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TAGS: GRE Instructor Joined: 10 Apr 2015
Posts: 2600
Followers: 96

Kudos [?]: 2800  , given: 41

What is the units digit of the product [#permalink]
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Expert's post 00:00

Question Stats: 12% (00:01) correct 87% (01:14) wrong based on 8 sessions
What is the units digit of the product $$(32^{28})(33^{47})(37^{19})$$?

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

Hint:
[Reveal] Spoiler:
There’s a fast approach and a slow approach
[Reveal] Spoiler: OA

_________________

Brent Hanneson – Creator of greenlighttestprep.com GRE Instructor Joined: 10 Apr 2015
Posts: 2600
Followers: 96

Kudos [?]: 2800  , given: 41

Re: What is the units digit of the product [#permalink]
1
KUDOS
Expert's post
GreenlightTestPrep wrote:
What is the units digit of the product $$(32^{28})(33^{47})(37^{19})$$?

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

Hint:
[Reveal] Spoiler:
There’s a fast approach and a slow approach

-----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

Cheers,
Brent
_________________

Brent Hanneson – Creator of greenlighttestprep.com VP Joined: 20 Apr 2016
Posts: 1087
WE: Engineering (Energy and Utilities)
Followers: 16

Kudos [?]: 938  , given: 226

Re: What is the units digit of the product [#permalink]
1
KUDOS
GreenlightTestPrep wrote:
What is the units digit of the product $$(32^{28})(33^{47})(37^{19})$$?

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

Hint:
[Reveal] Spoiler:
There’s a fast approach and a slow approach

Here,

Lets take the sequence of the unit digit that repeat itself:: Plz see the attach diag

The last digit of power of 2 repeat in a cycle of numbers – 2, 4, 8, 6

The last digit of power of 3 repeat in a cycle of numbers – 3, 9, 7, 1

The last digit of power of 4 repeat in a cycle of numbers – 4, 6

The last digit of power of 7 repeat in a cycle of numbers – 7 , 9 ,3 ,1

The last digit of power of 8 repeat in a cycle of numbers – 8, 4, 2, 6

The last digit of power of 9 repeat in a cycle of numbers – 9,1

Now to the ques.

$$(32^{28})(33^{47})(37^{19})$$

$$32^{28}$$ = last digit is 2 and to the power of 28. Since $$2^{power}$$ repeats after every 4th , hence we divide the power by 4 i.e $$\frac{28}{4}$$= 7 and reminder =0 . The last digit of $$32^{28}$$ = 6

Similarly for $$33^{47}$$ = here $$3^{power}$$ repeats after every 4th, so $$\frac{47}{4}$$= 11 and reminder 3, The last digit will be = 7

And for $$37^{19}$$ = here $$7^{power}$$ repeats after every 4th, so $$\frac{19}{4}$$ = 4 and reminder 3, The last digit will be = 3

Combining the digit = $$6*7*3 = 126$$ ,

the last digit for the whole equation is = 6
Attachments Unit digit.png [ 23.51 KiB | Viewed 299 times ]

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