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2011 AMC 10B Problem 10

P1-10~2 minPrint
Sequences & SeriesExponents, Logarithms & RadicalsBound the quantity above/below or estimate to pin it down

Consider the set of numbers {1,10,102,103,…,1010}\{1, 10, 10^2, 10^3, \ldots, 10^{10}\} . The ratio of the largest element of the set to the sum of the other ten elements of the set is closest to which integer?

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Key insight

The ten smaller numbers add up to 1,111,111,111. Nine times that is 9,999,999,999, just 1 short of the largest number, so the ratio is almost exactly 9.

Solution

1. The largest number is 1010=10,000,000,00010^{10} = 10{,}000{,}000{,}000.
2. The other ten numbers are 1,10,100,…,1091, 10, 100, \ldots, 10^9. Adding them, each power of 1010 puts a 11 in a different place:
1+10+100+⋯+1,000,000,000=1,111,111,111.1 + 10 + 100 + \cdots + 1{,}000{,}000{,}000 = 1{,}111{,}111{,}111.
3. We want 10,000,000,000÷1,111,111,11110{,}000{,}000{,}000 \div 1{,}111{,}111{,}111. Try 99:
9×1,111,111,111=9,999,999,999.9 \times 1{,}111{,}111{,}111 = 9{,}999{,}999{,}999.
4. That is only 11 less than 10,000,000,00010{,}000{,}000{,}000. So the ratio is a tiny bit more than 99, about 9.0000000099.000000009.
5. The closest whole number is 99.

The answer is (B) 9\boxed{\textbf{(B)}\ 9}.

Why this works

When each number is 1010 times the one before, the last number is bigger than all the others put together, but not by a factor of 1010: the earlier numbers add up to about one ninth of the last one. The quick check with 1,111,111,111×91{,}111{,}111{,}111 \times 9 shows it without any formulas.

Alternative approach

Try it on a smaller set first: for {1,10,100}\{1, 10, 100\} the ratio is 100÷11≈9.1100 \div 11 \approx 9.1; for {1,10,100,1000}\{1, 10, 100, 1000\} it is 1000÷111≈9.011000 \div 111 \approx 9.01. The ratio gets closer and closer to 99 as the set grows, so for eleven numbers it is essentially 99.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The ten smaller elements sum to
1+10+102+⋯+109=1,111,111,111=1010−19.1 + 10 + 10^2 + \cdots + 10^9 = 1{,}111{,}111{,}111 = \frac{10^{10} - 1}{9}.

The requested ratio is
1010(1010−1)/9=9⋅10101010−1,\frac{10^{10}}{(10^{10} - 1)/9} = \frac{9 \cdot 10^{10}}{10^{10} - 1},
which is slightly more than 99 (the denominator is a hair less than 101010^{10}) and far closer to 99 than to 1010. Numerically, 1010/1,111,111,111≈9.00000000910^{10}/1{,}111{,}111{,}111 \approx 9.000000009.

The answer is (B) 9\boxed{\textbf{(B)}\ 9}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.