Difference between revisions of "1991 AJHSME Problems/Problem 4"
5849206328x (talk | contribs) (Created page with '==Problem== If <math>991+993+995+997+999=5000-N</math>, then <math>N=</math> <math>\text{(A)}\ 5 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 20 \qquad \text{…') |
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<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 \\ | ||
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{{AJHSME box|year=1991|num-b=3|num-a=5}} | {{AJHSME box|year=1991|num-b=3|num-a=5}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 22:53, 8 October 2014
Problem
If 
, then 
Solution
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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