1 2 3 5 4 1 2 4 3 5 1 2 4 5 3 ⋮
The above numbers illustrate some of the ways to arrange the five distinct digits 1, 2, 3, 4, 5.
If I list out all the possible numbers, how many of them can be expressed as N × N for some integer N ?
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That's the same i thought, quite a simple and fast way to solve this. However, to ensure everyone understands, you could clarify the using of divisibility rules and, why the numbers divisible by 3 but not by 9 are not square numbers (I think it isn't obvious for someone not so used to number theory).
Theorem
All perfect squares have digital roots of either 1 , 4 , 7 , or 9 . It must be noted, however, that the converse, (all numbers with digital roots of 1 , 4 , 7 , or 9 are perfect squares) is not true.
Proof
By the division algorithm, every integer n is of the form 9 k + r , where 0 ≤ r < 9 . So n ≡ r ( m o d 9 ) and hence n 2 ≡ r 2 ( m o d 9 ) .
Since r ≡ r − 9 ( m o d 9 ) ,
0 2 ( ± 1 ) 2 ( ± 2 ) 2 ( ± 3 ) 2 ( ± 4 ) ≡ 0 ( m o d 9 ) ≡ 1 ( m o d 9 ) ≡ 4 ( m o d 9 ) ≡ 0 ( m o d 9 ) ≡ 7 ( m o d 9 )
Thus, n 2 is congruent to 0 , 1 , 4 , or 7 , so its digital root is 1 , 4 , 7 , or 9 .
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Now suppose all possible 5-digit combinations of 1 2 3 4 5 were perfect squares. Their digital roots would all be 1 + 2 + 3 + 4 + 5 = 1 5 → 1 + 5 = 6 . Since the only digital roots of perfect squares are 1 , 4 , 7 , or 9 , we conclude by contradiction that 1 2 3 4 5 and all of its 5-digit combinations cannot be perfect squares.
There's a simpler solution.
Hint: If mod 3 = 0, then mod 9 = 0 too!
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Divisibility rules of 3 and 9 tell us that all of the numbers are divisible by 3 , but not by 9 , hence 0 of them are square numbers.