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13^x + 12^x = n, where x is a positive integer
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24 Oct 2017, 01:22
24 Oct 2017, 01:22
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Question Stats:
45% (01:34) correct
54% (01:30) wrong based on 74 sessions
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\(13^x + 12^x = n\), where x is a positive integer
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.
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
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18 May 2018, 01:17
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.
Please clarify?
Please clarify?
Re: 13^x + 12^x = n, where x is a positive integer
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18 May 2018, 03:25
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
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
