Closest Sum Selection
Which of the following is closest to $$10^0 + 10^1 + 10^2 + 10^3 + 10^4 + 10^5?$$
A) $10^0$.
B) $10^1$.
C) $10^2$.
D) $10^3$.
E) $10^4$.
F) $10^5$.
G) $10^6$.
Write your answer as A, B, C, D, E, F, or G.
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
Converting the sum to a decimal gives $$
10^0 + 10^1 + 10^2 + 10^3 + 10^4 + 10^5 = 111,\!111.$$
The numbers $ 10^0, $ $ 10^1, $ $ 10^2, $ $ 10^3, $ $ 10^4, $ and $ 10^5 $ are all less than $ 111,\!111, $ so they all lie to the left of $ 111,\!111 $ on the number line. Since $ 10^5 $ is greater than $ 10^0, $ $ 10^1, $ $ 10^2, $ $ 10^3, $ and $ 10^4, $ it lies further to the right than these five powers of $ 10 $ on the number line, so it is closer to $ 111,\!111 $ than these five powers of $ 10 $. Therefore, either $ 10^5 $ or $ 10^6 $ have to be the number among the answer choices that is closest to $ 111,\!111 $.
Rounding $ 111,\!111 $ to the nearest million gives $ 0, $ so $ 111,\!111 $ is at least $ 500,\!000 $ away from $ 10^6 $.
Note that $ 10^5 = 100,\!000 $ and $$111,\!111 - 100,\!000 = 11,\!111,$$which is much less than $ 500,\!000, $ so $ 10^5 $ is closer to $ 111,\!111 $ than $ 10^6 $. So our final answer is $ \boxed{F} $.
Freeflows