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TAGS: Senior Manager Joined: 20 May 2014
Posts: 282
Followers: 18

Kudos [?]: 50  , given: 220

13^x + 12^x = n, where x is a positive integer [#permalink]
1
KUDOS 00:00

Question Stats: 53% (01:31) correct 46% (01:17) wrong based on 28 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)
Followers: 0

Kudos [?]: 79 , given: 99

Re: 13^x + 12^x = n, where x is a positive integer [#permalink]
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. Director  Joined: 07 Jan 2018
Posts: 644
Followers: 7

Kudos [?]: 601  , given: 88

Re: 13^x + 12^x = n, where x is a positive integer [#permalink]
1
KUDOS
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

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

Kudos [?]: 5 , given: 1

Re: 13^x + 12^x = n, where x is a positive integer [#permalink]
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
Followers: 0

Kudos [?]: 5 , given: 1

Re: 13^x + 12^x = n, where x is a positive integer [#permalink]
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 [?]: 5 , given: 1

Re: 13^x + 12^x = n, where x is a positive integer [#permalink]
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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