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 #1
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27*a mod 40 =17, solve for a

 

a == 40n + 11, where n=0, 1, 2, 3........etc.

 

The smallest "a" =11 and the 2nd smallest "a" =40+11 =51

 Dec 7, 2020
 #2
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Thanks for trying but the answer is incorrect

 Dec 7, 2020
 #3
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Sorry, just don't believe you !! Please give your "correct" answer.

Guest Dec 7, 2020
 #4
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Oh okay the answer is 62

 Dec 7, 2020
 #5
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You wanted the SUM of the smallest(11) and the 2nd smallest(51). Couldn't you ADD [11 + 51] = 62 ???!!! 

Guest Dec 7, 2020
 #6
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Oh sorry I did not add you answer

 Dec 7, 2020

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