1. Use Euclid’s division algorithm to find the HCF of:
(i) 135 and 225
As there is no reminder after third step
So, HCF = 45
(ii) 196 and 38220
So, HCF = 196
(iii) 867 and 255
So, HCF = 51
2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Let us assume q = 1
Then, 6q+1 = 7
6q+3 = 9
6q+5 = 11
It is clear that 7, 9 and 11 are consecutive negative integers.
3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
To get the solution we need to find the HCF of 616 and 32
So, HCF = 8
So, the maximum number of column is 8
4. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
Let us start with the smallest square number 4
4 = 3m+1
Next square number is 9
9 = 3m
5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Let us start with the smallest cube number 8
Next number is 27
27 = 9m
Next number is 64
64 = 9m + 1
Next number is 125
125 = 9m + 8
6. Express each number as a product of its prime factors:
7. Find the LCM and HCF of the following pairs of integers and verify that
LCM × HCF = product of the two numbers.
(i) 26 and 91
Solution: Factorization of numbers gives following:
(ii) 510 and 92
(iii) 336 and 54
8. Find the LCM and HCF of the following integers by applying the prime factorization method.
(i) 12, 15 and 21
(ii) 17, 23 and 29
Since all are prime numbers so HCF = 1
(iii) 8, 9 and 25
HCF = 1
9. Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution: As you know
10. Check whether 6n can end with the digit 0 for any natural number n.
Solution: As you know for getting 0 at unit’s place of a number there should be at least a 5 and a 2 in the factor.
Now, 6 is having 2 as one of its factors so we need a 5 to fulfill the condition. If n = 5 or product of 5 then there will be 0 at unit’s place.
11. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Solution: As per fundamental theorem of arithmetic every composite number is a product of primes.
The first example will be an even number because two odd numbers equal an even number when added. So, it is a composite number.
The second example will be a product of 5, hence a composite number.
As they are not prime numbers so it is obvious that they are composite numbers.
12. 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?
Solution: We need to calculate the LCM to find the answer.
After 36 minutes they will be together at the starting point.
13. Prove that
Solution: Let us prove it by the method of contradiction. Let us assume that
is rational. Let us assume that = ,
where a and b are co-prime that is they have not common factors other than 1.
Or, b = a
Or, b²5= a²
So, a² is divisible by 5
And a is also divisible by 5.
So a= 5c for some integer c.
Substituting the value of a from above equation we get
b²5 = 25c²
Or, b²= 5c²
This means that b² and b is divisible by 5.
So, a and b have common factors other than 1.
This contradicts our earlier assumption of a and b being co-prime.
The contradiction came because is an irrational number.
14. Prove that 3 +2
Solution: The addition of an irrational number and a rational number will always give an irrational number as answer. Other steps will be same as the previous question.
15. Prove that the following are irrationals :
(iii) 6 +
Solution: When an irrational number is added to, deleted from, multiplied by or divided by a rational number, the result is always an irrational number.