CAT Online Class – Interesting one from Number Systems

Question
Actually, this is not a CAT preparation question. This is way tougher than what one can expect to see in CAT. I saw it on some Olympiad paper. But this is a wonderful question to think about one very interesting property. Anyway, let us get to the question –

A six-digit number has digits ‘abcabd’, where a, b, c, d take values from 0 to 9. Further we know that d = c + 1 and that this number is a perfect square. Find the number. If there is more than one number possible, find all such numbers.

Answer
183184, 328329, 528529, 715716

Explanation
This is a fabulous question. It bears repeating that this is far far tougher than what we are likely to see in CAT. Nevertheless we will look at this question in order to discuss one key idea.

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. Or, 183184 is a perfect square. 428 * 428.

Now, all we need to do is check out all other variants. But having said that, even this can be a very time-consuming process. So, let us see if we can fine-tune this a touch. If one number is a multiple of 143, the other number should be a multiple of 7. So, the first number  + 2 or first number -2 should be a multiple of 7. If we say the first number were k, then either k + 2 or k – 2 should be a multiple of 7. In that case, we are through. or, k divided by 7 should give us a remainder of 2 or 5.

Now, 143 divided by 7 gives us a remainder of 3. This is why this does not work

2 * 143 divided by 7 will give us a remainder of 6. This does not work either
3 * 143 divided by 7 will give us a remainder of 9, which is same as 2. This works. This is what gave us the solution 183184
4 * 193 gives us a remainder of 12, which is same as 5. This should also work. Let us check this out. 4 * 143 = 572. 574 is a multiple of 7. Or, 572 * 574 will be a multiple of 1001.
572 * 574 = 328328. or, 328329 = 573 *573. We have got our second possibility!!

183184 and 328329 are both perfect squares.

Now, let us move to 5 * 143. This divided by 7 gives us a remainder of 15, which is same as 1. This does not work either

Let us move to 6 * 143. This divided by 7 gives us a remainder of 18, which is same as 4. This does not work either

7 * 143 = 1001. That is a 4-digit number, so we do not have to worry about this.

So, we have got two solutions so far. With the numbers being divided as product of 143 and of 7. Let us move to the combination of the numbers being broken as product of 91 and 11.

Now, let us go directly to the remainder approach. 91 divided by 11 gives us a remainder of 3. We need to find some multiple of 91 that on division by 11 gives us a remainder of either 2 or 9. (Why? scroll up to see this. We are looking for numbers that differ by 2 which between them account for all factors of 1001)

1 * 91 divided by 11 will give us a remainder of 3. This does not work.
2 * 91 divided by 11 will give us a remainder of 6. This does not work.
3 * 91 divided by 11 will give us a remainder of 9. This should work. Let us check this out. 273 is a multiple of 91, 275 is a multiple of 11. 273 * 275 should be a multiple of 1001. 273 * 275 = 75075. 75076 is 274 * 274. But this does not count because it has only 5 digits. Close, but no cigar. Let us take this further

4 * 91 divided by 11 will give us a remainder of 12, which is same as 1 This does not work.
5 * 91 divided by 11 will give us a remainder of 15, which is same as 4 This does not work either
6 * 91 divided by 11 will give us a remainder of 18, which is same as 7 This does not work either
7 * 91 divided by 11 will give us a remainder of 21, which is same as 10 This does not work either
8 * 91 divided by 11 will give us a remainder of 24, which is same as 2. Hello, do we have our third number here?!

8 * 91 = 728. 726 is a multiple of 11. 728 * 726 should be a multiple of 1001. 726 * 728 = 528528. 528529 = 727*727. So, our third number is 528529.

Let us go further, but quicker from now on.

9 * 91 divided by 11 will give us a remainder of 27, which is same as 5 This does not work.
10 * 91 divided by 11 will give us a remainder of 30, which is same as 8 This does not work either.

We do not need to worry about 11 * 91 as that is 1001.

Now, let us move on to numbers that break as 77p * 13q.

We need a multiple of 77 that when divided by 13 gives a remainder of either +2 or + 11. 77 divided by 13 gives a remainder of 12, or -1.

77 * 2, on division by 13 will give us a remainder of -2. This should work. But this will be a waste of time as it will result in a 5-digit number.

So, let us evaluate 77 * 3. This gives a remainder of -3. Not good enough.
Straight away, we can sense that the next number that will work for us is 77 * 11, where we get a remainder of -11, which is nothing but + 2.

77 * 11 = 847. 845 is a multiple of 13. 845 *847 is a multiple of 1001. 847 * 845 = 715715. 846 *846 = 715716 is our fourth number. 

No other possibility exists.

There are 4 possible numbers – 183184, 328329, 528529 and 715716. Fantabulous question to nail down the fact that 1001 = 7 * 11 * 13. 

Since you have been brilliant and diligent enough to get through this far to the solution, I am going to leave you with another nugget. ‘abcdabcd’ = ‘abcd’ * 10001.

This 10001 is not prime. It is a product of two primes. Which two? It is not for nothing that we at 2iim say CAT preparation can be fun.

Again, for the nth time, let me reiterate that the above question is way tougher than what we will find in CAT. If you want questions (lots of them) at or around CAT level, visit the questionbank.

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