In any 4 consecutive natural numbers:
Two numbers will be divisible by 2, one of which will be divisible by 4, at least one number will be divisible by 3.
These facts give us a lower bound of 2 x 4 x 3 = 24 for the solution.
But 24 is also the product of the first 4 consecutive natural numbers, so it is also an upper bound.
Let the least multiple of 7 be x, which when divided by 90 leaves the remainder 4. Then, x is of the form 90k + 4
Now, the minimum value of k for which 90k + 4 divisible by 7 is 4
? x = 90 x 4 + 4
= 364
L . C . M of 2, 4, 6, 8, 10 and 12 is 120
So the bells will toll together after 120 second i,e 2 minutes. In 30 min. they will together in (30/2) + 1 times, i.e 16 times.
Greatest possible length of each plank = H.C.F of 42, 49, 63 = 7 m
Biggest measure = H.C.F. of 403, 434 and 465
= 31 kg
L.C.M of 6, 7, 8, 9, 12 is 504
So, the bells will toll together after 504 sec.
In hour, they will toll together
= (60 x 60) / 504 times
= 7 times
Since 2, 3, 5, 17 are prime numbers and
The given expression = (23)20 x (3 x 5)24 x (17)15
= 260 x 324 x 524 x 1715
So the total number of prime factors in the given expression is (60 + 24 + 24 + 15 ) = 123.
Interval of changes = (L . C . M of 48, 72, 108) sec. = 432 sec.
So, the lights will simultaneously changes after every 432 seconds i,e 7 min. 12 sec.
So, the next simultaneously changes will take place at 8 : 27 : 12 hrs.
Given, length of four metal roads are 78, 104, 117 and 169 cm.
Now,
78 = 13 x 2 x 3,
104 = 13 x 2 x 2 x 2,
117 = 13 x 3 x 3,
169 = 13 x 13
Length of each piece of rod as long as possible, HCF = 13 CM
? Number of piece = 6 + 8 + 9 + 13 = 36
Gardening group meets once in 2 days, electronics group meets once in 3 days, chess group meets once in 4 days, yachting group meets once in 5 days and the photography group meets once in 6 days.
If they meet on the same day at one time, then the next time they will meet on same day again will be the LCM of 2, 3, 4, 5 and 6 which is equal to 60.
Hence, within 180 days all the five groups will meet on the same day = 180/60 = 3 times.
The time at which all the three persons meet will be the LCM of the time taken by each person individually to complete one round.
252 = 22 x 71 x 91
308 = 22 x 71 x 111
198 = 21 x 91 x 111
? LCM of 252, 308 and 198 = 22 x 71 x 91 x 111 = 2772
So, A, B and C will again meet at the starting point in 2772s i.e., 46 min and 12s
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