Difference between revisions of "1985 AJHSME Problems/Problem 2"

Revision as of 21:28, 1 February 2023

Problem

$90+91+92+93+94+95+96+97+98+99=$

$\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045$

Solution 1

To simplify the problem, we can group 90’s together: $90 + 91 + ... + 98 + 99 = 90 \cdot 10 + 1 + 2 + 3 + ... + 8 + 9$ .

$90\cdot10=900$ , and finding $1 + 2 + ... + 8 + 9$ has a trick to it.

Rearranging the numbers so each pair sums up to 10, we have: $(1 + 9)+(2+8)+(3+7)+(4+6)+5$ . $4\cdot10+5 = 45$ , and $900+45=\boxed{\text{B} 945}$ .

Solution 2

We can express each of the terms as a difference from 100 and then add the negatives using $\frac{n(n+1)}{2}$ to get the answer. $\begin{align*} (100-10)+(100-9)+\cdots + (100-1) &= 100 \times 10 -(1+2+\cdots +9+10)\\ &= 1000 - 55\\ &= 945 \rightarrow \boxed{\text{B}} \end{align*}$

Solution 3

Instead of breaking the sum and then rearranging, we can start by rearranging: $\begin{align*} 90+91+92+\cdots +98+99 &= (90+99)+(91+98)+(92+97)+(93+96)+(94+95) \\ &= 189+189+189+189+189 \\ &= 945\rightarrow \boxed{\text{B}} \end{align*}$

Solution 4

We can use the formula for finite arithmetic sequences.

It is $\frac{n}{2}\times$ ( $a_1+a_n$ ) where $n$ is the number of terms in the sequence, $a_1$ is the first term and $a_n$ is the last term.

Applying it here:

$\frac{10}{2} \times (90+99) = 945 \rightarrow \boxed{B}$

Video Solution

https://youtu.be/1NtsgKc6mXs

~savannahsolver

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.

@@ Line 7: / Line 7: @@
 ==Solution 1==
-One possibility is to simply add them. However, this can be time-consuming, and there are other ways to solve this problem.
+To simplify the problem, we can group 90’s together: <math>90 + 91 + ... + 98 + 99 = 90 \cdot 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: <math>900</math>. So how do we find <math>1 + 2 + ... + 8 + 9</math>?
+<math>90\cdot10=900</math>, and finding <math>1 + 2 + ... + 8 + 9</math> has a trick to it.
-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.
+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>.
-is <math>\boxed{\text{B}}</math>
 ==Solution 2==

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