Difference between revisions of "1978 AHSME Problems/Problem 4"
(Created page with "== Problem 4 == If <math>a = 1,~ b = 10, ~c = 100</math>, and <math>d = 1000</math>, then <math>(a+ b+ c-d) + (a + b- c+ d) +(a-b+ c+d)+ (-a+ b+c+d)</math> is equal to <math...") |
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==Solution 1== | ==Solution 1== | ||
Adding all four of the equations up, we can see that it equals | Adding all four of the equations up, we can see that it equals | ||
− | <cmath>3(a+b+c+d)</cmath> | + | <cmath>3(a+b+c+d)</cmath> |
− | This is equal to <math>3(1111) = \boxed{\textbf{(C) }3333}</math> | + | This is equal to <math>3(1111) = \boxed{\textbf{(C) }3333}</math> ~awin |
+ | |||
+ | ==See Also== | ||
+ | {{AHSME box|year=1978|num-b=3|num-a=5}} | ||
+ | {{MAA Notice}} |
Latest revision as of 11:00, 13 February 2021
Problem 4
If , and , then is equal to
Solution 1
Adding all four of the equations up, we can see that it equals This is equal to ~awin
See Also
1978 AHSME (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 • 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.