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

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\(4^4\) - Cyclicity of 4:

\(4^1 = 4\)

\(4^2 = 16\)

\(4^3 =\) __\(4\)

\(4^4 =\) __\(6\): Remainder 6

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\(5^4 =\) __\(5\): Remainder 5

\(6^4 =\) __\(6\): Remainder 6

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\(7^4\) - Cyclicity of 7:

\(7^1=7\)

\(7^2 =\) __\(9\)

\(7^3\) =__\(3\)

\(7^4 =\) __\(1\): Remainder 1

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\(8^4\) - Cyclicity of 8:

\(8^1 = 8\)

\(8^2 =\)__\(4\)

\(8^3 =\) __\(2\)

\(8^4 =\) __\(6\): Remainder 6

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\(9^4\) - 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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