Difference between revisions of "1985 AJHSME Problems/Problem 2"
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<math>\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045</math> | <math>\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045</math> | ||
| − | == | + | ==Solutions== |
===Solution 1=== | ===Solution 1=== | ||
| − | + | One possibility is to simply add them. However, this can be time-consuming, and there are other ways to solve this problem. | |
| − | |||
We find a simpler problem in this problem, and simplify -> <math>90 + 91 + ... + 98 + 99 = 90 \times 10 + 1 + 2 + 3 + ... + 8 + 9</math> | We find a simpler problem in this problem, and simplify -> <math>90 + 91 + ... + 98 + 99 = 90 \times 10 + 1 + 2 + 3 + ... + 8 + 9</math> | ||
| − | We know <math>90 \times 10</math>, that's easy | + | We know <math>90 \times 10</math>, that's easy: <math>900</math>. So how do we find <math>1 + 2 + ... + 8 + 9</math>? |
We rearrange the numbers to make <math>(1 + 9) + (2 + 8) + (3 + 7) + (4 + 6) + 5</math>. You might have noticed that each of the terms we put next to each other add up to 10, which makes for easy adding. <math>4 \times 10 + 5 = 45</math>. Adding that on to 900 makes 945. | We rearrange the numbers to make <math>(1 + 9) + (2 + 8) + (3 + 7) + (4 + 6) + 5</math>. You might have noticed that each of the terms we put next to each other add up to 10, which makes for easy adding. <math>4 \times 10 + 5 = 45</math>. Adding that on to 900 makes 945. | ||
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90+91+92+\cdots +98+99 &= (90+99)+(91+98)+(92+97)+(93+96)+(94+95) \\ | 90+91+92+\cdots +98+99 &= (90+99)+(91+98)+(92+97)+(93+96)+(94+95) \\ | ||
&= 189+189+189+189+189 \\ | &= 189+189+189+189+189 \\ | ||
| − | &= 945\rightarrow \boxed{\text{ | + | &= 945\rightarrow \boxed{\text{B}} |
\end{align*}</cmath> | \end{align*}</cmath> | ||
| + | |||
| + | ===Solution 3=== | ||
| + | |||
| + | We can use the formula for finite arithmetic sequences. | ||
| + | |||
| + | It is <math>\frac{n}{2}\times</math> (<math>a_1+a_n</math>) where <math>n</math> is the number of terms in the sequence, <math>a_1</math> is the first term and <math>a_n</math> is the last term. | ||
| + | |||
| + | Applying it here: | ||
| + | |||
| + | <math>\frac{10}{2} \times (90+99) = 945 \rightarrow \boxed{B}</math> | ||
==See Also== | ==See Also== | ||
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{{AJHSME box|year=1985|num-b=1|num-a=3}} | {{AJHSME box|year=1985|num-b=1|num-a=3}} | ||
| + | |||
| + | {{MAA Notice}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
Revision as of 18:44, 24 November 2020
Problem
Solutions
Solution 1
One possibility is to simply add them. However, this can be time-consuming, and there are other ways to solve this problem.
We find a simpler problem in this problem, and simplify -> 
We know 
, that's easy: 
. So how do we find 
?
We rearrange the numbers to make 
. You might have noticed that each of the terms we put next to each other add up to 10, which makes for easy adding. 
. Adding that on to 900 makes 945.
945 is 
Solution 2
Instead of breaking the sum and then rearranging, we can start by rearranging:
Solution 3
We can use the formula for finite arithmetic sequences.
It is 
(
) where 
is the number of terms in the sequence, 
is the first term and 
is the last term.
Applying it here:
See Also
| 1985 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
