How many perfect squares(2,4,6,9)?

How many perfect squares use the digits 2 , 4 , 6 , 9 2, 4, 6, 9 at most once?

As an explicit example, 4 and 49 satisfy the conditions of the problem.

other problem by me


The answer is 4.

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2 solutions

All square numbers end with either 0 , 1 , 4 , 5 , 6 0, 1, 4, 5, 6 or 9 9

Square numbers only end with 4 4 or 9 9 according to the instructions given as we are only allowed to use 2 , 4 , 6 , 9 2, 4, 6, 9

This will help us speed up the search for these "special numbers"

We can use the following formula to get the last two digits of any squared number \text{We can use the following formula to get the last two digits of any squared number }

z = ( 50 × x ) + y \boxed{z = (50 × x) + y}

Where z is equivalent to the number we want to square \text{Where }z\text{ is equivalent to the number we want to square}

And the number y , which when squared, has the same last two digits as the number we wish to calculate \text{And the number }y\text{, which when squared, has the same last two digits as the number we wish to calculate}

An example \text{An example}

We want to know the last two digits of 27 1 2 271^2

So, we rearrange the number like this:

50 × 5 + 21 50 × 5 + 21

It is said that the last two digits of 27 1 2 271^2 is equal to the last two digits of 2 1 2 21^2

Checking, we see that it is!

27 1 2 = 734 41 271^2 = 734\boxed{41} , and 2 1 2 = 4 41 21^2 = 4\boxed{41}


We can use this to prove that there are no more solutions to this question, as we know that most square numbers do not end with the numbers 2 , 4 , 6 , 9 2,4,6,9 , thereby excluding most of the "presumed solutions"

After checking all the square numbers, you must be forced to accept that...

4 , 9 , 49 and 64 are the only square numbers containing either 2, 4, 6 or 9 4, 9, 49 \text{ and }64 \text{ are the only square numbers containing }\boxed{\text{either 2, 4, 6 or 9}}

In fact, we also know what the last 2 digits of square numbers must be, which allows us to more easily verify that there are no further solutions.

Calvin Lin Staff - 11 months, 3 weeks ago

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Thanks for the information @Calvin Lin !

A Former Brilliant Member - 11 months, 3 weeks ago

9 = 3 \surd 9 = 3 , 4 = 2 \surd 4 = 2 , 49 = 7 \surd 49 = 7 and 64 = 8 \surd 64 = 8 .

There is no 3 3 -digit or 4 4 -digit square numbers.

Therefore, the answer is 4 \fbox 4 .

Really? Are there no 3-digit or 4-digit perfect squares?

Vinayak Srivastava - 11 months, 3 weeks ago

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Yes. I tried every permuation.

A Former Brilliant Member - 11 months, 3 weeks ago

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