M40-84

Question

List A consists of  n  consecutive positive integers:  \{1, 2, 3, ..., n\} . If the list contains exactly 99 multiples of 9, what is the maximum number of multiples of 3 it can contain?

A. 296
B. 297
C. 298
D. 299
E. 300

adityakaregamba · Accepted Answer

last multiple of 9 would be 9 * 99 = 891 
but the list can still have 8 more values 
892
893
894
895
896
897
899
it cant have 900 as a value in the list since then it violate the question because it says it can max 99 multiple of 9 if we include 900 then the list will have 100 multiple of 9.

from the above numbers the max number which is divisible byf 3 is 897.

897/3 = 299

therefor number of multiple of 3 would be 299

all the best

Bunuel · Answer

Official Solution:

List A consists of  n  consecutive positive integers:  \{1, 2, 3, ..., n\} . If the list contains exactly 99 multiples of 9, what is the maximum number of multiples of 3 it can contain?

A. 296
B. 297
C. 298
D. 299
E. 300

We aim to maximize the number of multiples of 3 in the list while ensuring there are exactly 99 multiples of 9. To achieve this, we can design a list where  n  is the closest multiple of 3 that is less than the 100th multiple of 9. The 100th multiple of 9 is 900, so  n= 897 . In this scenario, the list will contain exactly 99 multiples of 9 and the maximum number of multiples of 3 possible.
 
In this setup, the number of multiples of 3 in the list would be:
 
 \frac{last \ multiple \ of \ 3 \ in \ the \ list - first \ multiple \ of \ 3 \ in \ the \ list}{3} +1 = \frac{897 - 3}{3} +1 = 299 .

Answer: D

Dream009 · Answer

How I solved it was like 99*9 would be the last term. So terms divisible by 3 is 99*9/3 = 297
What is wrong with this logic @Bunel

Bunuel · Answer

You should have read the solution more carefully. Your logic assumes the largest multiple of 9 (99*9 = 891)  is  the last term in the list, but the question only says there are exactly 99 multiples of 9. You can extend the list beyond 891 without adding a 100th multiple of 9 by including numbers that are multiples of 3 but not of 9 (such as 894, 897). This is why the correct count is 299, not 297.

Bunuel · Answer

Also, please review a similar problem to ensure you fully grasp the concept being tested: https://gmatclub.com/forum/list-k-consi ... 98239.html

adarsh2209 · Answer

I like the solution - it’s helpful.

Goldenfuture · Answer

idea is 99*9/3 + (9-1)/3= answer
Why did we do 9-1/3 - because that will prevent it to be the 100th digit but still give me extra 3s

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