Difference between revisions of "1985 AHSME Problems/Problem 7"
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==Solution== | ==Solution== | ||
| − | The rightmost part of the expression is <math>c+d</math>, so <math>b-c+d</math> would be grouped as <math>b-(c+d)</math>, and thus the whole expression would be grouped as <math>a\div (b-(c+d)) = | + | The rightmost part of the expression is <math>c+d</math>, so <math>b-c+d</math> would be grouped as <math>b-(c+d)</math>, and thus the whole expression would be grouped as <math>a\div (b-(c+d)) = \boxed{\text{(E)} \ \frac{a}{b-c-d}}</math>. |
==See Also== | ==See Also== | ||
{{AHSME box|year=1985|num-b=6|num-a=8}} | {{AHSME box|year=1985|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 19:05, 19 March 2024
Problem
In some computer languages (such as APL), when there are no parentheses in an algebraic expression, the operations are grouped from right to left. Thus, 
in such languages means the same as 
in ordinary algebraic notation. If 
is evaluated in such a language, the result in ordinary algebraic notation would be
Solution
The rightmost part of the expression is 
, so 
would be grouped as 
, and thus the whole expression would be grouped as 
.
See Also
| 1985 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
