Difference between revisions of "1993 AHSME Problems/Problem 7"
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Therefore <math>\frac{R_{24}}{R_4} = \frac{ 10^{24}-1 }{10^4-1}</math>. | Therefore <math>\frac{R_{24}}{R_4} = \frac{ 10^{24}-1 }{10^4-1}</math>. | ||
− | + | We can recognize this is also the formula for the sum of a geometric series <math>1+10^4 + (10^4)^2 + \dots + (10^4)^5 = 1+ 10^4 + 10^8 + \dots + 10^{20}</math> | |
Now the 1's place has a 1, but the 10's, 100's and 1,000's place have 0's. The 10,000's place has a 1, but the <math>10^5</math>, <math>10^6</math> and <math>10^7</math> places have 0's. Between successive 1's in the decimal expansion, there are three 0's, which gives <math>5\times 3=15</math> zeros altogether. | Now the 1's place has a 1, but the 10's, 100's and 1,000's place have 0's. The 10,000's place has a 1, but the <math>10^5</math>, <math>10^6</math> and <math>10^7</math> places have 0's. Between successive 1's in the decimal expansion, there are three 0's, which gives <math>5\times 3=15</math> zeros altogether. |
Latest revision as of 21:28, 27 May 2021
Problem
The symbol stands for an integer whose base-ten representation is a sequence of ones. For example, , etc. When is divided by , the quotient is an integer whose base-ten representation is a sequence containing only ones and zeroes. The number of zeros in is:
Solution
Note .
Therefore .
We can recognize this is also the formula for the sum of a geometric series
Now the 1's place has a 1, but the 10's, 100's and 1,000's place have 0's. The 10,000's place has a 1, but the , and places have 0's. Between successive 1's in the decimal expansion, there are three 0's, which gives zeros altogether.
The answer is
See also
1993 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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