Question 2

# Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.    (i) 26 and 91     (ii) 510 and 92     (iii) 336 and 54

(i) 26 and 91

First we have to find the L.C.M and H.C.F of 26, 91 using fundamental theorem of arithmetic,
26 = 13 × 2
91 = 13 × 7
For L.C.M - list all the prime factors (only once) of 26, 91 with their greatest power.
L.C.M (26, 91) = 13 × 2 × 7 = 182
For H.C.F – write all the common factors (only once) with their smallest exponent.
H.C.F (26, 91) = 13
Verification :   L.C.M (26, 91) × H.C.F (26, 91) = 26 × 91
182 × 13 = 2366,       => 2366 = 2366
Hence, proved.

(ii) 510 and 92

First we have to find the L.C.M and H.C.F of 510, 92 using fundamental theorem of arithmetic,
510 = 2 × 3 × 5 × 17
92 = 2 × 2 × 23
For L.C.M - list all the prime factors (only once) of 510, 92 with their greatest power.
L.C.M (510, 92) = 2 × 2 × 3 × 5 × 17 × 23 = 23460
For H.C.F – write all the common factors (only once) with their smallest exponent.
H.C.F (510, 92) = 2
Verification :  L.C.M (510, 92) × H.C.F (510, 92) = 510 × 92
23460 × 2 = 46920,
=>  46920 = 46920
Hence, proved.

(iii) 336 and 54

First we have to find the L.C.M and H.C.F of 336, 54 using fundamental theorem of arithmetic,
336 = 2 × 2 × 2 × 2  × 3 × 7 = 24 × 3 × 7
54 = 2 × 3 × 3 × 3 = 2 × 33
For L.C.M - list all the prime factors (only once) of 336, 54 with their greatest power.
L.C.M (336, 54) = 3024
For H.C.F – write all the common factors (only once) with their smallest exponent.
H.C.F (336, 54) = 2 × 3 = 6
Verification : L.C.M (336, 54) × H.C.F (336, 54) = 336 × 54
3024 × 6 = 18144                    =>
18144 = 18144
Hence, proved.