Define HCF of two positive integers and find the HCF of the following pairs of numbers:

Solution:

(i) We need to find H.C.F. of 32 and 54.

By applying division lemma

$54=32 \times 1+22$

Since remainder $\neq 0$, apply division lemma on 32 and remainder 22

$32=22 \times 1+10$

Since remainder $\neq 0$, apply division lemma on 22 and remainder 10

$22=10 \times 2+2$

Since remainder $\neq 0$, apply division lemma on 10 and remainder 2

$10=2 \times 5+0$

Therefore, H.C.F. of 32 and 54 is 2

(ii) We need to find H.C.F. of 18 and 24.

By applying division lemma

$24=18 \times 1+6$

Since remainder $\neq 0$, apply division lemma on divisor 18 and remainder 6

$18=6 \times 3+0$

Therefore, H.C.F. of 18 and 24 is 6

(iii) We need to find H.C.F. of 70 and 30.

By applying Euclid’s Division lemma

$70=30 \times 2+10$

Since remainder $\neq 0$, apply division lemma on divisor 30 and remainder 10

$30=10 \times 3+0 .$

Therefore, H.C.F. of 70 and $30=10$

(iv) We need to find H.C.F. of 56 and 88.

By applying Euclid’s Division lemma

$88=56 \times 1+32$

Since remainder $\neq 0$, apply division lemma on 56 and remainder 32

$56=32 \times 1+24$

Since remainder $\neq 0$, apply division lemma on 32 and remainder 24

$32=24 \times 1+8 .$

Since remainder $\neq 0$, apply division lemma on 24 and remainder 8

$24=8 \times 3+0$

Therefore, H.C.F. of 56 and $88=8$.

(v) We need to find H.C.F. of 475 and 495.

By applying Euclid’s Division lemma

$495=475 \times 1+20$

Since remainder $\neq 0$, apply division lemma on 475 and remainder 20

$475=20 \times 23+15$

Since remainder $\neq 0$, apply division lemma on 20 and remainder 15

$20=15 \times 1+5$

Since remainder $\neq 0$, apply division lemma on 15 and remainder 5

$15=5 \times 3+0$

Therefore, H.C.F. of 475 and $495=5$.

(vi) We need to find H.C.F. of 75 and 243.

By applying Euclid’s Division lemma

$243=75 \times 3+18$

Since remainder $\neq 0$, apply division lemma on 75 and remainder 18

$75=18 \times 4+3$

Since remainder $\neq 0$, apply division lemma on divisor 18 and remainder 3

$18=3 \times 6+0$

Therefore, H.C.F. of 75 and $243=3$.

(vii) We need to find H.C.F. of 240 and 6552.

By applying Euclid’s Division lemma

$6552=240 \times 27+72$

Since remainder $\neq 0$, apply division lemma on divisor 240 and remainder 72

$240=72 \times 3+24$

Since remainder $\neq 0$, apply division lemma on divisor 72 and remainder 24

$72=24 \times 3+0$

Therefore, H.C.F. of 240 and $6552=24$.

(viii) We need to find H.C.F. of 155 and 1385.

By applying Euclid’s Division lemma 

$1385=155 \times 8+145$

Since remainder $\neq 0$, apply division lemma on divisor 155 and remainder 145

$155=145 \times 1+10$

Since remainder $\neq 0$, apply division lemma on divisor 145 and remainder 10

$145=10 \times 14+5$

Since remainder $\neq 0$, apply division lemma on divisor 10 and remainder 5

$10=5 \times 2+0 .$

Therefore, H.C.F. of 155 and $1385=5$.

(ix) We need to find H.C.F. of 100 and 190.

By applying Euclid’s division lemma

$190=100 \times 1+90$

Since remainder $\neq 0$, apply division lemma on divisor 100 and remainder 90

$100=90 \times 1+10 .$

Since remainder $\neq 0$, apply division lemma on divisor 90 and remainder 10

$90=\times 10 \times 9+0$

Therefore, H.C.F. of 100 and $190=10$.

(x) We need to find H.C.F. of 105 and 120.

By applying Euclid’s division lemma 

$120=105 \times 1+15$

Since remainder $\neq 0$, apply division lemma on divisor 105 and remainder 15

$105=15 \times 7+0$

Therefore, H.C.F. of 105 and $_{120}=15$.