Pedagogy of Mathematics - Part II (Numbers and Sequence - Ex 2.1), Numbers and Sequence, Maths, IX std Maths, Samacheerkalvi maths, II year B.Ed., Pedagogy, Mathematics, Euclid's Division Lemma, Euclid's Division algorithm,

1. PEDAGOGY OF
MATHEMATICS – PART II
BY
Dr. I. UMA MAHESWARI
Principal
Peniel Rural College of Education,Vemparali,
Dindigul District
iuma_maheswari@yahoo.co.in

2. X STD
Ex 2.1

25. Answer:
The positive integers when divided by 3 leaves remainder 2.
By Euclid’s division lemma a = bq + r, 0 ≤ r < b.
Here a = 3q + 2, where 0 ≤ q < 3, a leaves remainder 2 when divided by 3.
∴ 2, 5, 8, 11 ……………..

26. Answer:
Here a = 532, b = 21
Using Euclid’s division algorithm
a = bq + r
532 = 21 × 25 + 7
Number of completed rows = 21
Number of flower pots left over = 7

27. Solution:
Let n – 1 and n be two consecutive positive integers.Then their product is (n – 1)n.
(n – 1)(n) = n2 – n.
We know that any positive integer is of the form 2q or 2q + 1 for some integer q.
So, following cases arise.
Case I.When n = 2q.
In this case, we have
n2 – n = (2q)2 – 2q = 4q2 – 2q = 2q(2q – 1)
⇒ n2 – n = 2r, where r = q(2q – 1)
⇒ n2 – n is divisible by 2.

28. Answer:
Let the positive integer be a, b, and c
We know that by Euclid’s division lemma
a = bq + r
a = 13q + 9 ….(1)
b = 13q + 7 ….(2)
c = 13q + 10 ….(3)
Add (1) (2) and (3)
a + b + c = 13q + 9 + 13q + 7 + 13q + 10
= 39q + 26
a + b + c = 13 (3q + 2)
This expansion will be divisible by 13
∴ a + b + c is divisible by 13

29. Solution:
Let x be any integer.
The square of x is x2.
Let x be an even integer.
x = 2q + 0
then x2 = 4q2 + 0
When x be an odd integer
When x = 2k + 1 for some integer k.
x2 = (2k + 1 )2
= 4k2 + 4k + 1
= 4k (k + 1) + 1
= 4q + 1
where q = k(k + 1) is some integer.
Hence it is proved.

30. (i) 340 and 412
Answer:
To find the HCF of 340 and 412 using Euclid’s division algorithm.We get
412 = 340 × 1 + 72
The remainder 72 ≠ 0
Again applying Euclid’s division algorithm to the division of 340
340 = 72 × 4 + 52
The remainder 52 ≠ 0
Again applying Euclid’s division algorithm to the division 72 and remainder 52 we get
72 = 52 × 1 + 20
The remainder 20 ≠ 0
Again applying Euclid’s division algorithm 52 = 20 × 2 + 12

31. The remainder 12 ≠ 0
Again applying Euclid’s division algorithm
20 = 12 × 1 + 8
The remainder 8 ≠ 0
Again applying Euclid’s division algorithm
12 = 8 × 1 + 4
The remainder 4 ≠ 0
Again applying Euclid’s division algorithm
8 = 4 × 2 + 0
The remainder is zero
∴ HCF of 340 and 412 is 4

32. (ii) 867 and 255
Answer:
To find the HCF of 867 and 255 using
Euclid’s division algorithm.We get
867 = 255 × 3 + 102
The remainder 102 ≠ 0
Using Euclid’s division algorithm
255 = 102 × 2 + 51
The remainder 51 ≠ 0
Again using Euclid’s division algorithm
102 = 51 × 2 + 0
The remainder is zero
∴ HCF = 51
∴ HCF of 867 and 255 is 51

33. (iii) 10224 and 9648
Answer:
Find the HCF of 10224 and 9648 using Euclid’s division algorithm.We get
10224 = 9648 × 1 + 576
The remainder 576 ≠ 0
Again using Euclid’s division algorithm
9648 = 576 × 16 + 432
The remainder 432 ≠ 0
Using Euclid’s division algorithm
576 = 432 × 1 + 144
The remainder 144 ≠ 0
Again using Euclid’s division algorithm
432 = 144 × 3 + 0
The remainder is 0
∴ HCF = 144
The HCF of 10224 and 9648 is 144

34. (iv) 84,90 and 120
Answer:
Find the HCF of 84, 90 and 120 using Euclid’s division
algorithm
90 = 84 × 1 + 6
The remainder 6 ≠ 0
Using Euclid’s division algorithm
4 = 14 × 6 + 0
The remainder is 0
∴ HCF = 6
The HCF of 84 and 90 is 6
Find the HCF of 6 and 120
120 = 6 × 20 + 0
The remainder is 0
∴ HCF of 120 and 6 is 6
∴ HCF of 84, 90 and 120 is 6

35. Solution:
The required number is the H.C.F. of the
numbers.
1230 – 12 = 1218,
1926 – 12 = 1914
First we find the H.C.F. of 1218 & 1914 by
Euclid’s division algorithm.
1914 = 1218 × 1 + 696
The remainder 696 ≠ 0.

36. Again using Euclid’s algorithm
1218 = 696 × 1 + 522
The remainder 522 ≠ 0.
Again using Euclid’s algorithm.
696 = 522 × 1 + 174
The remainder 174 ≠ 0.
Again by Euclid’s algorithm
522 = 174 × 3 + 0
The remainder is zero.
∴The H.C.F. of 1218 and 1914 is 174.
∴The required number is 174.

37. Answer:
Find the HCF of 32 and 60
60 = 32 × 1 + 28 ….(1)
The remainder 28 ≠ 0
By applying Euclid’s division lemma
32 = 28 × 1 + 4 ….(2)
The remainder 4 ≠ 0
Again by applying Euclid’s division lemma
28 = 4 × 7 + 0….(3)
The remainder is 0

38. HCF of 32 and 60 is 4
From (2) we get
32 = 28 × 1 + 4
4 = 32 – 28
= 32 – (60 – 32)
4 = 32 – 60 + 32
4 = 32 × 2 -60
4 = 32 x 2 + (-1) 60
When compare with d = 32x + 60 y
x = 2 and y = -1
The value of x = 2 and y = -1

39. Solution:
Let a (+ve) integer be x.
x = 88 × y + 61
61 = 11 × 5 + 6 (∵ 88 is multiple of 11)
∴ 6 is the remainder. (When the number is divided by 88 giving the
remainder 61 and when divided by 11 giving the remainder 6).

40. Answer:
Let the consecutive positive integers be x and x + 1.
The two number are co – prime both the numbers are divided by 1.
If the two terms are x and x + 1 one is odd and the other one is even.
HCF of two consecutive number is always 1.
Two consecutive positive integer are always coprime.
FundamentalTheorem of Arithmetic
Every composite number can be written uniquely as the product of power of prime is called
fundamental theorem of Arithmetic.