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Difficulty:
25%
(medium)
Question Stats:
74% (01:17) correct
26%
(01:44)
wrong
based on 77
sessions
History
Date
Time
Result
Not Attempted Yet
Find the unit digit of (abc)^k where a, b and c are the digits of a three digit positive integer abc.
1. c is prime and even.
2. k leaves a remainder of 1 when divided by 4.
1. c is prime and even.
2. k leaves a remainder of 1 when divided by 4.
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
Celestial09
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Posts: 214
Hi,
IMO is should be C
st 1.. c is 2. nothing about k is known.. hence insufficient
st2 k could be anything with 1 as a remainder.. hence.. nothing could be known as we don't know c. insufficient.
combining both... c is 2 and k leaves remainder of 1 when divides by 4. remember.. 2 ha cyclicity fo 4.. hence unit digit is 2.
Thanks
Celestial
kudos for a correct solution.
IMO is should be C
st 1.. c is 2. nothing about k is known.. hence insufficient
st2 k could be anything with 1 as a remainder.. hence.. nothing could be known as we don't know c. insufficient.
combining both... c is 2 and k leaves remainder of 1 when divides by 4. remember.. 2 ha cyclicity fo 4.. hence unit digit is 2.
Thanks
Celestial
kudos for a correct solution.
Statement 1: \(c\) is prime and even. \(c = 2\). I don't know anything about \(a, b, or\) \(k\). Insufficient.
Statement 2: \(k\) leaves a remainder of \(1\) when divided by \(4\). Possible options for \(k\) are \(0, 5, 9, 13, ...\)I don't know anything about \(a, b, or\) \(c\). Insufficient.
Statement 1& 2:
Combining yields
\((ab2)^1\) and the units digit here is \(2\)
\((ab2)^5\) and the units digit here is \(2\) (\(2*2*2*2*2\) \(= 3\)2)
\((ab2)^9\) and the units digit here is \(2\) (I'm not writing this one all the way out. \(2^9 = 51\)2)
I'm not going to test \((ab2)^{13}\) since I see that I have three answers where the unit digit is \(2\).
The question asks me to find the units digit of \((abc)^k\) and in all of the cases it is \(2\). Therefore C is sufficient.
Answer: C
Statement 2: \(k\) leaves a remainder of \(1\) when divided by \(4\). Possible options for \(k\) are \(0, 5, 9, 13, ...\)I don't know anything about \(a, b, or\) \(c\). Insufficient.
Statement 1& 2:
Combining yields
\((ab2)^1\) and the units digit here is \(2\)
\((ab2)^5\) and the units digit here is \(2\) (\(2*2*2*2*2\) \(= 3\)2)
\((ab2)^9\) and the units digit here is \(2\) (I'm not writing this one all the way out. \(2^9 = 51\)2)
I'm not going to test \((ab2)^{13}\) since I see that I have three answers where the unit digit is \(2\).
The question asks me to find the units digit of \((abc)^k\) and in all of the cases it is \(2\). Therefore C is sufficient.
Answer: C
Quick soln
Statement 1: c known as 2 since only even prime but k unknown. reject.
Statement 2: k=4N+1. c unknown. reject.
Combined statement: 2^(4n+1) always gives 2 at the units place. solvable. Thus C.
---------------
Few advanced concepts which one may like to know (not at all required though) are Fermat's Little Theorem, Euler's totient function & Eueler's totient theorem. Highly unlikely to be tested but a very useful tool for remainder type questions.
Statement 1: c known as 2 since only even prime but k unknown. reject.
Statement 2: k=4N+1. c unknown. reject.
Combined statement: 2^(4n+1) always gives 2 at the units place. solvable. Thus C.
---------------
Few advanced concepts which one may like to know (not at all required though) are Fermat's Little Theorem, Euler's totient function & Eueler's totient theorem. Highly unlikely to be tested but a very useful tool for remainder type questions.
And what if k is a negative integer, like -5?
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