Tag: be my multiple, i'll be your factor

Questions Related to be my multiple, i'll be your factor

The 1st three common multiple of numbers $12,8,16 $ are

maths be my multiple, i'll be your factor co-prime numbers lcm lowest common multiple (l.c.m.)

  1. $12,24,36$

  2. $8,16,24$

  3. $16,32,48$

  4. $48,96,144$

Show answer
Correct Option: D
Explanation:
$12 = 2^{2}\times3$
$8 = 2^{3}$
$16 = 2^{4}$

$\Rightarrow$ LCM of $12,8,16 = 2^{4}\times3 = 48$

$\therefore 48$ is the least common multiple of $12,8,16$. Thus, all multiples of $48$ are common multiples of $12,8$ and $16$.

$\therefore$ First three common multiples $= 48,96,144$

Find first five common multiples of $1,2$ and $3$.

maths be my multiple, i'll be your factor co-prime numbers lcm lowest common multiple (l.c.m.)

  1. $2,4,8,10,20$

  2. $3,6,12,30,60$

  3. $6,12,18,24,30$

  4. $1,2,3,4,5$

Show answer
Correct Option: C
Explanation:
$\Rightarrow$ LCM of $1,2,3 = 1\times2\times3 = 6$

$\therefore 6$ is the least common multiple of $1,2,3$. Thus, all multiples of $6$ are common multiples of $1,2$ and $3$.

$\therefore$ First five common multiples $= 6,12,18,24,30$

Select the correct option.
The HCF and the LCM of $12, 21, 15$ respectively are

maths be my multiple, i'll be your factor co-prime numbers lcm lowest common multiple (l.c.m.)

  1. $3, 140$

  2. $12, 420$

  3. $3, 420$

  4. $420, 3$

Show answer
Correct Option: A
Explanation:

Numbers $= 12, 15, 21$


$12 =  2 \times 2 \times 3$


$15 = 3 \times 5$

$21 = 3 \times 7$

HCF = Product of smallest power of each common prime factor $= 3' = 3$
LCM = Product of greatest power of each prime factor 

$2^2 \times 3 \times 5 \times 7 = 4 \times 3 \times 5 \times 7 = 420$

$(C) \,\, 3, 420$

The greatest number which when subtracted from 5834, gives a number exactly divisible by each of 20, 28, 32 and 35 is

maths be my multiple, i'll be your factor co-prime numbers lcm lowest common multiple (l.c.m.)

  1. 1120

  2. 4714

  3. 5200

  4. 5600

Show answer
Correct Option: B
Explanation:

 Number which is exactly divisible by 20, 28, 32 and 35 should be the common multiple of all these.
$20 = 2^2 \times 5$
$28 = 2^2 \times 7 \Rightarrow 32 = 2^5$
35.=5 $\times $ 7
LCM = $2^5 \times 5 \times 7 = 1120$
Hence the greatest number that should be subtracted
=5834-1120 = 4714

GCF of $99$ and $100$ is __________

maths be my multiple, i'll be your factor proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

  1. $1$

  2. $2$

  3. $3$

  4. $4$

Show answer
Correct Option: A
Explanation:

when we divide $100$ with $99$ we will get remainder $1$ and hence next we divide 99 with $1$ which gives us remainder $0$.


therefore $1$ is the HCF or GCF of $99 $ and $100$