Laws of divisibilities comes in pairs

12354 12435 12453 12354 \\ 12435 \\ 12453 \\ \vdots

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 N\times N for some integer N ? N?

0 1 2 3

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2 solutions

Jesse Nieminen
Aug 9, 2017

Divisibility rules of 3 3 and 9 9 tell us that all of the numbers are divisible by 3 3 , but not by 9 9 , hence 0 \boxed{0} of them are square numbers.

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).

Tarmo Taipale - 3 years, 9 months ago
Zach Abueg
Aug 6, 2017

Theorem

All perfect squares have digital roots of either 1 1 , 4 4 , 7 7 , or 9 9 . It must be noted, however, that the converse, (all numbers with digital roots of 1 1 , 4 4 , 7 7 , or 9 9 are perfect squares) is not true.

Proof

By the division algorithm, every integer n n is of the form 9 k + r 9k + r , where 0 r < 9 0 \leq r < 9 . So n r ( m o d 9 ) n \equiv r \pmod 9 and hence n 2 r 2 ( m o d 9 ) n^2 \equiv r^2 \pmod 9 .

Since r r 9 ( m o d 9 ) r \equiv r - 9 \pmod 9 ,

0 2 0 ( m o d 9 ) ( ± 1 ) 2 1 ( m o d 9 ) ( ± 2 ) 2 4 ( m o d 9 ) ( ± 3 ) 2 0 ( m o d 9 ) ( ± 4 ) 7 ( m o d 9 ) \displaystyle \begin{aligned} 0^2 & \equiv 0 \pmod{9} \\ \left(\pm 1\right)^2 & \equiv 1 \pmod{9} \\ \left(\pm 2\right)^2 & \equiv 4 \pmod{9} \\ \left(\pm 3\right)^2 & \equiv 0 \pmod{9} \\ \left(\pm 4\right) & \equiv 7 \pmod{9} \end{aligned}

Thus, n 2 n^2 is congruent to 0 0 , 1 1 , 4 4 , or 7 7 , so its digital root is 1 1 , 4 4 , 7 7 , or 9 9 .

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Now suppose all possible 5-digit combinations of 12345 12345 were perfect squares. Their digital roots would all be 1 + 2 + 3 + 4 + 5 = 15 1 + 5 = 6 1 + 2 + 3 + 4 + 5 = 15 \rightarrow 1 + 5 = 6 . Since the only digital roots of perfect squares are 1 1 , 4 4 , 7 7 , or 9 9 , we conclude by contradiction that 12345 12345 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!

Pi Han Goh - 3 years, 10 months ago

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