EXPRESS HCF OF TWO NUMBERS IN LINEAR COMBINATION

Problem 1 :

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

Solution :

237 > 81

237 = 81 x 2 + 75

81 = 75 x 1 + 6

75 = 6 x 12 + 3

6 = 3 x 2 + 0

HCF of (81 and 237) is 3.

Linear combination :

237 = 81x + 237y

Problem 2 :

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

65m + 117n

Solution :

117 > 65

117 = 65 x 1 + 52

65 = 52 x 1 + 13

52 = 13 x 4 + 0

HCF of (117 and 65) is 13.

13 = 65(2) + 117(-1)

HCF = 65m + 117n

m = 2 and n = -1

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 :

First let us find HCF of 210 and 55.

210 > 55

210 = 55 ⋅ 3 + 45

55 = 45 ⋅ 1 + 10

45 = 10 ⋅ 4 + 5

10 = 5 ⋅ 2 + 0

HCF of 210 and 55 is 5.

HCF = 210x + 55y

5 = 210x + 55y

x = 1, then y = 

5 = 210(1) + 55y

5 - 210 = 55y

55y = -205

y = -205/55

Since we get decimal, we can apply some other value for x.

x = 5, then y =

5 = 210(5) + 55y

5 = 1050 + 55y

5 - 1050 = 55y

-1045 = 55y

y = -1045/55

y = -19

So, values of x and y are 5 and -19 respectively. By applying some more values for x, we will get different values for y.

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 :

The sum of two numbers = 528

Highest common factor = 33

Let the required numbers be 33x and 33y

33x + 33y = 528

x + y = 528/33

x + y = 16

Co-primes with the sum which is 16 are 

(1, 15), (3, 13), (5, 11) (7, 9)

So, there are such 4 pairs.

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 :

Numbers with highest common factor 15 must contain 15 as a factor.

Now multiples of 15 between 40 and 100 are 45, 60, 75 and 90

Number pairs with highest common factor 15 as (45, 60) (45, 75) (60, 75) and (75, 90)

highest common factor of (60, 90) is 30 and that of (45, 90) is 45.

Clearly there are 4 such pairs.

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 :

Out of the given numbers, the two with highest common factor is 12 and difference 12 are 84 and 96.

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 :

4665 - 1305 ==> 3360 

6905 - 4665 ==> 2240

6905 - 1305 ==> 5600

Highest common factor = 2 x 2 x 2 x 2 x 2 x 5 x 7

= 1120

Sum of the digits = 1 + 1 + 2 + 0

= 4

So, option a is correct.

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 :

First number = 77

Let a be the second number

Product of two numbers = product of lcm and hcf

77 x a = 11 x 693

a = (11 x 693) / 77

a = 99

So, the required number is 99.