Question

A man had $ 32 $ , $ 48 $ , $ 64 $ beads in $ 3 $ different bags. The least number of beads were blue in each bag. Each bag has the same number of blue beads. How many blue colour beads were in each bag:
A. $ 20 $
B. $ 10 $
C. $ 16 $
D. $ 24 $

Answer 1

Hint: Take the Highest common factor of the number of beads in each bag to take the common factor of all the number of beads which will be the number of blue beads in each bag.

Complete step-by-step answer:
The H.C.F. of any group of numbers is the highest common factor of all the numbers in the group of the numbers. The H.C.F. of group of numbers divides all the numbers of the group.
As the number of blue beads in each bag are the same. So, the number of blue beads in each bag is the highest common factor of the number of beads in each bag.
Take the prime factorization of the number of beads in each bag:
 $
\Rightarrow 32 = {2^5} \\
\Rightarrow 48 = {2^4}.3 \\
\Rightarrow 64 = {2^6} \;
  $
The highest common factor out of the number of beads in each bag is equal to $ {2^4} = 16 $ .
So, the number of blue beads in each bag is equal to $ 16 $ .
So, the correct answer is “16”.

Note: Take the prime factorization of the numbers starting with $ 2 $ and taking the powers as $ 0 $ for those whose prime numbers are not involved in the prime factorization of the number to take out the highest common factor.