[GMAT math practice question]
f(n) and g(n) are defined as follows:
f(n) = {0 if n is even 1 if n is odd and g(n) = {0 if n is a multiple of 5 1 if n is not a multiple of 5
If h(n) is defined as (1 + f(n))(1 - g(n)), what is h(1) + h(2) + …. + h(1900)?
A. 550
B. 560
C. 570
D. 580
E. 590
f(n) and g(n) are defined as follows:
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
=>
If n is not a multiple of 5, then h(n) = (1 + f(n))(1 - g(n)) = (1 + f(n))(1 - 1) = 0.
If n is a multiple of 5, then h(n) = (1 + f(n))(1 - g(n)) = (1 + f(n))(1 - 0) = (1 + f(n)).
Thus, if n is a multiple of 10, we have h(n) = (1 + f(n)) = (1 + 0) = 1 since n is an even number and f(n) = 0.
If n has a remainder 5 when n is divided by 10, we have h(n) = (1 + f(n)) = 1 + 1 = 2 since n is an odd number.
Then we have h(10) = h(20) = … = h(1900) = 1 and h(5) = h(15) = … = h(1895) = 2.
The number of multiples of 10 is (1900 – 10)/10 + 1 = 1890/10 + 1 = 189 + 1 = 190.
The number of multiples that have a remainder 5 when they are divided by 10 is (1895 – 5)/10 + 1 = 1890/10 + 1 = 189 + 1 = 190.
Thus, we have
h(5) + h(10) + h(15) + h(20) + … + h(1895) + h(1990)
= h(5) + h(15) + … + h(1895) + h(10) + h(20) + … + h(1990)
= 2·190 + 190 = 570.
Therefore, C is the correct answer.
Answer: C
If n is not a multiple of 5, then h(n) = (1 + f(n))(1 - g(n)) = (1 + f(n))(1 - 1) = 0.
If n is a multiple of 5, then h(n) = (1 + f(n))(1 - g(n)) = (1 + f(n))(1 - 0) = (1 + f(n)).
Thus, if n is a multiple of 10, we have h(n) = (1 + f(n)) = (1 + 0) = 1 since n is an even number and f(n) = 0.
If n has a remainder 5 when n is divided by 10, we have h(n) = (1 + f(n)) = 1 + 1 = 2 since n is an odd number.
Then we have h(10) = h(20) = … = h(1900) = 1 and h(5) = h(15) = … = h(1895) = 2.
The number of multiples of 10 is (1900 – 10)/10 + 1 = 1890/10 + 1 = 189 + 1 = 190.
The number of multiples that have a remainder 5 when they are divided by 10 is (1895 – 5)/10 + 1 = 1890/10 + 1 = 189 + 1 = 190.
Thus, we have
h(5) + h(10) + h(15) + h(20) + … + h(1895) + h(1990)
= h(5) + h(15) + … + h(1895) + h(10) + h(20) + … + h(1990)
= 2·190 + 190 = 570.
Therefore, C is the correct answer.
Answer: C
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]