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Suppose that x is an integer that satisfies the following congruences:

3+x =2^2 (Mod 3^3)

5+x=3^2 (Mod 5^3)

7+x=5^2 (Mod 7^3)

 

What is the remainder when x is divided by 105?

 #1
avatar+352 
+1

105=3x7x5

 

3+x=4(mod 27)

5+x=9(mod125)

7+x=25(mod343)

 

x=1(mod27)

x=4(mod125)

x=18(mod343)

 

27k+1=x

3(9k)+1=x

 

x=1(mod3)

 

for the same reasons:

 

x=4(mod5)

x=4(mod7)

 

x=4(mod 35)

 

possible sollutions lower than 105:

 

4, 39, 74.

 

the only one that has a remainder 1 when divided by 3 is $\boxed{4}$

 Mar 3, 2021
edited by SparklingWater2  Mar 3, 2021
 #2
avatar+22 
+1

Tysm!!!!


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