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=\(\frac{44^4−44}{44}\)

=\(\frac{44^344-44}{44}\)

=\(\frac{44(44^3-1)}{44}\)

=\(44^3-1\)

=Since 4 has a cyclicity of 2. Then \(44^3\) will end in 4. Hence the unit digit should be 3.

D is the answer.
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Bunuel
\(\frac{44^4 − 44}{44} =\)


A. 16,432
B. 70,321
C. 83,244
D. 85,183
E. 93,155


\(\frac{44^4 − 44}{44} =\)

= \(\frac{44(44^3 -1)}{44}\)

= \(44^3 - 1\)

Unit digits of 44^3 is the same as 4^3 .

\(4^3 = 64.\)

Now deduct 1 from 4.

4-1 = 3.

The only option is D .
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Bunuel
\(\frac{44^4 − 44}{44} =\)


A. 16,432
B. 70,321
C. 83,244
D. 85,183
E. 93,155

Re-expressing the single fraction as two separate fractions, we have:

44^4/44 - 44/44

44^3 - 1

Instead of calculating the actual difference, let’s only calculate the units digit of the difference.

Recall that the pattern of units digits of 4 raised to positive integer exponents is 4, 6, 4, 6, … , so odd powers of 44 will end in 4, and even powers of 44 will end in 6.

Thus, we see that 44^3 ends in a 4, so 44^3 - 1 must have a units digit of 3. The only answer choice that is a number with a units digit of 3 is choice D.

Answer: D
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