2 | 24 - 36 - 40 -------------------- 2 | 12 - 18 - 20 -------------------- 2 | 6 - 9 - 10 ------------------- 3 | 3 - 9 - 5 ------------------- | 1 - 3 - 5 L.C.M. = 2 x 2 x 2 x 3 x 3 x 5 = 360.
So, the numbers 12 and 16.
L.C.M. of 12 and 16 = 48.
On dividing 2497 by 60, the remainder is 37.
∴ Number to be added = (60 - 37) = 23.
84 = 22 x 3 x 7
∴ H.C.F. = 22 x 3 = 12.
Then, their L.C.M. = 6x.
So, 6x = 48 or x = 8.
∴ The numbers are 16 and 24.
Hence, required sum = (16 + 24) = 40.
= 1008 + 7
= 1015
So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.
128352) 238368 ( 1 128352 --------------- 110016 ) 128352 ( 1 110016 ------------------ 18336 ) 110016 ( 6 110016 ------- x ------- So, H.C.F. of 128352 and 238368 = 18336. 128352 128352 ? 18336 7 Therefore, ------ = -------------- = -- 238368 238368 ? 18336 13
∴ Required number is of the form 840k + 3
Least value of k for which (840k + 3) is divisible by 9 is k = 2.
∴ Required number = (840 x 2 + 3) = 1683.
Then, their L.C.M. = 60x.
So, 60x = 2400 or x = 40.
∴ The numbers are (3 x 40), (4 x 40) and (5 x 40).
Hence, required H.C.F. = 40.
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