Real Numbers Test - 11

Using the result,
HCF × LCM = Product of two natural numbers
 ⇒ LCM (a, b) = 1800/12 = 150

Smallest prime number = 2 and smallest composite number = 4

∴ HCF (2, 4) = 2

Let us subtract 5 (the remainder) from each number in order to find their HCF.

245 - 5 = 240

1029 - 5 = 1024

Now, Let us find HCF  of 240  , 1024

1024 = 240 × 4 + 64

240 = 64 × 3 + 48

64 = 48 × 1 + 16

48 = 16 × 3 + 0

The largest number which divides 245 and 1029 leaving remainder 5 in each case is 16.

Let us subtract 1 from 33 and 3 from 75 in order to find their HCF

33 -1 = 32

75 - 3 = 72

To find HCF  of 32 , 72, 

72 = 32 × 2 + 8

32 = 8 × 4 + 0

8 is the largest number which divides 33 and 75, leaving remainders 1 and 3 respectively.

LCM (12, 15, 18, 27) = 540

Now, largest four digit number = 9999

∴ 9999 ÷ 540 = 18 × 540 + 279 (Remainder = 279)

Therefore, the largest number of 4 digits exactly divisible by 12, 15, 18 and 27 is 9999 – 279 = 9720

LCM (12, 15, 18, 27) = 540

Now, smallest four digit number = 1000

∴ 1000 ÷ 540 = 1 × 540 + 460 (Remainder = 460)

Therefore, the smallest number of 4 digits exactly divisible by 12, 15, 18 and 27 is 1000 + (540 – 460) = 1000 + 80 = 1080

We know that maximum number columns = HCF ( of 616 , 32)

Applying Euclid’s division algorithm to find HCF of two numbers

⇒ 616 = 32 × 19 + 8

⇒ 32 = 8 × 4 + 0 [Zero remainder]

∴ HCF(616, 32) = 8

Therefore, the maximum number of columns in which two groups can march is 8.