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

-----ASIDE-----------------------------------------
There are some "nice" numbers that, when raised to various powers, ALWAYS have the same units digit.
For example, the units digit of 70^n will be 0 FOR ALL POSITIVE INTEGER VALUES OF N
Likewise, the units digit of 91^n will be 1 FOR ALL POSITIVE INTEGER VALUES OF N
And the units digit of 86^n will be 6 FOR ALL POSITIVE INTEGER VALUES OF N
-----NOW ONTO THE QUESTION-----------------

Notice 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)
NOTE: 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 [applied above rule]
=(---6)^28 (---1)^19 [I'm focusing solely on the units of each product. So, I use "---" 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

Answer:

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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

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

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

6 x 7 x 3 =126 and the answer is D) 6

That's a valid approach, amynicole.
There's another approach that's much faster. Can you spot it?

Hint:

Cheers,
Brent

Hi Brent,

How does "notice that the largest exponent is equal to the sum of the other two exponents" Helps? All(2, 3, 7, 8) the 3 numbers have cycle of 4 repetitions.

Thanks
Nandish
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I also solved it by using cycle and took me around 2:30 minutes
But this hint is genius

32^28 * 33^47 * 37^19 = (32^28 * 33^28 )*(33^19 * 37 ^19) == (32 *33)^28 * [Rest Does not matter] as 6 has cycle of 1 and unit digit will always be 6.

Kudos to GMATPrepNow

Wow! This is cool. I did it the long way...hopefully I can remember this on the test

Posted from my mobile device

Cyclicity of number 2 = 4

So, \(32^{28}\) will have units digit 6

Cyclicity of number 3 = 4

So, \(33^{47} = 33^{44+3}\) will have units digit 7

Cyclicity of number 7 = 4

So, \(37^{19} = 37^{16+3}\) will have units digit 3

Thus, the units digit of \(32^{28}*33^{47}*37^{19} = 6*7*3 = 6\); Answer will be (D) 6
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Awesome Explanation GMATPrepNow
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\(2^4 = 6\)

So, \(32^{28} = 2^{140} = 2^{4*35} = \)Units digit \(6\)

\(3^4 = 1\) & \(3^3 = 7\)

So, \(33^{47} = 33^{4*11+3} = 1*7 =\) units digit \(7\)

\(7^4 = 1\) & \(7^3 = 3\)

So, \(37^{19} = 37^{4*4+3} =\) Units digit 3

Thus, units digit is \(6*7*3 = 6\), Answer is (D)
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