when t^4 is divided by 10, the remainder is r

Hi Fedemaravilla

IMO the answer should be \(3\)

Given \(t^4=10q+r\), where \(q\) is the quotient when \(t^4\) is divided by \(10\)

so this implies that \(r\) will be the units digit of \(t^4\).

Now any positive number raised to the power \(4\) can have only four possible unit's digit - 0, 1, 5 & 6. but it is given that \(t^4\) is not a multiple of \(10\), hence will not have \(0\) as its unit's digit

Hence \(r\) can take \(3\) values

Option A

Hi Bunuel - can you confirm the answer and provide more clarity?

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

A couple of concepts are tested in this question:

1. When an integer is divided by 10, the remainder is the units digit of the integer. This is discussed here: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2015/1 ... questions/

When t^4 is divided by 10, the remainder is r. So r must be the units digit of t^4.

Which all values can the units digit of a fourth power take?

2. A perfect square cannot end in 2/3/7/8. This post shows the squares with various units digits:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2015/1 ... -the-gmat/

Other 6 digits are possible. But since t is not a multiple of 10, t^2 will not end in a 0.
So t^2 can end in 1/4/5/6/9.
t^4 will be square of these digits.
..1^2 = ..1
..4^2 = ..6
..5^2 = ..5
..6^2 = ..6
..9^2 = ..1

So t^4 will end in 1/5/6 (the fourth powers are also shown in the post above)
Only 3 values are possible.

Answer (A)

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