Using division algorithm, we may express HCF of two numbers as linear combination.
Step 1 :
In the given numbers, choose the larger number one.
Step 2 :
Decompose the larger one by the multiple of smaller one.
Step 3 :
Step 2 will have the form,
Divided = Divisor x quotient + Remainder
Repeat the process until we receive remainder as 0.
At last by dividing which value, we get the remainder as 0, that can be considered as HCF of the given numbers.
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 :
HCF of 56 and 72
56 = 16 × 3 + 8
8 = 4 × 2 + 0
HCF of 56 and 72 is 4.
4 = 56x + 72y
If x = 1 4 = 56(1) + 72y 72y = 4 - 56 72y = - 52 y is not integer. |
If x = 2 4 = 56(2) + 72y 72y = 4 - 112 72y = - 108 y is not integer. |
56(4) + 72y
224 + 72y
72y = -224
y = -224/72
y = -3
So, x = 4
and y = -3.
So, x and y are not unique.
Problem 2 :
Express the HCF of 468 and 222 as 468x + 222y where x, y are integers in two different ways.
Solution :
HCF of 468 and 222
468 = 222 × 2 + 44
222 = 24 × 9 + 6
24 = 6 × 4 + 0
HCF of 468 and 222 is 6.
6 = 222 - (24 × 9)
Decomposing 24 as combination of 468 and 222.
6 = 222 - {(468 – 222 × 2) × 9}
6 = 222 - {468 × 9 – 222 × 18}
6 = 222 + (222 × 18) - (468 × 9)
6 = 222(1 + 18) – 468 × 9
6 = 222 × 19 – 468 × 9
6 = 468 × (-9) + 222 × 19
x = -9 and y= 19.
Problem 3 :
Express the HCF of 210 and 55 as 210x + 55y where x, y are integers in two different ways.
Solution :
HCF of 210 and 55
210 = 55 × 3 + 45
55 = 45 × 1 + 10
45 = 10 × 4 + 5
10 = 5 × 2 + 0
HCF of 210 and 55 is 5.
Given, 210x + 55y where x, y are integers.
5 = 210(5) + 55y
5 = 1050 + 55y
55y = -1045
y = -1045/55
y = -19
So, the
value of x and y is 5 and -19.
Problem 4 :
If the HCF of 408 and 1032 is expressible in the form of 1032m – 408 × 5. Find m.
Solution :
First let us find HCF of 408 and 1032.
1032 > 408
1032 = 408 × 2 + 216
408 = 216 × 1 + 192
216 = 192 × 1 + 24
192 = 24 × 8 + 0
HCF of 408 and 1032 is 24.
To find m :
1032m – 408 × 5 = 24
1032m – 2040 = 24
1032m = 24 + 2040
1032m = 2064
m = 2064/1032
m = 2
So, the value of m is 2.
Problem 5 :
If the HCF of 657and 963 is expressible in the form of 657n + 963 × (-15). Find n.
Solution :
First let us find HCF of 963 and 657.
963 > 657
963 = 657 ×1 + 306
657 = 306 × 2 + 45
306 = 45 × 6 + 36
45 = 36 × 1 + 9
36 = 9 × 4 + 0
HCF of 657 and 963 is 9.
To find n :
657n + 963 × (-15) = 9
657n – 14445 = 9
657n = 9 + 14445
657n = 14454
n = 14454/657
n = 22
So, the value of n is 22.
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