Question

Which of the following numbers have the greatest number of divisors?
A. 176
B. 182
C. 99
D. 101

Answer 1

Hint: We first find the prime factorisation of the given numbers. From those prime factorisations we find the factors of the numbers. We then find which number has the greatest number of divisors.

Complete step by step solution:
We first find the prime factorisation of the given numbers and using the primes we find the factors of the numbers.
The prime factorisation of 176 is
\[\begin{align}
  & 2\left| \!{\underline {\,
  176 \,}} \right. \\
 & 2\left| \!{\underline {\,
  88 \,}} \right. \\
 & 2\left| \!{\underline {\,
  44 \,}} \right. \\
 & 2\left| \!{\underline {\,
  22 \,}} \right. \\
 & 11\left| \!{\underline {\,
  11 \,}} \right. \\
 & 1\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}\]
The factors are $1,2,4,8,11,16,22,44,88,176$. It has 10 factors.
The prime factorisation of 182 is
\[\begin{align}
  & 2\left| \!{\underline {\,
  182 \,}} \right. \\
 & 7\left| \!{\underline {\,
  91 \,}} \right. \\
 & 13\left| \!{\underline {\,
  13 \,}} \right. \\
 & 1\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}\]
The factors are $1,2,7,13,14,26,91,182$. It has 8 factors.
The prime factorisation of 99 is
\[\begin{align}
  & 3\left| \!{\underline {\,
  99 \,}} \right. \\
 & 3\left| \!{\underline {\,
  33 \,}} \right. \\
 & 11\left| \!{\underline {\,
  11 \,}} \right. \\
 & 1\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}\]
The factors are $1,3,9,11,33,99$. It has 6 factors.
The number 101 is prime. So, it has only 2 factors 1 and 101.
The maximum number of factors is present in 176.

So, the correct answer is “Option A”.

Note: We need to remember that the GCF has to be only one number. It is the greatest possible divisor of all the given numbers. If the given numbers are prime numbers, then the GCD of those numbers will always be 1.
Therefore, if for numbers $x$ and $y$, the GCD is $a$ then the GCD of the numbers $\dfrac{x}{a}$ and $\dfrac{y}{a}$ will be 1.