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

Quiz of the week: AQQ256 - making square root within integer numbers world.

Good luck! And try to be among the firsts!

Kuo-Chang

P.S. Don't hesitate to share those weekly quizzes in EZ to colleagues or friends!

Updated text.
[edited by: emassa at 7:52 PM (GMT -4) on 22 Mar 2024]
• 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!