[GMAT math practice question]
What is the remainder when 9^n -1 is divided by 10?
1) n is a multiple of 2
2) n is a multiple of 3
5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
What is the remainder when 9^n -1 is divided by 10?
This topic has expert replies
-
Max@Math Revolution
- Elite Legendary Member
- Posts: 3991
- Joined: Fri Jul 24, 2015 2:28 am
- Location: Las Vegas, USA
- Thanked: 19 times
- Followed by:37 members
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
Math Revolution
The World's Most "Complete" GMAT Math Course!
Score an excellent Q49-51 just like 70% of our students.
[Free] Full on-demand course (7 days) - 100 hours of video lessons, 490 lesson topics, and 2,000 questions.
[Course] Starting $79 for on-demand and $60 for tutoring per hour and $390 only for Live Online.
Email to : [email protected]
-
Max@Math Revolution
- Elite Legendary Member
- Posts: 3991
- Joined: Fri Jul 24, 2015 2:28 am
- Location: Las Vegas, USA
- Thanked: 19 times
- Followed by:37 members
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.
The remainder when 9^n -1 is divided by 10 is the same as the units digit of 9^n -1. This is easily determined from the units digit of 9^n.
9^1 = 9, 9^2 = 81 ~ 1, 9^3 ~ 9, 9^4 ~ 81 ~ 1, ...
So, the units digits of 9n have period 2:
They form the cycle 9 -> 1.
Thus, 9^n has a units digit of 9, if n is an odd number and a units digit if 1, if n is an even number.
Thus, condition 1) is sufficient.
Condition 2) is not sufficient since a multiple of 3 can be either an even or an odd number.
Therefore, A is the answer.
Answer: A
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.
The remainder when 9^n -1 is divided by 10 is the same as the units digit of 9^n -1. This is easily determined from the units digit of 9^n.
9^1 = 9, 9^2 = 81 ~ 1, 9^3 ~ 9, 9^4 ~ 81 ~ 1, ...
So, the units digits of 9n have period 2:
They form the cycle 9 -> 1.
Thus, 9^n has a units digit of 9, if n is an odd number and a units digit if 1, if n is an even number.
Thus, condition 1) is sufficient.
Condition 2) is not sufficient since a multiple of 3 can be either an even or an odd number.
Therefore, A is the answer.
Answer: A
Math Revolution
The World's Most "Complete" GMAT Math Course!
Score an excellent Q49-51 just like 70% of our students.
[Free] Full on-demand course (7 days) - 100 hours of video lessons, 490 lesson topics, and 2,000 questions.
[Course] Starting $79 for on-demand and $60 for tutoring per hour and $390 only for Live Online.
Email to : [email protected]
-
deloitte247
- Legendary Member
- Posts: 2214
- Joined: Fri Mar 02, 2018 2:22 pm
- Followed by:5 members
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
Statement 1=> n is a multiple of 2.
This means that 'n' is even when '9' is raised to an even number.
Also, 10 is also an even number and when an even number is divided by even number, we have remainder = 0.
$$e.g\ if\ n=2;\ 9^2-1=80;\ \ \ \ \ \ So,\ \frac{80}{10}=8\ remainder\ 0$$
$$e.g\ if\ n=4;\ 9^4-1=6560;\ \ \ \ \ \ So,\ \frac{6560}{10}=656\ \ remainder\ 0$$
Hence, statement 1 is SUFFICIENT.
Statement 2=> n is a multiple of 3.
This means that 'n' can be either an even or odd number since multiples of 3 can either be even or odd (e.g 3, 6, 9, 12).
So, when 9 is raised to any number (even or odd) in the original expression, we should have a remainder of zero (0) for statement 2 to be sufficient.
$$e.g\ if\ n=3;\ 9^3-1=728;\ \ \ \ \ \ So,\ \frac{728}{10}=72\ \ remainder\ 8$$
$$e.g\ if\ n=6;\ 9^6-1=531441;\ \ \ \ \ \ So,\ \frac{531440}{10}=53144\ \ remainder\ 0$$
From the expression above, with n=odd, the remainder is not zero but with n=even, we have the remainder to be zero. Since the multiple of 3 cannot totally give us a remainder zero (0), then statement 2 is NOT SUFFICIENT.
Therefore, only statement 1 alone is SUFFICIENT. Hence, option A is the current answer.
This means that 'n' is even when '9' is raised to an even number.
Also, 10 is also an even number and when an even number is divided by even number, we have remainder = 0.
$$e.g\ if\ n=2;\ 9^2-1=80;\ \ \ \ \ \ So,\ \frac{80}{10}=8\ remainder\ 0$$
$$e.g\ if\ n=4;\ 9^4-1=6560;\ \ \ \ \ \ So,\ \frac{6560}{10}=656\ \ remainder\ 0$$
Hence, statement 1 is SUFFICIENT.
Statement 2=> n is a multiple of 3.
This means that 'n' can be either an even or odd number since multiples of 3 can either be even or odd (e.g 3, 6, 9, 12).
So, when 9 is raised to any number (even or odd) in the original expression, we should have a remainder of zero (0) for statement 2 to be sufficient.
$$e.g\ if\ n=3;\ 9^3-1=728;\ \ \ \ \ \ So,\ \frac{728}{10}=72\ \ remainder\ 8$$
$$e.g\ if\ n=6;\ 9^6-1=531441;\ \ \ \ \ \ So,\ \frac{531440}{10}=53144\ \ remainder\ 0$$
From the expression above, with n=odd, the remainder is not zero but with n=even, we have the remainder to be zero. Since the multiple of 3 cannot totally give us a remainder zero (0), then statement 2 is NOT SUFFICIENT.
Therefore, only statement 1 alone is SUFFICIENT. Hence, option A is the current answer.
