Question

What number am I ?
I am a two-digit even number.
I am a common multiple of 6 and 7.
I have a total of 8 factors.
A.35
B.42
C.36
D.84

Answer 1

Hint: To answer this type of logical question just suppose the unknown number as xy then factorise it as it has already said that it has two common multiple so consider these two numbers. Once a two digit even number is found just check if it contains 8 factors or not.

Complete step-by-step answer:
Let xy be a two digit number
Given xy have common multiple 6 and 7
\[
  { \Rightarrow xy = 6 \times 7 \times y \times z} \\
  { = 42 \times y \times z} \;
\]
Also given it has 8 factors.
Lets check for 42 first
\[
  {42 = 21 \times 2 = 2 \times 21} \\
  { = 6 \times 7 = 7 \times 6} \\
  { = 14 \times 3 = 3 \times 14} \\
  { = 1 \times 42 = 42 \times 1} \;
\]
In the above factorisation of 42 we have seen that it has 8 factors which are 2,21,6,7,14,3,1 and 42.
42 has 8 factors and a common multiple of 6 and 7.
Hence the required even number of two digits which is a common multiple of 6 and 7 and having a total 8 factors is 42.
Therefore option B is the correct answer.
So, the correct answer is “Option B”.

Note: Here in this solution we have considered 1 and 42 also two factors of 42 as in this question it is not mentioned that don’t consider 1 and 42 as factors of 42. So students don’t get confused