Difference between revisions of "2020 AMC 10A Problems/Problem 8"
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Split the even numbers and the odd numbers apart. If we group every 2 even numbers together and add them, we get a total of <math>50\cdot (-2)=-100</math>. Summing the odd numbers is equivalent is equivalent to summing the first 100 odd numbers, which is equal to <math>100^2=10000</math>. Adding these two, we obtain the answer of <math>\boxed{\text{(B) }9900}</math>. | Split the even numbers and the odd numbers apart. If we group every 2 even numbers together and add them, we get a total of <math>50\cdot (-2)=-100</math>. Summing the odd numbers is equivalent is equivalent to summing the first 100 odd numbers, which is equal to <math>100^2=10000</math>. Adding these two, we obtain the answer of <math>\boxed{\text{(B) }9900}</math>. | ||
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+ | We can break this entire sum down into <math>4</math> integer bits, in which the sum is <math>2x</math>, where <math>x</math> is the first integer in this bit. We can find that the first sum of every sequence is <math>4x-3</math>, which we plug in for the | ||
==See Also== | ==See Also== |
Revision as of 21:23, 31 January 2020
Problem
What is the value of
Solution
Split the even numbers and the odd numbers apart. If we group every 2 even numbers together and add them, we get a total of . Summing the odd numbers is equivalent is equivalent to summing the first 100 odd numbers, which is equal to . Adding these two, we obtain the answer of .
We can break this entire sum down into integer bits, in which the sum is , where is the first integer in this bit. We can find that the first sum of every sequence is , which we plug in for the
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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 | ||
All AMC 10 Problems and Solutions |
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