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

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

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09 Apr 2017, 07:49
1
Top Contributor
5
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Difficulty:

45% (medium)

Question Stats:

69% (01:39) correct 31% (01:31) wrong based on 179 sessions

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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:
There’s a fast approach and a slow approach

*kudos for all correct solutions

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Brent Hanneson – Founder of gmatprepnow.com

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Joined: 12 Sep 2015
Posts: 2713
Re: What is the units digit of the product (32^28) (33^47) (37^19)?  [#permalink]

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10 Apr 2017, 08:55
4
Top Contributor
3
GMATPrepNow 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

-----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

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Brent Hanneson – Founder of gmatprepnow.com

##### General Discussion
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Joined: 10 Dec 2016
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Re: What is the units digit of the product (32^28) (33^47) (37^19)?  [#permalink]

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09 Apr 2017, 09:24
2
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
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Joined: 12 Sep 2015
Posts: 2713
Re: What is the units digit of the product (32^28) (33^47) (37^19)?  [#permalink]

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09 Apr 2017, 14:11
1
Top Contributor
That's a valid approach, amynicole.
There's another approach that's much faster. Can you spot it?

Hint:
notice that the largest exponent is equal to the sum of the other two exponents

Cheers,
Brent
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Brent Hanneson – Founder of gmatprepnow.com

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Re: What is the units digit of the product (32^28) (33^47) (37^19)?  [#permalink]

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09 Apr 2017, 16:55
GMATPrepNow wrote:
That's a valid approach, amynicole.
There's another approach that's much faster. Can you spot it?

Hint:
notice that the largest exponent is equal to the sum of the other two exponents

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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GMAT 1: 730 Q47 V42
GPA: 3.95
Re: What is the units digit of the product (32^28) (33^47) (37^19)?  [#permalink]

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09 Apr 2017, 19:40
2
GMATPrepNow wrote:
That's a valid approach, amynicole.
There's another approach that's much faster. Can you spot it?

Hint:
notice that the largest exponent is equal to the sum of the other two exponents

Cheers,
Brent

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
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Posts: 89
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GMAT 1: 740 Q49 V42
GPA: 3.49
Re: What is the units digit of the product (32^28) (33^47) (37^19)?  [#permalink]

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10 Apr 2017, 10:58
1
Wow! This is cool. I did it the long way...hopefully I can remember this on the test

Posted from my mobile device
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Re: What is the units digit of the product (32^28) (33^47) (37^19)?  [#permalink]

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10 Apr 2017, 11:17
2
GMATPrepNow 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

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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Re: What is the units digit of the product (32^28) (33^47) (37^19)?  [#permalink]

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11 Apr 2017, 07:24
GMATPrepNow wrote:
GMATPrepNow 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

-----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

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Awesome Explanation GMATPrepNow
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Re: What is the units digit of the product (32^28) (33^47) (37^19)?  [#permalink]

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17 Jul 2018, 21:59
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