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01 Sep 2022, 15:34
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D
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List A consist 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
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A. 296
B. 297
C. 298
D. 299
E. 300
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21 Mar 2024, 01:58
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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
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
General Discussion
02 Sep 2022, 01:31
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It is given List A contains 99 multiples of 9.
1st multiple of 9 = 9*1
2nd multiple of 9 = 9*2
99th multiple of 9 = 9*99 = 891
Therefore set consists of integers from 1 to at least 891.
This set can also contain Integers up to 899, so that the 100th multiple of 99 ('900') is not included in the set.
Because we are required to find maximum Number of multiples of 3 in this set, considering the List A to be
{1,2,3,4,........,899}
we find that maximum multiple of 3 --> 897 = 3*299
Hence, maximum number of multiples of 3 in List A containing exactly 99 multiples of 9 = 299
Answer: D
1st multiple of 9 = 9*1
2nd multiple of 9 = 9*2
99th multiple of 9 = 9*99 = 891
Therefore set consists of integers from 1 to at least 891.
This set can also contain Integers up to 899, so that the 100th multiple of 99 ('900') is not included in the set.
Because we are required to find maximum Number of multiples of 3 in this set, considering the List A to be
{1,2,3,4,........,899}
we find that maximum multiple of 3 --> 897 = 3*299
Hence, maximum number of multiples of 3 in List A containing exactly 99 multiples of 9 = 299
Answer: D
