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What is the unique three-digit positive integer  satisfying 100x = 11 (mod 997)

 Jun 30, 2021
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11 + 997a = 100x

Since 11 + 997a needs to be divisable by 100, the last 2 digits need to be 00. 

11 + 997a = 0 (mod 100)

11 + 97a = 0 (mod 100)

a = 37

 

11 + 997(37) = 100x

36900 = 100x

x = 369

 

=^._.^=

 Jun 30, 2021

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