Fedemaravilla wrote:
when t^4 is divided by 10, the remainder is r. If t can be any positive integer that is not a multiple of 10, then there are how many different possible values for r?
(A) Three
(B) Four
(C) Six
(D) Nine
(E) Ten
If a number is divided by 10, the remainder is the last digit of the number.
We need the last digits of 1 through 9 to the 4th power. The number can't be a multiple of 10, and two+ digit numbers end in 1 through 9.
I have not memorized the last digit of all single integers' powers of 4.
So I listed them quickly, using cyclicity if needed. Well under a minute for the problem
\(1^4 = 1\): Remainder 1
\(2^4 = 16\): Remainder 6
\(3^4 = 81\): Remainder 1
---------------------------------
\(4^4\) - Cyclicity of 4:
\(4^1 = 4\)
\(4^2 = 16\)
\(4^3 =\) __\(4\)
\(4^4 =\) __\(6\): Remainder 6
-----------------------------------
\(5^4 =\) __\(5\): Remainder 5
\(6^4 =\) __\(6\): Remainder 6
-----------------------------------
\(7^4\) - Cyclicity of 7:
\(7^1=7\)
\(7^2 =\) __\(9\)
\(7^3\) =__\(3\)
\(7^4 =\) __\(1\): Remainder 1
------------------------------------
\(8^4\) - Cyclicity of 8:
\(8^1 = 8\)
\(8^2 =\)__\(4\)
\(8^3 =\) __\(2\)
\(8^4 =\) __\(6\): Remainder 6
-------------------------------------
\(9^4\) - Cyclicity of 9:
\(9^1 = 9\)
\(9^2\) = __\(1\)
\(9^3\)=__\(9\)
\(9^4\)=__\(1\): Remainder 1
There are three remainders when a number to the fourth power is divided by 10: 1, 5, and 6.
Answer A
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