Difference between revisions of "2018 AMC 10A Problems/Problem 1"
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== Solution == | == Solution == | ||
− | + | <math>2+1=3</math>, so the reciprocal is <math>\frac{1}{3}</math>. Summing <math>1</math> and repeating leads to <math>\frac{3}{4}</math>. Rinse and repeat to get <math>\frac{4}{7}</math>. Adding <math>1</math> to that is <math>\frac{11}{7}</math> (B) | |
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+ | -Welp... | ||
== See Also == | == See Also == |
Revision as of 20:01, 8 February 2018
Problem
What is the value of
Solution
, so the reciprocal is . Summing and repeating leads to . Rinse and repeat to get . Adding to that is (B)
-Welp...
See Also
2018 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
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All AMC 10 Problems and Solutions |
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