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

+2
57
2
+22

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
+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
+22
+1

Tysm!!!!