TIFR CSE 2021 | Part A | Question: 7

Question

Let $d$ be the positive square integers (that is, it is a square of some integer) that are factors of $20^{5} \times 21^{5}$. Which of the following is true about $d$?

• A. $50\leq d< 100$
• B. $100\leq d< 150$
• C. $150\leq d< 200$
• D. $200\leq d< 300$
• E. $300\leq d$

Answer 1

Soln:-

Comments

nice

Answer 2

Answer $(E)$

$= 20^5 \times 21^5$

$= (2^2 \times 5^1)^5 \times (7 \times 3)^5$

$= 2^{10} \times 5^5 \times 7^5 \times 3^5$

Now to form the numbers which are perfect squares, for each of $2,3,5,7$ we should pick ONLY even powers.

So the numbers possible to be present in the perfect square are $\{2^2,2^4,2^6,2^8,2^{10},3^2,3^4,5^2,5^4,7^2,7^4\}$

Now, we can take any subset of this and multiply the numbers present in that subset to get the perfect square number. For example, if we pick the subset $\{2^4, 7^2\}$, then the number will be $2^4 \times 7^2 = 784 = (28)^2$

Total subsets possible = $2^{11}$

But, we don’t have to consider the empty set (as $0$ is only factor of itself and question also asks for positive numbers only).

So the answer should be $2^{11} – 1 = 2047$.

Comments

Total subsets will not be 2^11.

Because if you count in this way you can take both 2^6 and 2^8 but 2 can be at most

power of 10. So you OVERCOUNT here. I think we have to check each case from here.

If one checks it, he will get the d value between 150 and 200.

So Option C is correct.