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# 13^x + 12^x = n, where x is a positive integer

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Senior Manager
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13^x + 12^x = n, where x is a positive integer [#permalink]  24 Oct 2017, 01:22
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Question Stats:

45% (01:31) correct 54% (01:15) wrong based on 37 sessions
$$13^x + 12^x = n$$, where x is a positive integer

 Quantity A Quantity B The number of distinct numbers that can be the units digit of n 4

A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given

Kudos for correct solution.
[Reveal] Spoiler: OA
Manager
Joined: 29 Nov 2017
Posts: 190
Location: United States
GRE 1: Q142 V146
WE: Information Technology (Computer Software)
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Kudos [?]: 93 [0], given: 99

Re: 13^x + 12^x = n, where x is a positive integer [#permalink]  18 May 2018, 01:17
I think option A should be the answer since we raise 13 to different first few positive powers I got different units therefore same when we are adding 13^x +12^x then we are ought to get more than 4 types of unit digit.

Active Member
Joined: 07 Jan 2018
Posts: 694
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Kudos [?]: 769 [1] , given: 88

Re: 13^x + 12^x = n, where x is a positive integer [#permalink]  18 May 2018, 03:25
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No need to raise 13 and 12 to any digits. Since we are interested in finding the unit digits only we can work with 3 and 2.
We should also keep in mind that the unit digits of any number will reveal a pattern when raised to certain numbers.
For eg.

$$3^1 =3$$
$$3^2 = 9$$
$$3^3 = 27$$
$$3^4 = 81$$

$$3^5 = 243$$
$$3^6 = 729$$

You can notice that the units digit of $$3^5$$ is same as $$3^1$$ and units digit of $$3^2$$ is same as $$3^6$$
Hence the units digits repeat in an interval of 4
Similar is the case for the powers of 2

$$2^1 =2$$
$$2^2 = 4$$
$$2^3 = 8$$
$$2^4 = 16$$
From here on the units digit of powers of 2 will repeat.

Now we just need to add the respective units digit of 2 and 3 for a given digit of power
For x = 1; 3+2 = 5
for x = 2; 9+4 = 13 or, 3
for x = 3; 7+8 = 15 or, 5
for x = 4; 1+6 = 7

since 5 is repeated we count it as 1.
Therefore there are in total 3 distinct unit digits
option B
_________________

This is my response to the question and may be incorrect. Feel free to rectify any mistakes
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Intern
Joined: 15 May 2018
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Kudos [?]: 6 [0], given: 1

Re: 13^x + 12^x = n, where x is a positive integer [#permalink]  22 May 2018, 05:19
IshanGre wrote:
I think option A should be the answer since we raise 13 to different first few positive powers I got different units therefore same when we are adding 13^x +12^x then we are ought to get more than 4 types of unit digit.

Intern
Joined: 15 May 2018
Posts: 38
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Kudos [?]: 6 [0], given: 1

Re: 13^x + 12^x = n, where x is a positive integer [#permalink]  09 Jul 2018, 18:38
Bunuel wrote:
$$13^x + 12^x = n$$, where x is a positive integer

 Quantity A Quantity B The number of distinct numbers that can be the units digit of n 4

A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given

Kudos for correct solution.

Intern
Joined: 15 May 2018
Posts: 38
Followers: 0

Kudos [?]: 6 [0], given: 1

Re: 13^x + 12^x = n, where x is a positive integer [#permalink]  09 Jul 2018, 18:39
Bunuel wrote:
$$13^x + 12^x = n$$, where x is a positive integer

 Quantity A Quantity B The number of distinct numbers that can be the units digit of n 4

A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given

Kudos for correct solution.

Re: 13^x + 12^x = n, where x is a positive integer   [#permalink] 09 Jul 2018, 18:39
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