Suggest a shortest method to solve it

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Suggest a shortest method to solve it

by Jeevanantham » Fri Jun 29, 2012 1:47 am
If n = (33)^43 + (43)^33, what is the unit digit of n ?

A) 0

B) 2

C) 4

D) 6

E) 8

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by Anurag@Gurome » Fri Jun 29, 2012 2:01 am
Jeevanantham wrote:If n = (33)^43 + (43)^33, what is the unit digit of n?
Unit's digit of powers of 3 has the following pattern
  • Unit's digit of 3^(Multiple of 4) = 1
    Unit's digit of 3^(Multiple of 4 + 1) = 3
    Unit's digit of 3^(Multiple of 4 + 2) = 9
    Unit's digit of 3^(Multiple of 4 + 3) = 7
Unit's digit of 33^43 = unit's digit of (4*8 + 1)^43 = 3
Unit's digit of 43^33 = unit's digit of (4*10 + 3)^43 = 7

Hence, unit's digit of n = unit's digit of (7 + 3) = 0

The correct answer is A.
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by vikrambansal » Sun May 17, 2015 5:35 am
Unit's digit of 33^43 = unit's digit of (4*8 + 1)^43 = 3
Unit's digit of 43^33 = unit's digit of (4*10 + 3)^43 = 7
If 43 = 4*10 + 3 then why not 33 = 3*10 + 3? Then the answer would be 7 + 7 = 14 or 4 (C)

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by GMATGuruNY » Sun May 17, 2015 6:05 am
vikrambansal wrote:
Unit's digit of 33^43 = unit's digit of (4*8 + 1)^43 = 3
Unit's digit of 43^33 = unit's digit of (4*10 + 3)^43 = 7
If 43 = 4*10 + 3 then why not 33 = 3*10 + 3? Then the answer would be 7 + 7 = 14 or 4 (C)
3¹ --> units digit of 3.
3² --> units digit of 9. (Since the product of the preceding units digit and 3 = 3*3 = 9.)
3³ --> units digit of 7. (Since the product of the preceding units digit and 3 = 9*3 = 27.)
3� --> units digit of 1. (Since the product of the preceding units digit and 3 = 7*3 = 21.)
From here, the units digits will repeat in the same pattern: 3, 9, 7, 1.
The units digit repeat in a CYCLE OF 4.
Implication:
When an integer with a units digit of 3 is raised to a power that is a multiple of 4, the units digit will be 1.

Thus:
33�� and 43³² each have a units digit of 1.
From here, the cycle of units digits will repeat: 3, 9, 7, 1...
Thus:
33�¹ and 43³³ each have a units digit of 3.
33�² has a units digit of 9.
33�³ has a units digit of 7.

Result:
Since n = 33�³ + 43³³, n ---> (units digit of 3) + (units digit of 7) = units digit of 0.

The correct answer is A.
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by Brent@GMATPrepNow » Sun May 17, 2015 7:15 am
If you're interested, we have a free video on finding the units digit of large powers: https://www.gmatprepnow.com/module/gmat- ... ts?id=1031

I also wrote an article on the same topic: https://www.gmatprepnow.com/articles/uni ... big-powers

The article ends with two additional questions to practice with.

Cheers,
Brent
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by vikrambansal » Sun May 17, 2015 8:37 am
Thanks to both of you. I realized that it's not derivation of no. 33 (10*3 + 3 or 8*4 + 1) but POWER 33.

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by andymal » Sun May 17, 2015 7:47 pm
Thanks all. I fully understand after the breakdown. This is a great example of a deeper question that comes to mind on quant questions like this. What would give you the clues or the signals to think to break it down? What should one be thinking, or what's the thought process to know how to unpack a question like this?