Difference between revisions of "2022 MMATHS Problems/Problem 1"
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==Solution 1== | ==Solution 1== | ||
We solve everything in terms of <math>a</math>. <math>b = 20 - a</math> and <math>c = 2022 - a</math>. Therefore, <math>20-a + 2022-a = 22</math>. Solving for <math>a</math>, we get that <math>a</math> = 1010. Since <math>c = 2022-a, c = 1012</math>. Since <math>b = 20-a, b = -990</math>. Computing <math>\frac {1010-(-990)}{1012-1010}</math> gives us the answer of <math>\boxed {1000}</math>. | We solve everything in terms of <math>a</math>. <math>b = 20 - a</math> and <math>c = 2022 - a</math>. Therefore, <math>20-a + 2022-a = 22</math>. Solving for <math>a</math>, we get that <math>a</math> = 1010. Since <math>c = 2022-a, c = 1012</math>. Since <math>b = 20-a, b = -990</math>. Computing <math>\frac {1010-(-990)}{1012-1010}</math> gives us the answer of <math>\boxed {1000}</math>. | ||
~Arcticturn | ~Arcticturn | ||
Revision as of 19:58, 18 December 2022
Solution 1
We solve everything in terms of 
. 
and 
. Therefore, 
. Solving for , we get that = 1010. Since 
. Since 
. Computing 
gives us the answer of 
.
~Arcticturn
