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Difference between revisions of "1985 AJHSME Problem 2"

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The sum of each pair is <math>189</math> and there are <math>5</math> pairs, so the sum is <math>945</math> and the answer is <math>\boxed{\textbf{(B)}\ 945}</math>.
 
The sum of each pair is <math>189</math> and there are <math>5</math> pairs, so the sum is <math>945</math> and the answer is <math>\boxed{\textbf{(B)}\ 945}</math>.
  
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==Cheap Solution==
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==Solution 3 (Cheap and Quick)==
 
We know that <math>10(90) = 900</math> and <math>10(100) = 1000.</math> Quick estimation reveals that this sum is in between these two numbers, so the only answer available is <math>\boxed{\textbf{(B)}\ 945}</math>.
 
We know that <math>10(90) = 900</math> and <math>10(100) = 1000.</math> Quick estimation reveals that this sum is in between these two numbers, so the only answer available is <math>\boxed{\textbf{(B)}\ 945}</math>.
  
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==Solution 3==
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==Solution <math>3\frac{1}{2}</math>==
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You see that <math>90+91+92+93+94+95+96+97+98+99</math> is equal to <math>0+1+3+4+5+6+7+8+9+900</math>. You can use the formula <math>\frac{n(n+1)}{2}</math> to get 45+900=945
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You see that <math>90+91+92+93+94+95+96+97+98+99</math> is equal to <math>0+1+3+4+5+6+7+8+9+900</math>. You can use the formula <math>\frac{n(n+1)}{2}</math> to get <math>45+900=\boxed{\textbf{(B)}\ 945}</math>.
  
 
==Video Solution==
 
==Video Solution==

Latest revision as of 08:40, 22 March 2025

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

We can add as follows: \[90+91+92+93+94+95+96+97+98+99= 10(90) +1+2+3+4+5+6+7+8+9 = 900 + 45 = \boxed{945}\] The answer is $\boxed{\textbf{(B)}\ 945}$.

Solution 2

Pair the numbers like so: \[(90+99)+(91+98)+(92+97)+(93+96)+(94+95)\] The sum of each pair is $189$ and there are $5$ pairs, so the sum is $945$ and the answer is $\boxed{\textbf{(B)}\ 945}$.

Solution 3 (Cheap and Quick)

We know that $10(90) = 900$ and $10(100) = 1000.$ Quick estimation reveals that this sum is in between these two numbers, so the only answer available is $\boxed{\textbf{(B)}\ 945}$.

Solution $3\frac{1}{2}$

You see that $90+91+92+93+94+95+96+97+98+99$ is equal to $0+1+3+4+5+6+7+8+9+900$. You can use the formula $\frac{n(n+1)}{2}$ to get $45+900=\boxed{\textbf{(B)}\ 945}$.

Video Solution

https://youtu.be/1NtsgKc6mXs

~savannahsolver

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