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What is the modulo 5 remainder of the sum 1+2+3+4+5+...+120+121+122+123??

 Apr 1, 2021
 #1
avatar+121004 
+1

Sum of 1st   120  integers   =   (120) (121)  /  2    =   60 * 121

 

(60 * 121)  mod 5  = 0       since 60 is evenly divisiblle  by  5

 

121 + 122 + 123  =

 

366

 

366 mod 5  =  1

 

Remainder =  1

 

cool cool cool

 Apr 1, 2021
 #2
avatar+53 
+1

Thank you!

QuestionableBean  Apr 1, 2021
 #3
avatar+121004 
0

Welcome....!!!

 

 

cool cool cool

CPhill  Apr 1, 2021

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