M7MBA wrote:If n = (33)^43 + (43)^33 what is the units digit of n?

A. 0
B. 2
C. 4
D. 6
E. 8

3¹ --> units digit of 3.
3² --> units digit of 9. (Since the product of the preceding units digit and 3 = 3*3 = 9.)
3³ --> units digit of 7. (Since the product of the preceding units digit and 3 = 9*3 = 27.)
3� --> units digit of 1. (Since the product of the preceding units digit and 3 = 7*3 = 21.)
From here, the units digits will repeat in the same pattern: 3, 9, 7, 1.
The units digit repeat in a CYCLE OF 4.
Implication:
When an integer with a units digit of 3 is raised to a power that is a multiple of 4, the units digit will be 1.

Thus:
33�� and 43³² each have a units digit of 1.
From here, the cycle of units digits will repeat: 3, 9, 7, 1...
Thus:
33�¹ and 43³³ each have a units digit of 3.
33�² has a units digit of 9.
33�³ has a units digit of 7.

Result:
Since n = 33�³ + 43³³, n ---> (units digit of 3) + (units digit of 7) = units digit of 0.

The correct answer is A.

M7MBA wrote:If n = (33)^43 + (43)^33 what is the units digit of n?

A. 0
B. 2
C. 4
D. 6
E. 8

Since we are asked to determine only the units digit of n, we can rewrite the expression as:

n = 3^43 + 3^33

Let's now determine the units digit of 3^43 and 3^33

Let's start by evaluating the pattern of the units digits of 3^n for positive integer values of n. That is, let's look at the pattern of the units digits of powers of 3. When writing out the pattern, notice that we are ONLY concerned with the units digit of 3 raised to each power.

3^1 = 3

3^2 = 9

3^3 = 7

3^4 = 1

3^5 = 3

The pattern of the units digit of powers of 3 repeats every 4 exponents. The pattern is 3-9-7-1. In this pattern, all positive exponents that are multiples of 4 will produce a 1 as its units digit. Thus:

3^44 has a units digit of 1.

Therefore, 3^43 has a units digit of 7.

and

3^32 has a units digit of 1.

Therefore, 3^33 has a units digit of 3.

Thus, n has a units digit of 0, since the units digit of 7 + 3 = 10 is 0.

Answer: A