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 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 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>. | ||
| − | == Solution 2 == | + | == Solution 2 (bashy) == |
| − | 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 <math>50</math> bits in the entire sequence is <math>1+2+3+\cdots+50=1275</math>, so then we can plug it into the first term of every sequence equation we got above <math>4(1275)-3(50)=4950</math>, and so the sum of every bit is <math>2x</math>, and we only found the value of <math>x</math>, the sum of the sequence is <math>4950\cdot2=\boxed{(B)9900}</math>. | + | 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 <math>50</math> bits in the entire sequence is <math>1+2+3+\cdots+50=1275</math>, so then we can plug it into the first term of every sequence equation we got above <math>4(1275)-3(50)=4950</math>, and so the sum of every bit is <math>2x</math>, and we only found the value of <math>x</math>, the sum of the sequence is <math>4950\cdot2=\boxed{(B)9900}</math>.-middletonkids |
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| − | -middletonkids | ||
== Solution 3 == | == Solution 3 == | ||
Revision as of 18:01, 1 February 2020
Problem
What is the value of
Solution 1
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 to summing the first 100 odd numbers, which is equal to 
. Adding these two, we obtain the answer of 
.
Solution 2 (bashy)
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 
bits in the entire sequence is 
, so then we can plug it into the first term of every sequence equation we got above 
, and so the sum of every bit is , and we only found the value of , the sum of the sequence is 
.-middletonkids
Solution 3
Another solution involves adding everything and subtracting out what is not needed. The first step involves solving 
. To do this, we can simply multiply 
and 
and divide by 
. This will get us 
. The next step involves subtracting out the numbers with minus signs. We actually have to do this twice, because we need to take out the numbers we weren’t supposed to add and then subtract them from the problem. Doing a quick calculation, we can see that from to , incrementing by , there are numbers that we have to subtract. To do this we can do times 
divided by , and then we can multiply by , because we are counting by fours, not ones. Our answer will be 
, but remember, we have to do this twice. As a result, we will get 
. Finally, we just have to do 
, and our answer is .
— 
Video Solution
~IceMatrix
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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