Floor function

Algebra Level 4

Find the last 7 digits (from left to right) of 10 9999 1 0 3333 + 2016 \left\lfloor \dfrac { { 10 }^{ 9999 } }{ 10^{ 3333 }+2016 } \right\rfloor .

Notation : \lfloor \cdot \rfloor denotes the floor function .


The answer is 4064255.

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1 solution

1 0 9999 1 0 3333 + 2016 = 1 0 9999 + 201 6 3 201 6 3 1 0 3333 + 2016 = ( 1 0 3333 + 2016 ) ( 1 0 6666 2016 1 0 3333 + 201 6 2 ) 201 6 3 1 0 3333 + 2016 = 1 0 6666 2016 1 0 3333 + 201 6 2 201 6 3 1 0 3333 + 2016 = 1 0 6666 2016 1 0 3333 + 201 6 2 1 = 999...999 3329 digits 7984 000...000 3333 digits + 4064256 1 = 999...999 3329 digits 7984 000...000 3326 digits 4064255 7 digits \begin{aligned} \left \lfloor \frac {10^{9999}}{10^{3333}+2016} \right \rfloor & = \left \lfloor \frac {10^{9999}+2016^3-2016^3}{10^{3333}+2016} \right \rfloor \\ & = \left \lfloor \frac {(10^{3333}+2016)(10^{6666}-2016\cdot10^{3333}+2016^2)-2016^3}{10^{3333}+2016} \right \rfloor \\ & = \left \lfloor 10^{6666}-2016\cdot10^{3333} + 2016^2 - \frac {2016^3}{10^{3333}+2016} \right \rfloor \\ & = 10^{6666}-2016\cdot10^{3333} + 2016^2 -1 \\ & = \underbrace{999...999}_{3329 \text{ digits}}7984\underbrace{000...000}_{3333 \text{ digits}} + 4064256 - 1 \\ & = \underbrace{999...999}_{3329 \text{ digits}}7984\underbrace{000...000}_{3326 \text{ digits}} \underbrace{\boxed{4064255}}_{7 \text{ digits}} \end{aligned}

Hi Sir, it would be good if you could put 201 6 2 = ( 2000 + 16 ) 2 = ( 4000000 + 64000 + 2 56 ) = 4064256 2016^2=(2000+16)^2=(4000000+64000+2​56)=4064256 .

Joshua Chin - 4 years, 11 months ago

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Oh, I just used a calculator for that. I think it is not the main purpose of the solution.

Chew-Seong Cheong - 4 years, 11 months ago

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