# MAT question How do we do this? Is there a property of square numbers that would help?

Exactly one of these five numbers is a square number. Which one?
(a) 99,999,999, (b) 123,333,333, (c) 649,485,225, (d) 713,291,035, (e) 987,654,000.
(edited 3 weeks ago) Original post by Sasuto
How do we do this? Is there a property of square numbers that would help?

Exactly one of these five numbers is a square number. Which one?
(a) 99,999,999, (b) 123,333,333, (c) 649,485,225, (d) 713,291,035, (e) 987,654,000.

a) is 1 off 100,000,000 which is a square number (even number of zeros), so that cant be a square
b) divisble by 3 but not 9 (digit sum) so not square
d) has 5^1 in prime factorisation (not a multiple of 25) so not square
e) has 5^3 in the prime factorisation so not square

so c) Original post by Sasuto
How do we do this? Is there a property of square numbers that would help?

Exactly one of these five numbers is a square number. Which one?
(a) 99,999,999, (b) 123,333,333, (c) 649,485,225, (d) 713,291,035, (e) 987,654,000.

As a start, what are the possible last digits for a square number?
You might also want to think about the highest power of 5 dividing each number, and if that's compatible with it being square. Original post by mqb2766
a) is 1 off 100,000,000 which is a square number (even number of zeros), so that cant be a square
b) divisble by 3 but not 9 (digit sum) so not square
d) has 5^1 in prime factorisation (not a multiple of 25) so not square
e) has 5^3 in the prime factorisation so not square

so c)

I see, so I can also deduce info from its prime factorisation. So generalising d and e, if a number has a prime factor that is a square, the number is also a square? Original post by Sasuto
I see, so I can also deduce info from its prime factorisation. So generalising d and e, if a number has a prime factor that is a square, the number is also a square?

You should probably know this, but a square number has prime factors with even exponents. Otherwise when you root it (half the exponents), the result will not be an integer. Original post by DFranklin
As a start, what are the possible last digits for a square number?
You might also want to think about the highest power of 5 dividing each number, and if that's compatible with it being square.

I see that there is a recurring pattern of 1,4,9,6,5,6,9,4,1,0... which eliminates b. I know it definitely can't be a.

Ok, does your second point about dividing by the highest power of 5 apply to other numbers, eg dividing by the highest power of 3. Or does it only work for 5? Original post by mqb2766
You should probably know this, but a square number has prime factors with even exponents. Otherwise when you root it (half the exponents), the result will not be an integer.

Oh of course