+0  
 
+1
946
1
avatar+586 

What is the sum of the smallest and second-smallest positive integers \(a\) satisfying the congruence

 

 \(27a\equiv 17 \pmod{40}~?\)

 Jul 26, 2019
 #1
avatar
+1

27a = 17 (mod 40)
27 mod 40 = 27 + 13
27*2 mod 40 =      14
27*5 mod 40 =      15
27*8 mod 40 =      16
27*11 mod 40=      17
a = 40m + 11, where m =0, 1, 2, 3......etc.
The smallest value of "a" =[40*0 + 11] = 11
The 2nd smallest value of "a" =[40*1 + 11] = 51

The sum  =11 + 51 = 62

 Jul 26, 2019

4 Online Users