Difference between revisions of "1985 AJHSME Problems/Problem 2"
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Rearranging the numbers so each pair sums up to 10, we have: | Rearranging the numbers so each pair sums up to 10, we have: | ||
− | <cmath>(1 + 9)+(2+8)+(3+7)+(4+6)+5</cmath>. <math>4\cdot10+5 = 45</math>, and <math>900+45=\boxed{\text{B} 945}</math>. | + | <cmath>(1 + 9)+(2+8)+(3+7)+(4+6)+5</cmath>. <math>4\cdot10+5 = 45</math>, and <math>900+45=\boxed{\text{(B)}~945}</math>. |
==Solution 2== | ==Solution 2== | ||
− | We can express each of the terms as a difference from 100 and then add the negatives using <math>\frac{n(n+1)}{2}</math> to get the answer. | + | We can express each of the terms as a difference from <math>100</math> and then add the negatives using <math>\frac{n(n+1)}{2}=1+2+3+\cdots+(n-1)+n</math> to get the answer. |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | (100-10)+(100-9)+\cdots + (100-1) &= 100 \ | + | (100-10)+(100-9)+\cdots+(100-1) &= 100\cdot10 -(1+2+\cdots+9+10)\\ |
&= 1000 - 55\\ | &= 1000 - 55\\ | ||
− | &= | + | &= \boxed{\text{(B)}~945} |
\end{align*}</cmath> | \end{align*}</cmath> | ||
==Solution 3== | ==Solution 3== | ||
− | Instead of breaking the sum | + | Instead of breaking the sum then rearranging, we can rearrange directly: |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | 90+91+92+\cdots +98+99 &= | + | 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 \\ | ||
− | &= | + | &= \boxed{\text{(B)}~945} |
\end{align*}</cmath> | \end{align*}</cmath> | ||
==Solution 4== | ==Solution 4== | ||
− | + | The finite arithmetic sequence formula states that the sum in the sequence is equal to <math>\frac{n}{2}\cdot(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 the formula, we have: | |
− | + | <cmath>\frac{10}{2}\cdot(90+99)=\boxed{\text{(B)}~945}</cmath> | |
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==Video Solution== | ==Video Solution== |
Revision as of 21:54, 1 February 2023
Problem
Solution 1
To simplify the problem, we can group 90’s together: .
, and finding has a trick to it.
Rearranging the numbers so each pair sums up to 10, we have: . , and .
Solution 2
We can express each of the terms as a difference from and then add the negatives using to get the answer.
Solution 3
Instead of breaking the sum then rearranging, we can rearrange directly:
Solution 4
The finite arithmetic sequence formula states that the sum in the sequence is equal to where is the number of terms in the sequence, is the first term and is the last term.
Applying the formula, we have:
Video Solution
~savannahsolver
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.