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Location: Canada
What is the units digit of the product (32^28) (33^47) (37^19)?
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Updated on: 13 Nov 2019, 17:16
Updated on: 13 Nov 2019, 17:16
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Difficulty:
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Question Stats:
71% (01:39) correct
29%
(01:55)
wrong
based on 543
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:
*kudos for all correct solutions
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
Originally posted by BrentGMATPrepNow on 09 Apr 2017, 07:49.
Last edited by BrentGMATPrepNow on 13 Nov 2019, 17:16, edited 2 times in total.
Last edited by BrentGMATPrepNow on 13 Nov 2019, 17:16, edited 2 times in total.
GMAT Club Legend
Joined: 12 Sep 2015
Posts: 6803
Given Kudos: 799
Location: Canada
What is the units digit of the product (32^28) (33^47) (37^19)?
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Updated on: 29 Sep 2020, 13:06
Updated on: 29 Sep 2020, 13:06
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GMATPrepNow
What is the units digit of the product (32^28)(33^47)(37^19)?
A) 0
B) 2
C) 4
D) 6
E) 8
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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Originally posted by BrentGMATPrepNow on 10 Apr 2017, 08:55.
Last edited by BrentGMATPrepNow on 29 Sep 2020, 13:06, edited 1 time in total.
Last edited by BrentGMATPrepNow on 29 Sep 2020, 13:06, edited 1 time in total.
Re: What is the units digit of the product (32^28) (33^47) (37^19)?
[#permalink]
09 Apr 2017, 09:24
09 Apr 2017, 09:24
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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
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
