[Guest]

Determine all possible base ten perfect square(s) of the form PQPQRQPR, where each of the capital letters in bold represents a different digit from 1 to 9.

Note: While a solution is trivial with the aid of a computer program, show how to derive it without one.

[Guest]

Not quite there....

If X is the square root of PQPQRQPR then the following applies.

The lowest and largest numbers that fit the pattern PQPQRQPR are 12123213 and 98987897, making X a number between 3482 and 9949.

Therefore X is a 4 digit number ABCD where B, C, and D are any digits between 0 and 9, and A is a digit between 3 and 9

If D = 0 then R = 0, therefore D is not 0.

If D = 1 or 9 then R = 1 and P is 0, 4 or 8. Since 0 is not allowed, P is 4 or 8.

Therefore if D = 1 or 9, X² is one of

4Q4Q1Q41 where Q is 2, 3, 5, 6, 7, 8, 9

8Q8Q1Q81 where Q is 2, 3, 4, 5, 6, 7, 9

If D = 2 or 8 then R = 4 and P must be even. Since 0 is not allowed P is 2, 6 or 8

Therefore if D = 2 or 8, X² is one of

2Q2Q4Q24 where Q is 1, 3, 5, 6, 7, 8, 9

6Q6Q4Q64 where Q is 1, 2, 3, 5, 7, 8, 9

8Q8Q4Q84 where Q is 1, 2, 3, 5, 6, 7, 9

If D = 3 or 7 then R = 9 and QP must be divisible by 4.

Therefore if D = 3 or 7, X² is one of

4Q4Q9Q29 where Q is odd (1, 3, 5 or 7)

8Q8Q9Q49 where Q is even (2 or 6)

4Q4Q9Q69 where Q is odd (1, 3, 5 or 7)

8Q8Q9Q89 where Q is even (2, 4, or 6)

If D = 4 or 6 then R = 6 and P must be odd.

Therefore if D = 4 or 6, X² is one of

1Q1Q6Q16 where Q is 2, 3, 4, 5, 7, 8, 9

3Q3Q6Q36 where Q is 1, 2, 4, 5, 7, 8, 9

5Q5Q6Q56 where Q is 1, 2, 3, 4, 7, 8, 9

7Q7Q6Q76 where Q is 1, 2, 3, 4, 5, 8, 9

9Q9Q6Q96 where Q is 1, 2, 3, 4, 5, 7, 9

If D = 5 then PR = 25 and Q = 0 or 2, or RQ = 06 or 56. Since 0 is not allowed and Q cannot = P, then Q = 2.

Therefore, if D = 5, X² is 26265625

This gives a total 84 possible values for PQPQRQPR. I know it would be a simple thing to check out each of the 84 numbers and see which have integer square roots, but I was hoping to find some more tricks to narrow down the search. I'm still working on it.

Edited July 20, 2009 by cpotting

[Guest]

If D = 5 then PR = 25 and Q = 0 or 2, or RQ = 06 or 56. Since 0 is not allowed and Q cannot = P, then Q = 2.

should read

If D = 5 then PR = 25 and Q = 0 or 2, or RQ = 06 or 56. Since 0 is not allowed and Q cannot = P, then Q = 2 6.

[Guest]

A little further along...

Okay, all perfect squares give a result of 0 or 1 when modulus 3 is applied. So for any perfect square Y, Y or (Y-1) will be divisible by 3. Additionally the sum of the digits of any number that is divisible by 3 is also divisible by 3.

Therefore P+Q+P+Q+R+Q+P+R % 3 = 0 or (P+Q+P+Q+R+Q+P+R - 1) % 3 = 0

Therefore 3P + 3Q + 2R % 3 = 0 or (3P + 3Q + 2R - 1) % 3 = 0. 3P and 3Q are already divisible by 3 so...

2R % 3 = 0 or 1

Looking at the previous possibilities we can see that

D cannot be 0

If D = 1 or 9 then R = 1 and 2R % 3 = 2.

So D cannot be 1

If D = 2 or 8 then R = 4 and 2R % 3 = 2.

So D cannot be 2 or 8

If D = 3 or 7 then R = 9 and 2R % 3 = 0.

So, from my previous post, if D = 3 or 7, X² is one of

4Q4Q9Q29 where Q is odd (1, 3, 5 or 7)

8Q8Q9Q49 where Q is even (2 or 6)

4Q4Q9Q69 where Q is odd (1, 3, 5 or 7)

8Q8Q9Q89 where Q is even (2, 4, or 6)

If D = 4 or 6 then R = 6 and 2R % 3 = 0.

Again, from my previous post, if D = 4 or 6, X² is one of

1Q1Q6Q16 where Q is 2, 3, 4, 5, 7, 8, 9

3Q3Q6Q36 where Q is 1, 2, 4, 5, 7, 8, 9

5Q5Q6Q56 where Q is 1, 2, 3, 4, 7, 8, 9

7Q7Q6Q76 where Q is 1, 2, 3, 4, 5, 8, 9

9Q9Q6Q96 where Q is 1, 2, 3, 4, 5, 7, 9

and if D = 5 then X² is 26265625

So now I am down to 49 possible answers, and still looking.

[Guest]

I noticed some typos in my previous posts (including the number of possible answers). So I thought I would post a consolidated summary of where I have gotten to so far.

If X is the square root of PQPQRQPR then the following applies.

The lowest and largest numbers that fit the pattern PQPQRQPR are 12123213 and 98987897, making X a number between 3482 and 9949.

Therefore X is a 4 digit number ABCD where B, C, and D are any digits between 0 and 9, and A is a digit between 3 and 9

All perfect squares give a result of 0 or 1 when modulus 3 is applied. So for any perfect square Y, Y or (Y-1) will be divisible by 3. Additionally the sum of the digits of any number that is divisible by 3 is also divisible by 3.

Therefore P+Q+P+Q+R+Q+P+R % 3 = 0 or 1

Therefore 3P + 3Q + 2R % 3 = 0 or 1. 3P and 3Q are already divisible by 3 so...

2R % 3 = 0 or 1

The square of a number ending in 0 ends in 0...

So if D = 0 then R = 0. 0 is not allowed.

Therefore D is not 0.

The square of a number ending in 1 or 9 ends in 1...

So if D = 1 or 9 then R = 1 and 2R % 3 = 2.

Therefore D cannot be 1

The square of a number ending in 2 or 8 ends in 4...

So, if D = 2 or 8 then R = 4 and 2R % 3 = 2.

Therefore D cannot be 2 or 8

The square of a number ending in 3 or 7 ends in two digits divisible by 4...

And numbers divisible by 4 end in an odd digit followed by 0, 4, or 8, or an even digit followed by 2 or 6.

So D = 3 or 7 then R = 9, and QP must be divisible by 4, and 2R % 3 = 0.

Therefore if D = 3 or 7, X² is one of

2Q2Q9Q29 where Q is odd (1, 3, 5 or 7)

4Q4Q9Q49 where Q is even (2, 6 or 8)

6Q6Q9Q69 where Q is odd (1, 3, 5 or 7)

8Q8Q9Q89 where Q is even (2, 4, or 6)

The square of a number ending in 4 or 6 ends in an odd digit followed by a 6...

So if D = 4 or 6 then R = 6 and P must be odd, and 2R % 3 = 0.

Therefore if D = 4 or 6, X² is one of

1Q1Q6Q16 where Q is 2, 3, 4, 5, 7, 8, or 9

3Q3Q6Q36 where Q is 1, 2, 4, 5, 7, 8, or 9

5Q5Q6Q56 where Q is 1, 2, 3, 4, 7, 8, or 9

7Q7Q6Q76 where Q is 1, 2, 3, 4, 5, 8, or 9

9Q9Q6Q96 where Q is 1, 2, 3, 4, 5, 7, or 8

The square of a number ending in 5 ends in 025, 225 or 625...

So if D = 5 then PR = 25 and Q = 0, 2 or 6. Q = 0 is not allowed. If Q = 2 then P would also be 2 which is not allowed.

Therefore, if D = 5, X² is 26265625

This gives a total 50 possible values for PQPQRQPR.

[Pickett]

So, you may be interested in noting that the OP stated that the digits are between 1 and 9...not 0 and 9...so, hopefully that will help you narrow down your possible answers a bit :c)

[Guest]

So, you may be interested in noting that the OP stated that the digits are between 1 and 9...not 0 and 9...so, hopefully that will help you narrow down your possible answers a bit :c)

Yes, I've already accounted for that fact. But thanks, nonetheless. This one has been irking at me like an itch between the shoulder blades! There has to be a way to eliminate more of the 50 possible answers without having to check the square root of each. I'm still working on it.

Edited July 28, 2009 by cpotting

[Guest]

and still more typos...

The square of a number ending in 3 or 7 ends in two digits divisible by 4...

should read

The square of a number ending in 3 or 7 ends in two digits divisible by 4 followed by a 9...