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Let N=233^{68}
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30 Aug 2026, 09:48
30 Aug 2026, 09:48
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Let \( N=233^{68}\) and \(P=149^{37} \)
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
Digit at unit place of (N+P) |
Digit at unit place of \(N^P\) |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
- Part of the project: GRE Daily Quant Challenge ALL 4 CORE QUANT AREAS
- Also replying to the unanswered questions
Re: Let N=233^{68}
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31 Aug 2026, 14:20
31 Aug 2026, 14:20
1
3^1= 3
3^2 = 9
3^3 = 7
3^4 = 1
Unit digit for 3 raised to different powers gives a repeating 4 cycle
So unit digit of 233^68 = 1
9^1 = 9
9^2 = 1
P = 149^37
Since 9 is raised to an odd power, unit digit of P = 9
N + P unit digit is 0
Now N ^P will give a minimum unit digit of 1, we dont need to find the value.
Answer is B
3^2 = 9
3^3 = 7
3^4 = 1
Unit digit for 3 raised to different powers gives a repeating 4 cycle
So unit digit of 233^68 = 1
9^1 = 9
9^2 = 1
P = 149^37
Since 9 is raised to an odd power, unit digit of P = 9
N + P unit digit is 0
Now N ^P will give a minimum unit digit of 1, we dont need to find the value.
Answer is B
gmatclubot
Re: Let N=233^{68} [#permalink]
31 Aug 2026, 14:20
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