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Can there be a perfect square ?
Problem 1Can there be a perfect square whose digits consist of exactly 4 ones, 4 twos and 4 zeros in any order?
Solution
Let N = be such a number.
Since the sum of digits of the number " N " 4*1 + 4*2 +4*0 = 4 + 8 + 0 = 12 is divisible by 3,
it implies that the number N itself is divisible by 3 (the "divisibility by 3 rule").
In turn, it implies that the number " n " itself is divisible by 3.
Then the number is divisible by 3^2 = 9; hence, the number N is divisible by 9.
But the sum of the digits of the number N, which was calculated above as 12, is not divisible by 9.
It contradicts to the "divisibility by 9 rule".
Hence, such a number N with assigned properties DOES NOT EXIST.
The proof is completed.
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On divisibility rules by 3 and by 9 see the lessons
- Divisibility by 3 rule
- Divisibility by 9 rule
in this site.
My other lessons in this site on divisibility numbers are
- Divisibility by 2 rule
- Divisibility by 3 rule
- Divisibility by 4 rule
- Divisibility by 5 rule
- Divisibility by 6 rule
- Divisibility by 9 rule
- Divisibility by 10 rule
- Divisibility by 11 rule
- Restore the omitted digit in a number in a way that the number is divisible by 9
- Restore the omitted digit in a number in a way that the number is divisible by 11
- Math circle level problems on divisibility numbers
- Math circle level problem on restoring digit in the product of two 16-digit numbers
- Math circle level problem on finding remainders
- OVERVIEW of Divisibility rules by 2, 3, 4, 5, 6, 9, 10 and 11
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