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Algebra Level 5

$\left \lfloor r+\frac{19}{100} \right \rfloor+\left \lfloor r+\frac{20}{100} \right \rfloor+\cdots+\left \lfloor r+\frac{91}{100} \right \rfloor=546.$

Given that $r$ is a real number that satisfy the equation above, find the value of $\lfloor 100r \rfloor$ .

The answer is 743.

2 solutions

Aakash Khandelwal
Nov 24, 2015

Let R denote greatest integer not exceeding r. Whereas f denotes fractional part of r. Hence r= R + f. Now add 19/100 to both sides and take [.] on both sides of equation.# Then equation looks like [r+19/100]= R +[f+19/100] . R is outside the [.] as its an integer .# Now framing such equations for 20/100, ... till 91/100 and adding up them all we get : 73R + all [.] terms = 546.# Now R can be 7 as rest all are not feasible. Then we find sum of all [.] terms must be 546-511=35. Also f is between 0 and 1. Hence each [.] can be either zero or one. Therfore for getting 35 f needs to be 0.43 or more decimal places after that and answer = 743 .

Dexter Woo Teng Koon
Dec 5, 2015

number of terms=91-19+1=73

546/73≈7.479

then 7(x)+8(73-x)=546

solved x = 38

so we know that [r+0.19]+...+[r+0.56]=7(38)

and [r+0.57]+...+[r+0.91]=8(35)

solved that 7.44>r≧ 7.43

so [100r]=743

I hope there are more explanation to improve my solution , i would appreciate it.

For completeness, you'll want to establish the arrangement of 38, 35 terms equal to 7, 8 respectively the only possible one. You've already established that there is only one way to partition 7 and 8 to get the desired sum, but is there an arrangement possible with numbers other than 7 and 8 (like 6 or 9)?

Matt O - 5 years, 5 months ago

Wait a minute

The answer is 743.

2 solutions

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