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Difference between revisions of "1991 AJHSME Problems/Problem 4"
m (→‎Solution)
 
Line 8: Line 8:
  
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
991+993+995+997+999=5000-N &\Rightarrow (1000-9)+(1000-7)+(1000-5)+(1000-3)+(1000-1) = 5000-N \\
+
991+993+995+997+999=5000-N \\ &\Rightarrow (1000-9)+(1000-7)+(1000-5)+(1000-3)+(1000-1) = 5000-N \\
 
&\Rightarrow 5\times 1000-(1+3+5+7+9) = 5000 -N \\
 
&\Rightarrow 5\times 1000-(1+3+5+7+9) = 5000 -N \\
 
&\Rightarrow 5000-25=5000-N \\
 
&\Rightarrow 5000-25=5000-N \\

Latest revision as of 22:53, 8 October 2014

Problem

If $991+993+995+997+999=5000-N$, then $N=$

$\text{(A)}\ 5 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 20 \qquad \text{(E)}\ 25$

Solution

\begin{align*} 991+993+995+997+999=5000-N \\ &\Rightarrow (1000-9)+(1000-7)+(1000-5)+(1000-3)+(1000-1) = 5000-N \\ &\Rightarrow 5\times 1000-(1+3+5+7+9) = 5000 -N \\ &\Rightarrow 5000-25=5000-N \\ &\Rightarrow N=25\rightarrow \boxed{\text{E}}. \end{align*}

See Also

1991 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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

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