Define HCF of two positive integers and find the HCF of the following pairs of numbers:(vii) 240 and 6552
(viii) 155 and 1385
(ix) 100 and 190
(x) 105 and 120
Answers
SOLUTION :
HCF (highest common factor) :
The largest positive integer that divides given two positive integers is called the Highest Common Factor of these positive integers.
(vii) Given : Two positive integers 240 and 6552.
Here, 6552 > 240
Let a = 6552 and b = 240
6552 = 240 x 27 + 72.
[By applying division lemma, a = bq + r]
Here, remainder = 72 ≠ 0, so take new dividend as 240 and new divisor as 72.
Let a = 240 and b= 72
240 = 72 x 3+ 24.
Here, remainder = 24 ≠ 0, so take new dividend as 72 and new divisor as 24.
Let a = 72 and b= 24
72 = 24 x 3 + 0.
Here, remainder is zero and divisor is 24.
Hence ,H.C.F. of 240 and 6552 is 24.
(viii) Given : Two positive integers 155 and 1385
Here, 1385 > 155
Let a = 1385 and b = 155
1385 = 155 x 8 + 145.
[By applying division lemma, a = bq + r]
Here, remainder = 145 ≠ 0, so take new dividend as 155 and new divisor as 145.
Let a = 155 and b= 145
155 = 145 x 1 + 10.
Here, remainder = 10 ≠ 0, so take new dividend as 145 and new divisor as 10.
Let a = 145 and b= 10
145 = 10 x 14 + 5.
Here, remainder = 5 ≠ 0, so take new dividend as 10 and new divisor as 5.
Let a = 10 and b= 5
10 = 5 x 2 + 0.
Here, remainder is zero and divisor is 5.
Hence ,H.C.F. of 155 and 1385 is 5.
(ix) Given : Two positive integers 100 and 190.
Here, 190 > 100
Let a = 190 and b = 100
190 = 100 x 1 + 90.
[By applying division lemma, a = bq + r]
Here, remainder = 90 ≠ 0, so take new dividend as 100 and new divisor as 90.
Let a = 100 and b= 90
100 = 90 x 1 + 10.
Here, remainder = 10 ≠ 0, so take new dividend as 90 and new divisor as 10.
Let a = 90 and b= 10
90 = 10 x 9 + 0.
Here, remainder is zero and divisor is 10..
Hence ,H.C.F. of 100 and 190 is 10.
(x) Given : Two positive integers 105 and 120.
Here, 120 > 105
Let a = 120 and b = 105
120 = 105 x 1 + 15.
[By applying division lemma, a = bq + r]
Here, remainder = 15 ≠ 0, so take new dividend as 105 and new divisor as 15.
Let a = 105 and b= 15
105 = 15 x 7 + 0.
Here, remainder is zero and divisor is 15..
Hence ,H.C.F. of 105 and 120 is 15.
HOPE THIS ANSWER WILL HELP YOU…
Answer:
(vii) 240 and 6552
Solution:
By applying Euclid’s Division lemma on 6552 and 240 we get,
6552 = 240 x 27 + 72.
Since remainder ≠ 0, apply division lemma on divisor 240 and remainder 72
240 = 72 x 3+ 24.
Since remainder ≠ 0, apply division lemma on divisor 72 and remainder 24
72 = 24 x 3 + 0.
Therefore, H.C.F. of 240 and 6552 is 24
(viii) 155 and 1385
Solution:
By applying Euclid’s Division lemma on 1385 and 155 we get,
1385 = 155 x 8 + 145.
Since remainder ≠ 0, apply division lemma on divisor 155 and remainder 145.
155 = 145 x 1 + 10.
Since remainder ≠ 0 apply division lemma on divisor 145 and remainder 10
145 = 10 x 14 + 5.
Since remainder ≠ 0, apply division lemma on divisor 10 and remainder 5
10 = 5 x 2 + 0.
Therefore, H.C.F. of 155 and 1385 is 5
(ix) 100 and 190
Solution:
By applying Euclid’s division lemma on 190 and 100 we get,
190 = 100 x 1 + 90.
Since remainder ≠ 0, apply division lemma on divisor 100 and remainder 90
100 = 90 x 1 + 10.
Since remainder ≠ 0, apply division lemma on divisor 90 and remainder 10
90 = 10 x 9 + 0.
Therefore, H.C.F. of 100 and 190 is 10
(x) 105 and 120
Solution:
By applying Euclid’s division lemma on 120 and 105 we get,
120 = 105 x 1 + 15.
Since remainder ≠ 0, apply division lemma on divisor 105 and remainder 15
105 = 15 x 7 + 0.
Therefore, H.C.F. of 105 and 120 is 15