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04 Sep 2022, 09:53
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D
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Difficulty:
75%
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Question Stats:
51% (01:45) correct
49%
(01:49)
wrong
based on 246
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List K consist of n consecutive positive integers: {1, 2, 3, ..., n}. If the list contains exactly 38 multiples of 13, what is the maximum number of multiples of 11 it can contain ?
A. 40
B. 44
C. 45
D. 46
E. 47
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A. 40
B. 44
C. 45
D. 46
E. 47
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04 Sep 2022, 13:10
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We know that the set consists of 38 multiples of 13. So the max number that can be there in the set is (13*39) - 1
Max number in the set = 507 - 1 = 506
So the set consists of consecutive integers from 1 to 506
The first multiple of 11 in the set = 11
The last multiple of 11 in the set = 506
Number of multiples = \(\frac{(506-11)}{11}+1\) = 45 + 1 = 46
IMO D
Max number in the set = 507 - 1 = 506
So the set consists of consecutive integers from 1 to 506
The first multiple of 11 in the set = 11
The last multiple of 11 in the set = 506
Number of multiples = \(\frac{(506-11)}{11}+1\) = 45 + 1 = 46
IMO D
21 Mar 2024, 02:21
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Official Solution:
List K consist of n consecutive positive integers: \(\{1, 2, 3, ..., n\}\). If the list contains exactly 38 multiples of 13, what is the maximum number of multiples of 11 it can contain ?
A. 40
B. 44
C. 45
D. 46
E. 47
We aim to maximize the number of multiples of 11 in the list while ensuring there are exactly 38 multiples of 13. To achieve this, we can design a list where \(n\) is the closest multiple of 11 that is less than the 39th multiple of 13. The 39th multiple of 13 is \(39*13 = 507\). The closest multiple of 11 less than 507 is 506 (to find this, observe that 440 and 66 are both multiples of 11, hence \(440 + 66 = 506\), is also a multiple of 11.). Therefore, \(n= 506\). In this scenario, the list will contain exactly 38 multiples of 13 and the maximum number of multiples of 11 possible.
In this setup, the number of multiples of 11 in the list would be:
\(\frac{last \ multiple \ of \ 11 \ in \ the \ list - first \ multiple \ of \ 11 \ in \ the \ list}{11} +1 = \frac{506 - 11}{11} +1 = 46\).
Answer: D
List K consist of n consecutive positive integers: \(\{1, 2, 3, ..., n\}\). If the list contains exactly 38 multiples of 13, what is the maximum number of multiples of 11 it can contain ?
A. 40
B. 44
C. 45
D. 46
E. 47
We aim to maximize the number of multiples of 11 in the list while ensuring there are exactly 38 multiples of 13. To achieve this, we can design a list where \(n\) is the closest multiple of 11 that is less than the 39th multiple of 13. The 39th multiple of 13 is \(39*13 = 507\). The closest multiple of 11 less than 507 is 506 (to find this, observe that 440 and 66 are both multiples of 11, hence \(440 + 66 = 506\), is also a multiple of 11.). Therefore, \(n= 506\). In this scenario, the list will contain exactly 38 multiples of 13 and the maximum number of multiples of 11 possible.
In this setup, the number of multiples of 11 in the list would be:
\(\frac{last \ multiple \ of \ 11 \ in \ the \ list - first \ multiple \ of \ 11 \ in \ the \ list}{11} +1 = \frac{506 - 11}{11} +1 = 46\).
Answer: D
