KCC's Quizzes on AQQ256 about square roots in the digital world

N² = 12,345,678,987,654,321; what is N?

11² = 121

So, 110,000,000² = 12,100,000,000,000,000

12² = 144

So, 120,000,000² = 14,400,000,000,000,000

Therefore, the number must be between the two of these.

But the carry between units and 10’s in 11² is interesting because it makes the number a palindrome…

Let’s see if this is a pattern that repeats:

111² = 111 * 111

111² = 111 * 1 + 111 * 10 + 111 * 100 = 111 + 1110 + 111000 = 12321

1111² = 1111 + 11110 + 111100 + 1111000 = 1234321

…

111,111,111² = 12,345,678,987,654,321

N = 111111111

N = 10^[log(12345678987654321)/2]  ;My calculator had enough resolution to give an integer result.

To prove that N = 111111111 square it by hand calculation!

The reasoning will go like this. 

1*1= 1, 11*11=121, 111*111= 12321 and they are in Palindromic and it will go till the no of digits are there. 

For 12345678987654321 as per above logic the answer will be 111111111

My thought process…

I’m going to call K = 12345678987654321 for simplicity.

N has to be divisible by 9 since the sum of the numbers in K are divisible by 9.

Since the answer (K) ends in a 1 and it’s the square of a number, N must end in a 1 or 9 since those are the only two numbers whose square is a 1 (1 and 81).

I note that N has to lie within the bounds of 100,000,000 (which gives 17 digits as an answer), and 1,000,000,000,000 (lowest to give 18 digits), so it will be a 9 digit number.

I notice there is a similarity in K with the square of 11 (= 121) and 111 (12321) as the pattern of increasing numbers and decreasing numbers are shared.

Make a guess that the solution (N) may be made up of only the digit 1 (ie, 1111….) and since the solution must be divisible by 9, as well as be a 9 digit number, I guess the solution is…

N = 111,111,111

Plugging into N2 reveals the guess was correct!  N2 = 12345678987654321.

- Brian

1. N=111 111 111

2. You observe:

1²=1

11²=121 

111²=12321

1111²=1234321

You can prove by induction that:

111111111²=12345678987654321

1. 111,111,111

2. Intuitively, knowing how multiplication works with decimal places being 10^n, having 111,111,111 multiplied by itself would yield this number multiplied by itself and shifted decimal places to the left should yield this result.

1.  111111111

2.   Last digit of the square is a 1, so  the last digit of the root is either a 1 either a 9  (cannot be even, neither 3, 5, nor 7, since their last digit of those, squared, do not results in a 1 (or modulo 10, if you prefer) ). 

The second last digit of the square is a 2, to we either have, for the root,  ....11  or  ...89.
Indeed  11^2  = 121 = 21 modulo 100  and we also have 89^2 = 7921 = 21 modulo 100

We can eliminate the root like: 100c + 89 (c is an integer from 0 to 10), since, for the last three digits of the square,  we should have its modulo 1000 equals to 321  :

(100c + 89) ^2  = 321  mod 1000,  

which has no solution in non-negative integers c in the spedified range.

We are left with  (100c + 11)^2  = 321  mod 1000, giving  c = 1.

So the only possible root is ...111.  If we continue like that, we find the previously solution given  (in fact, I suspected the continuation with the ones, and short-cut the process with a numerical confirmation of my intuition).  

I started with 11 * 11 = 121, similar is 111 * 111 = 12321 In the middle is 3 => 3times 1. In the quiz is in the middle 9, thus N=111 111 111

111,111,111

(11)^=121

(111)^2=12321

Similar way we will get palindrome no 12345678987654321 

Bingo Martin! You get it and with a brilliant development and justifictaion!