To know the measuring cylinder that can fill all the given capacities, they must be divisible by the required number.

98,182,266 all are divisible by 14
So  14 liters is the largest cylinder that can fill all the given cylinders.

(or)

The other method takes HCF of all given capacities i.e 98, 182 and 266.

3.78 meters =378 cm = 2 × 3 × 3 × 3 × 7

5.25 meters=525 cm = 5 × 5 × 3 × 7

Hence common factors are 3 and 7

Hence LCM = 3 × 7 = 21

Hence largest size of square tiles that can be paved exactly with square tiles is 21 cm.

C.M. of 252, 308 and 198 = 2772.

So, A, B, and C will again meet at the starting point in 2772 sec. i.e., 46 minutes and 12 seconds.

L.C.M. of 6, 9, 15 and 18 is 90.

Let required number be 90k + 4, which is multiple of 7.

Least value of k for which (90k + 4) is divisible by 7 is k = 4.

 Required number = (90 x 4) + 4   = 364.

Product of two numbers = HCF * LCM
So 2nd number = 70*1050/210 =====>>> 350

L.C.M. of 5, 6, 7, 8 = 840.
∴ 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.

LCM of 27-18-12-15 is 540.
After dividing 9999 by 540 we get 279 remainder.
So answer will be 9999-279 = 9720

In this type of question, its obvious we need to calculate the HCF, trick is
HCF of (1661 – 10) and (2045 -13)
= HCF (1651, 2032) = 127

Let the required numbers be 33a and 33b.

Then 33a +33b= 528   =>   a+b = 16.

Now, co-primes with sum 16 are (1,15) , (3,13) , (5,11) and (7,9).

Therefore, Required numbers are  ( 33 x 1, 33 x 15), (33 x 3, 33 x 13), (33 x 5, 33 x 11), (33 x 7, 33 x 9)

The number of such pairs is 4

 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.

N = H.C.F. of (4665 – 1305), (6905 – 4665) and (6905 – 1305)

= H.C.F. of 3360, 2240 and 5600 = 1120.

Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

x + y = 80, xy = 5*275 = 1375
sum of reciprocals = 1/x + 1/y

= (x+y)/xy

= 80/1375 = 16/275

LCM of 2-4-6-8-10-12 is 120 seconds, that is 2 minutes.
Now 60/2 = 30
Adding one bell at the starting it will 30+1 = 31

Let the numbers 13a and 13b.

Then, 13a x 13b = 2028

 ab = 12.

Now, the co-primes with product 12 are (1, 12) and (3, 4).

[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]

So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).

Clearly, there are 2 such pairs.

Let the numbers be 3x, 4x and 5x.

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.

LCM of 8-24-36-54 will be
2*2*2*3*3*3 = 216