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Difficulty:
75%
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Question Stats:
48% (01:53) correct
52%
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wrong
based on 505
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What is the units digit of \(n^{(p+1)}\), where n is a positive integer and p is a prime number greater than 3?
(1) When p is divided by 3, the remainder is 2.
(2) When n is divided by 10, the tenths digit of the resulting number is 9.
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(1) When p is divided by 3, the remainder is 2.
(2) When n is divided by 10, the tenths digit of the resulting number is 9.
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Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
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Quote:
(1) tells us about the power. p can be 5 or 11 (or someother number, i didn't need to calculate). But without knowing anything about n, we can't say what n^(p+1) will be. So, it's not sufficient
(2) the tenths digit of n/10 is the unit digit of n. So, units digit of n is 9.
9^(anything odd) --> units digit is 9
9^(anything even) --> units digit is 1
Now, you may think this isn't sufficient, but let's look again.
We know p is a prime number > 3, so p has to be odd.
=> p+1 has to be even
so, we can determine exactly what the unit digit of 9^(even number) will be.
Hence statement 2 alone is sufficient. Answer B
Answer: Option B
Please check the video for the step-by-step solution.
GMATinsight's Solution
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Please check the video for the step-by-step solution.
GMATinsight's Solution
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Screenshot 2020-08-26 at 8.45.49 PM.png [ 1.02 MiB | Viewed 7270 times ]
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Shokhrukh22
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GMAT 1: 680 Q49 V33
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Statement 1: When p is divided by 3, the remainder is 2. - This statement is clearly insufficient, because it gives information only about p. We don't know anything about n from this statement.
Statement 2: When n is divided by 10, the tenths digit of the resulting number is 9. - Since the tenth digit of the resulting number is 9, we can be sure that the last digit of n is equal to 9. Now let's try some numbers. 9^1=9, 9^2=81, 9^3=9, 9^4=1. From this it is clear that the units digit of n will be 9 if p+1 is odd and 1 if p+1 is even. Since it is given that p is a prime number greater than 3, p+1 is even number. So the units digit of n^(p+1) is equal to 1. Sufficient. Answer is B.
Statement 2: When n is divided by 10, the tenths digit of the resulting number is 9. - Since the tenth digit of the resulting number is 9, we can be sure that the last digit of n is equal to 9. Now let's try some numbers. 9^1=9, 9^2=81, 9^3=9, 9^4=1. From this it is clear that the units digit of n will be 9 if p+1 is odd and 1 if p+1 is even. Since it is given that p is a prime number greater than 3, p+1 is even number. So the units digit of n^(p+1) is equal to 1. Sufficient. Answer is B.
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