Question : Find the L.C.M and H.C.F of the following pairs of integers and verify :
L.C.M \(\times\) H.C.F = Product of the two numbers
(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54
Solution :
(i) 26 and 91
26 = 2 \(\times\) 13 and 91 = 7 \(\times\) 13
\(\therefore\) L.C.M of 26 and 91 = 2 \(\times\) 7 \(\times\) 13 = 182
and H.C.F of 26 and 91 = 13
Now, 182 \(\times\) 13 = 2366 and 26 \(\times\) 91 = 2366
Hence, 182 \(\times\) 13 = 26 \(\times\) 91
(ii) 510 and 92
510 = 2 \(\times\) 3 \(\times\) 5 \(\times\) 17 and 92 = 2 \(\times\) 2 \(\times\) 23
\(\therefore\) L.C.M of 510 and 92 = 2 \(\times\) 2 \(\times\) 3 \(\times\) 5 \(\times\) 17 \(\times\) 23 = 23460
and H.C.F of 510 and 92 = 2
Now, 23460 \(\times\) 2 = 46920 and 510 \(\times\) 92 = 46920
Hence, 23460 \(\times\) 2 = 510 \(\times\) 92
(iii) 336 and 54
336 = 2 \(\times\) 2 \(\times\) 2 \(\times\) 2 \(\times\) 3 \(\times\) 7
and 54 = 2 \(\times\) 3 \(\times\) 3 \(\times\) 3
\(\therefore\) L.C.M of 336 and 54 = 2 \(\times\) 2 \(\times\) 2 \(\times\) 2 \(\times\) 3 \(\times\) 3 \(\times\) 3 \(\times\) 7 = 3024
and H.C.F of 336 and 54 = 2 \(\times\) 3 = 6
Now, 3024 \(\times\) 6 = 18144 and 336 \(\times\) 54 = 18144
Hence, 3024 \(\times\) 6 = 336 \(\times\) 54
