Three different containers contain different qualities of mixtures of milk and water, whose measurement are 403 kg, 434 kg and 465 kg. What biggest measure must be there to measure all the different qualities exactly?

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

H.C.F. of 21, 42, 56 = 7
Number of rows of mango trees, apple trees and oranges trees are 21/7 = 3, 42/7 = 6 and 56/7 = 8
? Required number of rows = (3 + 6 + 8 ) = 17

Let the number be 81a and 81b where a and b are two numbers prime to each other.
? 81a + 81b = 1215
? a + b = 1215/81 = 15

Now, find two numbers, whose sum is 15, the possible pairs are (14, 1) , (13, 2), (12, 3), (11, 4), (10, 5), (9, 6 ), (8, 7) of these the only pairs of numbers that are prime to each other are (14, 1) , (13, 2), (11, 4), and (8, 7).

Hence, the required number are
(14 x 81, 1 x 81), (13 x 81, 2 x 81), (11 x 81, 4 x 81), (8 x 81, 7 x 81)
or (1134, 81), (1053, 162), (891, 324), (648, 567)
So, there are four such pairs .

Let the number be 16a and 16b, where a and be are two number prime to each other.

? 16a x 16b = 7168
? ab = 28
Now, the pairs of number whose products is 28, are (28, 1) : (14, 2), (7,4)
14 and 2 which are not prime to each other should be rejected.
Hence, the required answer
= (448 + 16) + (112 + 64) = 640

The largest possible number of persons in a class is given by the H.C.F. of 391 and 323 i,e. 17

? No of classes of boys = 391/17 = 23 and
No. of classes of girls = 323 / 17 = 19.

Greatest possible length of each plank = H.C.F of 42, 49, 63 = 7 m

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.

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

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.

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.