# Square Root Day Promotional Event: Some Featured Responses

We recently had a promotional email giving discounts for answering questions. The response was very heartening. I was thrilled by the number of students who tried answering the question.

Quite a few students got the answers right and so it would be remiss of us not to publish some of the better solutions we have got.

The questions are as follows

1. There is only one 4-digit number of the form ‘aabb’, where a and b are digits that is a perfect square. Find that
2. There are a few square numbers of the form ‘abcabd’ where a, b, c, d are digits and d = c + 1. Give 2 of these.

Devdutta Agarwalla, Chinmaya Hegde, Devjyot, were among the first to get to the answers along with detailed solutions. Having said that, quite a few got it right and we offered the 49% discount to way more than 3 students that we had originally planned.

We had promised the 49% discount and did not have the heart to give it only to the first three students.

There were multiple solutions for the questions. I am outlining the best few

Best Solutions to the first question are as follows

From Devdutta Agarwalla:

‘aabb’ a 4 digit number, It must be between 322-992
so it is 1000 a + 100a + 10 b + b  = 1100 a + 11 b = 11(100 a + b).
Now for it to be a perfect square 100a + b has to be divisible by 11 and so a + b has to be divisible by 11 so a + b should be 11 as maximum sum of 2 digits is 18 and 100a+b has to be (11c)2
now a perfect square ends on 0, 1, 4. 5, 6, 9
But it cannot be zero then a> 9, it does not end with 11, 55,66 or 99.
So 44 is the only choice. So b= 4 and that gives a = 7
100 a + b = 704 = 11 * 64 so ans is 7744

From Chinmaya Hegde:

A number of the form AABB can be factored as A0B * 11
It is a perfect square and hence it should be divisible by 11 (contain another factor of 11).
Ergo, A0B is divisible by 11. According to the division rule of 11, A+B must be a multiple of 11.
Since A and B are digits, A+B=11.
So the various possibilities for A and B (in A0B, so that it is divisible by 11 and a perfect square) are:
A=2, B=9   209/11 = 19, it is not a perfect square
A=3, B=8   308/11 = 28, it is not a perfect square
A=4, B=7   407/11 = 37, it is not a perfect square
A=5, B=6   506/11 = 46, it is not a perfect square
A=6, B=5   605/11 = 55, it is not a perfect square
A=7, B=4   704/11 = 64, it is a perfect square…..(8square = 64) **
A=8, B=3   803/11 = 73, it is not a perfect square
A=9, B=2   902/11 = 82, it is not a perfect square
** the only perfect square is 64 and the corresponding A0B value is 704 (which is divisible by 11).

The third one in this list is my favourite. Why is this my favourite you ask? Because I also solved this question using the same method 🙂

From Devjyot Singh:

N2  is of the form ‘aabb’.
So, 11 | N2
11|N
N can take values 33, 44, 55, 66, 77, 88 or 99
Simple trial and error gives us 882 which is equal to 7744

The best solution (From Devdutta Agarwalla) was as follows:

abcabd is a perfect square.
So abcabd = x^2
Now abcabd = abcabc + 1.
So abcabc= x^2 – 1
abcabc= (x +1)(x – 1)
So on simplifying we get
abcabc = abc * 1001.
1001=7*11*13
So from the first hit and trial we get
Now, ‘abcabd’ is a perfect square. So, say, ‘abcabd’ = x^2
Now, ‘abcabd’ = ‘abcabc’ + 1. So, ‘abcabc’ = x^2 – 1

After this, we are halfway there to the solution

‘abcabc’ = (x +1) (x – 1)

Now, ‘abcabc’ = ‘abc’ * 1001. Or, ‘abcabc’ = ‘abc’ * 1001. This is the key idea that is very vital in a bunch of questions.

‘abc’ * 1001 = ‘abc * 7 * 11 * 13 = (x – 1) (x + 1)

So, between x -1 and x + 1, we need to account for 7, 11 and 13 as factors. Wherever this works, we are through. Now, x – 1 or x + 1, either number alone cannot be a multiple of 7, 11 and 13. So, one of these two should be a multiple of one of the three primes and the other should be a multiple of the other two.

So, of the two numbers one can be a multiple of 7 and the other 143. or,
one can be a multiple of 11 and the other 91. or,
one can be a multiple of 13 and the other 77,

After this we are down to trial and error; but a very scientific form of trial and error.

Let us say, we are picking two numbers such that one is a multiple of 7 and the other of 143. Since 143 is the far larger number, it helps to look for multiples of 143 and see if we can spot the scenario where the other number is a multiple of 7.

If one number were 143, the other would have to  be 145 or 141. Neither is a multiple of 7. This does not work.

If one number were 286, the other would have to  be 284 or 288. Neither is a multiple of 7. This does not work either.

If one number were 429 (143 * 3), the other would have to  be 427 or 429. 427 is a multiple of 7. Houston, we have an answer!

(x -1) * (x + 1) = 427 * 429 works. 427 * 429 = 183183 by taking 1001 into consideration and in the next hit and trial immediately we get 328329

The above solution was provided in gorgeous detail.

Interestingly, 4 numbers satisfy the condition given in the question. Most of the students had got 183184 and 328329.

The other two numbers are 715716 and 528529. Students got different combinations of two out of these 4, but surprisingly no one had mentioned all 4.

Perhaps, the thrill of getting two took some of the rigour away and prevented students from going for all the answers. May be next time, we will give additional discounts for getting all possible answers.

The answers to the verbal questions were even more fun. I have given a few best sentences below:

1. On a pitch which is turning SQUARE Virat Kohli played a SQUARE drive of a ball which spun a SQUARE km. Any other batsmen would’ve been SQUARED up with that delivery. ( all cricketing uses from Raveendra)
2. Thanks for the challenge Rajesh, I guess now we are square. (Shrey Shekhar)
3. But I think I am little late to revert and might not get the discount so I am back to square one (Shrey Shekhar)
4. A mission for the square mind. What could be a better way to celebrate square day than to square off with mysterious mind-boggling problems. 2IIM team has now and again proven to divide my doubts, multiply curiosity and square productivity.When my phone beeped a mail from 2IIM with a quest, I squared off the rest of my work as quickly as I could, grabbed a pen and set to work. Unfortunately, my solutions did not square with the requirements of this mission. I felt disappointed and frustrated but like all mathematicians and problem solvers, I was convinced that I proceeded right and had therefore arrived at a square answer. So I put my doubts aside, decided to mail my solution to Rajesh, hoping to win this contest, fair and square.:) (Shruthi Muralidharan)
5. They were ready to answer the questions square in the face. (Prathima Chivakula)

We also got a series of cute replies from students which were heartening. A great many had dropped a mail to broadly say “Hi, We loved the mail. Keep them coming.” The list included ones who were not preparing for CAT and students who had already decided to join an MBA in 2016. It was wonderful to know that the thrill in solving the question actually exceeded the joy in getting whatever discount was available.

We have resolved to keep the emails and questions coming thick and fast. From now on, I will replicate all the emails on this blog as well.