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?
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To solve this problem, let's break it down step by step.
In December 1900, the population in Atown is a perfect square. Let's assume the population in December 1900 is x, and we know that x is a perfect square.
In January 1901, 48 people moved in, and the population became 1 more than a perfect square. So, the population in January 1901 is (x + 48 + 1), which is (x + 49), and we know that (x + 49) is also a perfect square.
In February 1901, 48 more people moved in, and the population became a perfect square. So, the population in February 1901 is (x + 49 + 48), which is (x + 97), and we know that (x + 97) is also a perfect square.
We can set up two equations based on the information we have:
1. (x + 49) is a perfect square.
2. (x + 97) is a perfect square.
To find the value of x, we can start by looking at possible perfect square values for (x + 49). We can start with the smallest perfect square greater than 49, which is 64 (8^2). So, (x + 49) could be equal to 64.
If (x + 49) = 64, then x = 15.
Now we can substitute the value of x back into the second equation to check if it satisfies (x + 97) being a perfect square.
(x + 97) = (15 + 97) = 112.
112 is not a perfect square, so this value of x does not satisfy the conditions.
We can continue this process by trying the next perfect square greater than 64, which is 81 (9^2).
If (x + 49) = 81, then x = 32.
Now substituting the value of x back into the second equation:
(x + 97) = (32 + 97) = 129.
129 is not a perfect square either.
We can continue this process until we find a value of x that satisfies both equations.
Trying the next perfect square greater than 81, which is 100 (10^2).
If (x + 49) = 100, then x = 51.
Now substituting the value of x back into the second equation:
(x + 97) = (51 + 97) = 148.
148 is not a perfect square either.
By continuing this process, we can try the next perfect square, which is 121 (11^2).
If (x + 49) = 121, then x = 121 - 49 = 72.
Now substituting the value of x back into the second equation:
(x + 97) = (72 + 97) = 169.
169 is a perfect square! (13^2)
So, the population in Atown in December 1900 is x = 72.
Therefore, the population in Atown in December 1900 is 72.