HCF OF TWO NUMBERS USING EUCLID DIVISION LEMMA WORKSHEET

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Problem 1 :

If d is the HCF of 56 and 72, find x, y satisfying

d = 56x + 72y

Also show that x and y are not unique.

Solution

Problem 2 :

Express the HCF of 468 and 222 as 468x + 222y where x, y are integers in two different ways.

Solution

Problem 3 :

Express the HCF of 210 and 55 as 210x + 55y where x, y are integers in two different ways.

Solution

Problem 4 :

If the HCF of 408 and 1032 is expressible in the form of 1032m – 408 × 5. Find m.

Solution

Problem 5 :

If the HCF of 657and 963 is expressible in the form of 657n + 963 × (-15). Find n.

Solution

Problem 6 :

The product of two numbers if 2028 and their highest common factor is 13. The number of such pairs is  is

a) 1     b)  2     c)  3      d)  4

Solution

Problem 7 :

The greatest number which can divide 1356, 1868 and 2764 leaving the same remainder 12 in each case is 

a) 64     b)  124     c)  156      d) 260

Solution

Answer Key

1) x = 4 and y = -3.

2)  x = -9 and y= 19.

3)  x and y is 5 and -19.

4) m is 2.

5) n = 22

6) 2 pairs

7) 64

Problem 1 :

Find HCF of 81 and 237 and express it as a linear combination of 81 and 237.

Solution

Problem 2 :

Find the HCF of 65 and 117 and express it in the form

65m + 117n

Solution

Problem 3 :

If the HCF of 210 and 55 is expressible in the form of 

210x + 55y

Also show that x and y are not unique.

Solution

Problem 4 :

The sum of two numbers is 528 and their highest common factor is 33. The number of pairs of numbers satisfying the above conditions is 

a) 4    b)  6    c)  8    d)   12

Solution

Problem 5 :

The number of number pairs lying between 40 and 100 with their highest common factor as 15 is

a) 3    b)  4    c)  5    d)   6

Solution

Problem 6 :

The highest common factor of two numbers is 12 and their difference is 12. The numbers are 

a) 66, 78     b) 70, 82     c) 94, 106     d) 84, 96

Solution

Problem 7 :

Let N be the greatest number that will divide 1305, 4665 and 6905 leaving the same remainder in each case. Then sum of the digits in N is 

a)  4     b)  5    c)  6      d)  8

Solution

Problem 8 :

There are two numbers. HCF of both the numbers is 11, and their LCM is 693. If the first number is 77, find the second number?

a) 89      b) 56       c) 78         d) 99

Solution

Answer Key

1)  HCF is 3

2)  m = 2 and n = -1

3)  x and y are 5 and -19 

4) (1, 15), (3, 13), (5, 11) (7, 9) , there are such 4 pairs.

5) highest common factor 15 as (45, 60) (45, 75) (60, 75) and (75, 90)

Clearly there are 4 such pairs.

6) 84 and 96, option d

7) Sum of the digits is 4.

8) the required number is 99.

Problem 1 :

Use Euclid's division algorithm to find HCF of 441, 567, 693.

Solution

Problem 2 :

Using Euclid's division algorithm, find the largest number that divides 1251, 9377 and 15628 leaving remainders 1, 2 and 3 respectively.

Solution

Problem 3 :

Using Euclid's division algorithm find the HCF of 9828 and 14742

Solution

Problem 4 :

Using Euclid's division algorithm, find which of the following pairs of numbers are coprime.

(i)  231, 396     (ii)  847, 2160

Solution

Problem 5 :

Use Euclid's division algorithm to show of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Solution

Problem 6 :

Use Euclid's division algorithm to show of any positive integer is either of the form 9q, 9q + 1 or 9q + 8 for some integer q.

Solution

Answer Key

1) HCF of 693, 567 and 441 is 63.

2) the largest number which divides 1251, 9377 and 15628 leaving remainders 1, 2 and 3 respectively is 625

3) HCF of (14742 and 9828) is 4914.

4) i)  HCF of (396, 231) is 33.

ii) HCF of 847 and 2160 is 1. So, these two are coprimes.

5) 

Let a be any positive integer, we know any positive integers is in the form 3q, 3q + 1 or 3q + 2

When a = 3q, then

a2 = (3q)2

= 9q2

= 3 (3q2)

Let 3q2 = m

= 3m

When a = 3q + 1, then 

a2 = (3q + 1)2

= (3q)2 + 2(3q) (1) + 12

= 9q2 + 6q + 1

= 3(3q2 + 2q) + 1

= 3m + 1

When a = 3q + 2, then 

a2 = (3q + 2)2

= (3q)2 + 2(3q) (2) + 22

= 9q2 + 12q + 4

= 3(3q2 + 4q) + 3 + 1

= 3(3q2 + 4q + 1) + 1

= 3m + 1

Here a2 is of form of 3m or 3m + 1.

6) 

When x = 3q then 

x3 = (3q)3

= 27q3

= 9(3q3)

= 9m

When x = 3q + 1 then 

x3 = (3q + 1)3

= (3q)3 + 3 (3q)2(1) + 3 (3q) (1)2 + 13

27q3 + 27q2 + 9q + 1

= 9q(3q2 + 3q + 1) + 1

= 9m + 1

When x = 3q + 2 then 

x3 = (3q + 2)3

= (3q)3 + 3 (3q)2(2) + 3 (3q) (2)2 + 23

27q3 + 54q2 + 36q + 8

= 9q(3q2 + 6q + 4) + 8

= 9m + 8

Hence x3 is either of the form 9m, 9m + 1 or 9m + 8.

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