Required number = (L.C.M. of 12, 15, 20, 54) + 8
= 540 + 8
= 548.

(OR)

Given numbers are 12, 15, 20, 54

To Find: least no. which is divided by given nos. and leaves remainder 8

Least no which is divisible by all given no is LCM of all nos.

LCM means the least common multiple.

First, we find LCM of 12, 15, 20, 54  by prime factorization method

12 = 2 × 2 × 3

15 = 3 × 5

20 = 2 × 2 × 5

54 = 2 × 3 × 3 × 3

LCM ( 12, 15 , 20 , 54 ) = 2 × 2 × 3 × 3 × 3 × 5 = 540

To find the required no. we add 8 to LCM

⇒ Required No. = 540 + 8 = 548

Thus, 548 is the least no which when divided by 12 15 20 and 54 leaves in each case a remainder of 8.

First number * second number = LCM * HCF
52*Second number = 520 * 4
Second number = (520*4)/52
= 40
Then the other number is 40.

Take LCM of 3, 4, 5, 6, 7, and 8. It will be = 840.

Now, 10000/840 = 760 (remainder).

840 – 760 = 80. If we add 80 in the given number, the number is exactly divisible by 840.

So, the required number is 10080.

For any this type of question, remember
Product of two numbers = Product of their HCF and LCM
So Other number = 693×11/99 = 77

LCM = 1260

1260 + 7 = 1267

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.

Required number = H.C.F. of (1657 – 6) and (2037 – 5)

= H.C.F. of 1651 and 2032 = 127.

First number * Second number = LCM * HCF
Let the second number be x.
10x = 30*5
x = 150/10 = 15

Greatest three digit number = 999
LCM of 6, 9 and 12

LCM = 2 x 2 x 3 x 3 = 36

On dividing 999 by 36,

Remainder = 27.

∴ The greatest three digit number divisible by 6, 9, and 12

= (999 – 27) = 972

As per the question, the required number is (972 + 3) = 975.

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.

Total Cards = 52. Numbered Cards = (2, 3, 4, 5, 6, 7, 8, 9, 10) 9 from each suit 4 × 9 = 36
P (E) = 36/52 = 9/13

Or

If you count an ace as a number card there are 40 in a pack so p(get a number card) = 40/52 = 10/13.

If aces don’t count then there are 9 x 4 = 36 in a pack so p(get a number card) = 36/52 = 9/13.

If we want to pack the drinks in the least number of cans possible, then each can should contain the maximum numbers of liters possible.As each can contains the same number liters of a drink, the number of liters in each can is a comman factor for 80,144 and 368; and it is also the highest such factor, as we need to store the maximum number of liters in each can.

So, the number of liters in each can  = HCF of 80,144 and 368 = 16 liters.

Now, number of cans of Maaza = 80/16 = 5

Number of cans of Pepsi = 144/16 = 9

Number of cans of Sprite = 368/16 = 23

Thus, the total number of cans required = 5 + 9 + 23 = 37