The population in a town called Atown in December 1900 is a perfect square. In January 1901, 48 people moved in, and the population is 1 more than a perfect square. In February 1901, 48 more people moved in, and the population is a perfect square. Assuming that no one was born, died or moved out during these months, what is the population of Atown in December 1900?
a^2 + 48 = b^2
b^2 + 48 = c^2
since (n+1)^2 = n^2+2n+1, 2n+1 must be less than 48, so n<24
If a is odd, it must end in 1 or 9 for a^2+48 to be a square. If a or b is greater than 24, the next possible greater square is more than 48 away.
1^2+48 = 7^2
11^2+48 = 13^2
but these don't end in 1 or 9, so adding 48 cannot produce a square
so, a is even and must end in 6
Hmmm. No good.
Sorry. I don't see a solution.
Got it!
Initial population is 23^2, which = 529.
23^2 + 48 = 577 = 576 + 1, so
23^2 + 48 = 24^2 + 1, and
23^2 + 96 = 625, so
23^2 = 25^2
Did it by exhaustion with Excel. Check back later if I can do it algebraicly!
I mean 23^2 + 96 = 25^2
To find the population of Atown in December 1900, we need to examine the given information step by step.
Let's represent the population in December 1900 as a perfect square number. Suppose the population is x^2, where x is an integer.
In January 1901, 48 people moved in, and the population became 1 more than a perfect square. Therefore, we can write this as (x^2 + 48) = (y^2 + 1), where y is also an integer.
In February 1901, 48 more people moved in, and the population became a perfect square. So, we can write this as (x^2 + 48 + 48) = z^2, where z is an integer.
To solve these equations and find the value of x, we can use the process of elimination. First, let's subtract the second equation from the third equation:
(x^2 + 48 + 48) - (x^2 + 48) = z^2 - (y^2 + 1)
96 = z^2 - y^2 + 1
After simplifying, we get:
95 = z^2 - y^2
Now, let's express 95 as the difference of two squares:
95 = (z + y)(z - y)
We want to find two factors of 95 that have a difference of 2 (because we are looking for a pair of squares that differ by 2, as mentioned in the problem). By examining the factors of 95, we find that 95 = 19 * 5.
So, we have two possible sets of equations:
Case 1: z + y = 19 and z - y = 5
Case 2: z + y = 95 and z - y = 1
Let's solve these cases separately:
Case 1:
Adding the two equations results in:
2z = 24
z = 12
Substituting the value of z back into either equation gives us:
12 + y = 19
y = 7
Now, let's substitute the values of x, y, and z into the initial equation of February 1901:
(x^2 + 48 + 48) = z^2
x^2 + 96 = 12^2
x^2 = 144 - 96
x^2 = 48
x = √48
x = 4√3
Since the population is represented by a perfect square, and x is not an integer, we will discard this solution.
Moving on to Case 2:
Adding the two equations results in:
2z = 96
z = 48
Substituting the value of z back into either equation gives us:
48 + y = 95
y = 47
Now, let's substitute the values of x, y, and z into the initial equation of February 1901:
(x^2 + 48 + 48) = z^2
x^2 + 96 = 48^2
x^2 = 2304 - 96
x^2 = 2208
x = √2208
x = 48√3
Since the population is represented by a perfect square, and x is an integer, this is a valid solution.
Therefore, the population of Atown in December 1900 is x^2 = (48√3)^2 = 2304 * 3 = 6912.