Reliance Jio LCM and HCF Quiz 1

The required number will be=

= {LCM of (30   45   65   78) – 8}

LCM of 30   45   65   78 =

Here the difference between the divisor and its corresponding remainder = 17 (35-18= 17 similarly 45-28= 17 and 55-38= 17)

Therefore the required number will be = { (LCM of 35 45 55) – 17}

Let the numbers be 12a and 12b

Then 12a*12b= 2160 or ab=

Therefore the values of co-primes a and b are (1,15) and (3,5)

Hence the two-digit numbers are (3*12, 5*12), ie 36, 60

Product of two numbers = LCM x HCF

Therefore, the other number = 42*1260/210= 252

79 is the highest factor of both the number that is common.

Let the numbers be 8x, 9x, 10x.
Their LCM = 360x
So 360x = 3600
x= 10
The number are (8*10), (9*10), (10*10)
HCF = 10

Let the required number be R, Then (1837-r-5) or (1832-r) must be exactly divisible by the LCM of 8, 12, 15

Now the LCM of 8, 12, 15 = 2*2*3*2*5= 120

Now (1832-r) when divided by 120 leaves the remainder (32-r)= 0 or r= 32

Hence the required number is 32

Least number = L.C.M of (12, 16, 18 and 20) + 2

= 720 +2 = 722

The necessary digit will be the HCF of 522, 1276 and 1624

Now 522= 2*3²*29; 1276= 2²*11*29 and

1624= 2³*7*29

Therefore HCF= 2*29= 58

The required number will be = {LCM of (30 45 65 78) – 8}

LCM of 30 45 65 78 = 5*3*13*2*3 = 1170

Hence the required number = 1170-8= 1162