer

Both 7×11×13+137 \times 11 \times 13 + 13 and 7×6×5×4×3×2×1+57 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 are composite numbers as they have factors other than 11 and themselves.

Common Mistakes
  • Arithmetic Error in Common Factors: Forgetting to include +1+1 when taking the common factor out, e.g., incorrectly writing 13×(7×11)13 \times (7 \times 11) instead of 13×(7×11+1)13 \times (7 \times 11 + 1).
  • Misunderstanding Composite Definition: Confusing prime factors with any factors. A number is composite as long as it has any integer factor other than 11 and itself.

More questions in Exercise 1.1

Q1

Express each number as a product of its prime factors:

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

Q2

Find the LCM and HCF of the following pairs of integers and verify that LCM×HCF=product of the two numbers\text{LCM} \times \text{HCF} = \text{product of the two numbers}.

(i) 26 and 91

(ii) 510 and 92

(iii) 336 and 54

Q3

Find the LCM and HCF of the following integers by applying the prime factorisation method.

(i) 12, 15 and 21

(ii) 17, 23 and 29

(iii) 8, 9 and 25

Q4

Given that HCF(306,657)=9\text{HCF}(306, 657) = 9, find LCM(306,657)\text{LCM}(306, 657).

Q5

Check whether 6n6^n can end with the digit 0 for any natural number nn.

Q6

Explain why 7×11×13+137 \times 11 \times 13 + 13 and 7×6×5×4×3×2×1+57 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 are composite numbers.

Q7

There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?

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