What is the Units Digit of the Product (32^28) (33^47) (37^19)? GMAT Problem Solving

Question: What is the units digit of the product (32^28) (33^47) (37^19)?

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

“What is the units digit of the product (32^28) (33^47) (37^19)?”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Quantitative Review”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.

Solution and Explanation:

Approach Solution 1:

Let us consider each numbers provided in the question one by one:

First is 32^28
for powers of 32, the units digit follows the pattern (^power - units digit):
^1 - 2
^2 - 4
^3 - 8
^4 - 6
^5 - 2 and the pattern repeats, every 4 powers
we are looking for the 28th power, 28/4 = 7 R0.
Therefore we use the 4th place in the pattern --> 6

Second is 33^47
for powers of 33, the units digit follows the pattern (^power - units digit):
^1 - 3
^2 - 9
^3 - 7
^4 - 1
^5 - 3 and the pattern repeats, every 4 powers
we are looking for the 47th power, 47/4 = 11 R3.
Therefore we use the 3rd place in the pattern --> 7

Third is 37^19
for powers of 37, the units digit follows the pattern (^power - units digit):
^1 - 7
^2 - 9
^3 - 3
^4 - 1
^5 - 7 and the pattern repeats, every 4 powers
we are looking for the 19th power, 19/4 = 4 R3.
Therefore we use the 3rd place in the pattern --> 3

Hence, we get:
6 x 7 x 3 =126 and the answer is 6

Correct Answer: D

Approach Solution 2:

There are some "nice" numbers that, when raised to various powers, have always the same units digit.
For example, the units digit of 70^n will be 0
Likewise, the units digit of 91^n will be 1
And the units digit of 86^n will be 6

Now, if we notice the question, we see that the exponent 47 is equal to the sum of the other two exponents (28 and 19)
So, it might be useful to take 33^47 and re-write it as (33^28)(33^19)
Later on, we'll apply a nice exponent rule that says "(a^n)(b^n) = (ab)^n"

We get: (32^28)(33^47)(37^19) = (32^28)(33^28)(33^19)(37^19)
= (32^28 x 33^28)(33^19 x 37^19)
= (32 x 33)^28 (33 x 37)^19
=(---6)^28 (---1)^19 [We are using "---" to represent the other digits]
=(----6)(----1) [When ----6 is raised to any power the units digit is always 6. The same applies to ----1]
= -------6

Correct Answer: D

Approach Solution 3:

We know that the exponent 47 is equal to the SUM of the other two exponents (28 and 19)
So, it might be useful to take 33^47 and rewrite it as (33^28)(33^19)
Applying the rule that says "(a^n)(b^n) = (ab)^n"

We get: (32^28)(33^47)(37^19) = (32^28)(33^28)(33^19)(37^19)
= (32^28 x 33^28)(33^19 x 37^19)
= (32 x 33)^28 (33 x 37)^19
=(---6)^28 (---1)^19 [Using "---" to represent the other digits]
=(----6)(----1)

As per numeric ----6 is raised to any power the units digit is always 6. The same applies to ----1

Correct Answer: D

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