In 1990, the population of Centerville was a perfect square. Ten years later, with an increase
of 100 people, the population was one more than a perfect square. Now, with another
increase of 100 people, the population is again a perfect square. What is the present
population of Centerville? Can there be more than one answer?
of 100 people, the population was one more than a perfect square. Now, with another
increase of 100 people, the population is again a perfect square. What is the present
population of Centerville? Can there be more than one answer?
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Hello,
In 1990, there were n² inhabitants in Centerville.
In 2000, The population increased by 100 people to m²+1 inhabitants.
With another increase of 100 people, the population would be p² inhabitants.
So we get the relations:
{ n² + 100 = m² + 1
{ n² + 200 = p²
where n, m and p are whole integers.
m² + 1 = n² + 100
m² - n² = 99
(m + n)(m - n) = 99
Since m and n are integers, m+n and m-n are also integers. This means that m-n and m+n are factors of 99.
The prime decomposition of 99 is:
99 = 3 × 3 × 11
So there are only three possibilities (considering that m is obviously greater than n):
99 = 99 × 1 →→→ m=50 and n=49
99 = 33 × 3 →→→ m=18 and n=15
99 = 11 × 9 →→→ m=10 and n=1
Let's check whether n²+200 is a perfect square:
If n=49, n² + 200 = 2401 + 200 = 2601 = 51²
If n=15, n² + 200 = 225 + 200 = 425 (not a perfect square)
If n=1, n² + 200 = 1 + 200 = 201 (not a perfect square)
So there is only one valid solution.
Thus Centerville had 2401=49² citizens in 1990.
It increased to 2501=50²+1 citizens in 2000.
If 100 more people settled there, they would number 2601=51² citizens.
And this is the unique possible solution.
Logically,
Dragon.Jade :-)
In 1990, there were n² inhabitants in Centerville.
In 2000, The population increased by 100 people to m²+1 inhabitants.
With another increase of 100 people, the population would be p² inhabitants.
So we get the relations:
{ n² + 100 = m² + 1
{ n² + 200 = p²
where n, m and p are whole integers.
m² + 1 = n² + 100
m² - n² = 99
(m + n)(m - n) = 99
Since m and n are integers, m+n and m-n are also integers. This means that m-n and m+n are factors of 99.
The prime decomposition of 99 is:
99 = 3 × 3 × 11
So there are only three possibilities (considering that m is obviously greater than n):
99 = 99 × 1 →→→ m=50 and n=49
99 = 33 × 3 →→→ m=18 and n=15
99 = 11 × 9 →→→ m=10 and n=1
Let's check whether n²+200 is a perfect square:
If n=49, n² + 200 = 2401 + 200 = 2601 = 51²
If n=15, n² + 200 = 225 + 200 = 425 (not a perfect square)
If n=1, n² + 200 = 1 + 200 = 201 (not a perfect square)
So there is only one valid solution.
Thus Centerville had 2401=49² citizens in 1990.
It increased to 2501=50²+1 citizens in 2000.
If 100 more people settled there, they would number 2601=51² citizens.
And this is the unique possible solution.
Logically,
Dragon.Jade :-)
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x, x+99, and x+200 must all be perfect squares. Then you open one of these tables http://www.naturalnumbers.org/PS-100.txt and find that the numbers must be 2401, 2500, and 2601.
Hope this helps,
Marco
Hope this helps,
Marco