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

 

 

b)  Find some integer that  is the multiplicative inverse to 160 modulo 1399.

 May 12, 2019
 #1
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a)  27a = 17  (mod 40)

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

So, the smallest and 2nd smallest positive integers that satisfy the congrurence are:

11 and (40 x 1) + 11 =51

 

b)  160^(-1) mod 1399 = 1058.

 May 12, 2019
edited by Guest  May 12, 2019
 #2
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sorry my congruence for a is:  27a = 17  (mod 40)

 May 12, 2019
 #3
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See a) above:

 May 12, 2019

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