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# Help 2 problems

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1) Suppose r is a real number for which $$\left\lfloor r + \frac{19}{100} \right\rfloor + \left\lfloor r + \frac{20}{100} \right\rfloor + \left\lfloor r + \frac{21}{100} \right\rfloor + \cdots + \left\lfloor r + \frac{91}{100} \right\rfloor = 546.$$Find $$\lfloor 100r \rfloor.$$

2) Find the sum of the squares of the solutions to $$\left| x^2 - x + \frac{1}{2008} \right| = \frac{1}{2008}.$$

Aug 7, 2019

#1
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2)

Solve for x:
2008 abs(x^2 - x + 1/2008) = 1

Divide both sides by 2008:
abs(x^2 - x + 1/2008) = 1/2008

Split the equation into two possible cases:
x^2 - x + 1/2008 = 1/2008 or x^2 - x + 1/2008 = -1/2008

Subtract 1/2008 from both sides:
x^2 - x = 0 or x^2 - x + 1/2008 = -1/2008

Factor x from the left hand side:
x (x - 1) = 0 or x^2 - x + 1/2008 = -1/2008

Split into two equations:
x - 1 = 0 or x = 0 or x^2 - x + 1/2008 = -1/2008

x = 1 or x = 0 or x^2 - x + 1/2008 = -1/2008

Subtract 1/2008 from both sides:
x = 1 or x = 0 or x^2 - x = -1/1004

x = 1 or x = 0 or x^2 - x + 1/4 = 125/502

Write the left hand side as a square:
x = 1 or x = 0 or (x - 1/2)^2 = 125/502

Take the square root of both sides:
x = 1 or x = 0 or x - 1/2 = 5 sqrt(5/502) or x - 1/2 = -5 sqrt(5/502)

x = 1 or x = 0 or x = 1/2 + 5 sqrt(5/502) or x - 1/2 = -5 sqrt(5/502)
x = 1 or x = 0 or x = 1/2 + 5 sqrt(5/502) or x = 1/2 - 5 sqrt(5/502)
Sum of squares= 1^2 + [1/2 + 5 sqrt(5/502)]^2 + [1/2 - 5 sqrt(5/502)]^2 =1003 / 502

Aug 8, 2019
#2
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1 -  r = 7.43

Floor[100 * 7.43] = 743.

Aug 8, 2019
#3
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1)   10920/803<=r<10940/803

Guest Aug 8, 2019