Question

The number of 6-digit numbers of them form ababab (in base 10) each of which is a product of exactly 6 distinct primes ?

Answer 1

Hint: This is a question from number theory. We have to solve them mostly through comparison. Prime numbers are those numbers whose factors are $1$ and itself. Now we know that $ababab$ is a $6$-digit number. So it can be expressed in expanded form. The expanded form of $ababab$ is ${10}^{5}a+{10}^{4}b+{10}^{3}a+{10}^{2}b+{10}^{1}a+b$ . Now we take all the $10s$ common and see what we get . Upon doing so, we write the number we get as a product of primes. After doing that, we draw out conditions for the remaining relation between $a,b$.

Complete step-by-step solution:
Now let us write $ababab$ in expanded form.
Upon doing so, we get the following :
$\Rightarrow ababab={10}^{5}a+{10}^{4}b+{10}^{3}a+{10}^{2}b+{10}^{1}a+b$ .
Now let us all the $a$ common and see what we get.
Upon doing so, we get the following :
$\begin{array}{rl}& \Rightarrow ababab={10}^{5}a+{10}^{4}b+{10}^{3}a+{10}^{2}b+{10}^{1}a+b\\ & \Rightarrow ababab=a({10}^5+{10}^3+{10}^1)+{10}^{4}b+{10}^{2}b+b\end{array}$
Now let us all the $b$ common and see what we get.
Upon doing so, we get the following :
$\begin{array}{rl}& \Rightarrow ababab={10}^{5}a+{10}^{4}b+{10}^{3}a+{10}^{2}b+{10}^{1}a+b\\ & \Rightarrow ababab=b({10}^4+{10}^2+1)+a({10}^5+{10}^3+{10}^1)\end{array}$
Now let us all the $10s$ common and see what we get.
Upon doing so, we get the following :
$\begin{array}{rl}& \Rightarrow ababab={10}^{5}a+{10}^{4}b+{10}^{3}a+{10}^{2}b+{10}^{1}a+b\\ & \Rightarrow ababab=b({10}^4+{10}^2+1)+a({10}^5+{10}^3+{10}^1)\\ & \Rightarrow ababab=b({10}^4+{10}^2+1)+10a({10}^4+{10}^2+1)\end{array}$
Now let us all the ${10}^4+{10}^2+1$ common and see what we get.
Upon doing so, we get the following :
$$\begin{array}{rl}& \Rightarrow ababab={10}^{5}a+{10}^{4}b+{10}^{3}a+{10}^{2}b+{10}^{1}a+b\\ & \Rightarrow ababab=b({10}^4+{10}^2+1)+a({10}^5+{10}^3+{10}^1)\\ & \Rightarrow ababab=b({10}^4+{10}^2+1)+10a({10}^4+{10}^2+1)\\ & \Rightarrow ababab={10}^4+{10}^2+1(10a+b)\\ & \Rightarrow ababab=10101(10a+b)\\ & \Rightarrow ababab=3\times 7\times 13\times 37(10a+b)\end{array}$$
Now $10a+b$ must be a product of primes. It has to lie between $10-100$ since a number below $10$ would give me $5$-digit number and a number above $100$ would give me a $7$-digit number.
So now let us fix one prime number.
Let one prime number be $2$. So the other prime number must be greater than $2$. This would be our lower limit. We need to set our upper limit. Our upper limit would be $50$ since $2\times 50=100$. So now, we should point out all the prime numbers between $2$ and $50$ excluding $3,7,13,37$.That would be $10$.
Now, let us fix another prime number.
Let the other prime number be $5$. So the other prime number must be greater than $5$. This would be our lower limit. We need to set our upper limit. Our upper limit would be $20$ since $5\times 20=100$. So now, we should point out all the prime numbers between $5$ and $20$ excluding $3,7,13,37$. That would be $3$.
So the total numbers would be $10+3=13$.
$\therefore$ The number of $6$-digit numbers of them form $ababab$( in base $10$) each of which is a product of exactly $6$ distinct primes is $13$. So the answer is option C.

Note: Even if we try to fix prime numbers greater than $5$ as the lower limit excluding $3,7,13,37$, we would get the same combination number again. So this would be a repetition. We would get more numbers than there are. Questions from Number theory appear pretty simple but the concepts involved in the solution are a little complex. To build up the intuition for every solution, we should be exposed to a lot of problems. So we should practice thoroughly.