What is the remainder when #9432^6664# is divided by #10# ?

3 Answers
Narad T.
Aug 8, 2017

The remainder is #=6#

Explanation:

We are going to use modular arithmetic

#9432-=2 mod10#

as #9432=9430+2# and #9430# is divisible by #10#

Therefore,

we must calculate

#9432^6664mod10=2^6664mod10#

#6664=2^12+2^11+2^9+2^3#

#2^1-=2 mod10#

#2^2-=2*2 mod10=4mod10#

#2^3-=2*4 mod10=8mod10#

#2^4-=4*4 mod10=6mod10#

#2^8-=6*6 mod10=6mod10#

#2^9-=(2^4)^2*2mod10=36*2mod10=2mod10#

#2^11-=(2^4)^2*2^3mod10=36*8mod10=8mod10#

#2^12-=(2^4)^3mod10=36*6mod10=6mod10#

Therefore,

#2^6664=2^(4096+2048+512+8)-=2^4096*2^2048*2^512*2^8=6*8*2*6mod10=6mod10#

Cesareo R.
Aug 8, 2017

#6#

Explanation:

#9432=2^3*3^2*131#

and

#2*3*131 equiv 6 mod 10# because #131 equiv 1 mod 10# so

the remainder is #6#

NOTE:

#9432^6664=((2 xx 3)^2)^6664 xx 131^6664 xx (2^4)^1666#

George C.
Aug 8, 2017

#6#

Explanation:

Since we are asking for the remainder when divided by #10# and #9432 = 943*10+2#, the remainder when dividing #9432^6664# by #10# is the same as that of dividing #2^6664# by #10#.

Let us look at the pattern of remainders for successive powers of #2#:

#2^0 = 1" "# gives remainder #1#

#2^1 = 2" "# gives remainder #2#

#2^2 = 4" "# gives remainder #4#

#2^3 = 8" "# gives remainder #8#

#2^4 = 16" "# gives remainder #6#

#2^5 = 32" "# gives remainder #2#

#2^6 = 64" "# gives remainder #4#

etc.

Notice that after #2^0#, we get the repeating pattern #2, 4, 8, 6#.

In particular every power that is a multiple of #4# yields a remainder #6#.

Then #6664 = 1666*4# is a multiple of #4#, so #2^6664# gives remainder #6# when divided by #10#.

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