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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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New post 09 Apr 2017, 06:49
1
Top Contributor
6
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

69% (01:54) correct 31% (01:58) wrong based on 191 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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Re: What is the units digit of the product (32^28) (33^47) (37^19)?  [#permalink]

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New post 10 Apr 2017, 07:55
5
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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

Answer:

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

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New post 09 Apr 2017, 08:24
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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Re: What is the units digit of the product (32^28) (33^47) (37^19)?  [#permalink]

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New post 09 Apr 2017, 13:11
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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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Re: What is the units digit of the product (32^28) (33^47) (37^19)?  [#permalink]

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New post 09 Apr 2017, 15:55
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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Re: What is the units digit of the product (32^28) (33^47) (37^19)?  [#permalink]

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New post 09 Apr 2017, 18:40
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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Re: What is the units digit of the product (32^28) (33^47) (37^19)?  [#permalink]

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New post 10 Apr 2017, 09:58
1
Wow! This is cool. I did it the long way...hopefully I can remember this on the test

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

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New post 10 Apr 2017, 10:17
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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New post 11 Apr 2017, 06:24
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

Answer:

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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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