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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]
Parents
• 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).

• 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).

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