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> | ||
− | ==Solution== | + | ==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. | ||
945 is <math>\boxed{\text{B}}</math> | 945 is <math>\boxed{\text{B}}</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | Instead of breaking the sum and then rearranging, we can start by rearranging: | ||
+ | <cmath>\begin{align*} | ||
+ | 90+91+92+\cdots +98+99 &= (90+99)+(91+98)+(92+97)+(93+96)+(94+95) \\ | ||
+ | &= 189+189+189+189+189 \\ | ||
+ | &= 945\rightarrow \boxed{\text{B}} | ||
+ | \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== | ||
− | + | {{AJHSME box|year=1985|num-b=1|num-a=3}} | |
+ | |||
+ | {{MAA Notice}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] |
Latest revision as of 02:13, 16 February 2021
Problem
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.