Question

AL NUMBERS
EXERCISE 1.
1. Use Euclid's division algorithm to find the HCF
(0) 135 and 225
(ii) 196 and 38220
2. Show that any positive odd integer is of the form
some integer
3. An army contingent of 616 members is to march b
a parade. The two groups are to march in the sa
maximum number of columns in which they can
Use Euclid's division lemma to show that the squ
the form 3m or 3m + 1 for some integer m.
THint : Letxbe any positive integer then it is of the
each of these and show that they can be rewritten
G. Use Euclid's division lemma to show that the cub
9m, 9m + 1 or m +8.​

Answer 1

Answer:

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Answer 2

Answer:

Answer:

(i) 135 and 225

Since 225 > 135, we apply the division lemma to 225 and 135 to obtain

225 = 135 * 1 + 90

Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain

135 = 90 * 1 + 45

We consider the new divisor 90 and new remainder 45 and apply the division lemma to obtain

90 = 2 * 45 + 0

Since the remainder is zero, then he process stops.

Since the divisor at this stage is 45, therefore the HCF of 135 and 225 is 45

(ii) 196 and 38220

Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain

38220 = 196 * 195 + 0

Since the remainder is zero, then he process stops.

Since the divisor at this stage is 196, therefore the HCF of 196 and 38220 is 196

(iii) 867 and 255

Since 867 > 255, we apply the division lemma to 867 and 255 to obtain

867 = 255 * 3 + 102

Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102 to obtain

255 = 102 * 2 + 51

We consider the new divisor 102 and new remainder 51 and apply the division lemma to obtain

102 = 51 * 2 + 0

Since the remainder is zero, then he process stops.

Since the divisor at this stage is 51, therefore the HCF of 867 and 255 is 51

this 1one answer

Let a be any positive integer and b = 6

Then by Euclid’s lemma, a = 6q + r for some integer q ≥ 0 and r = 0, 1, 2, 3, 4, 5 because

0 ≤ r < 6

Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5

Now, 6q + 1 = 2 * 3q + 1 = 2k1 + 1, where k1 is a positive integer.

6q + 3 = (6q + 2) + 1 = 2 * (3q + 1) + 1 = 2k2 + 1, where k2 is a positive integer.

6q + 5 = (6q + 4) + 1 = 2 * (3q + 2) + 1 = 2k3 + 1, where k3 is a positive integer.

Clearly, 6q + 1 or 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.

Therefore, 6q + 1 or 6q + 3, 6q + 5 are not exactly divisible by 2.

Hence, these expressions of numbers are odd numbers and therefore, any odd integer can be

expressed in the form 6q + 1 or 6q + 3 or 6q + 5.

2one answer

HCF(161, 32) will give the maximum number of columns in which they can march.

We can use Euclid’s lemma to find the HCF.

616 = 32 * 19 + 8

32 = 8 * 4 + 0

The HCF(616, 32) is 8.

Therefore, they can march in 8 columns each.

3one answer