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11 Feb 2021, 05:45
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
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Question Stats:
60% (02:30) correct
40%
(02:11)
wrong
based on 167
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You are given an unlimited supply of each of the digits 1, 2, 3 or 4. Using only these four digits, you construct n digit numbers. Such n digit numbers will be called LEGITIMATE if it contains the digit 1 either an even number times or not at all. Number of n digit legitimate numbers are
(A) \(2^n + 1\)
(B) \(2^{n+1} + 2\)
(C) \(2^{n+2} + 4\)
(D) \(2^{n–1}(2^n + 1)\)
(E) \(2^{n–1}(2^n + 2)\)
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
(A) \(2^n + 1\)
(B) \(2^{n+1} + 2\)
(C) \(2^{n+2} + 4\)
(D) \(2^{n–1}(2^n + 1)\)
(E) \(2^{n–1}(2^n + 2)\)
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
11 Feb 2021, 18:11
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If n=1, number of legitimate numbers = 3
Eliminate B, C and E
If n=2,
1. Numbers that don't contain 1 = 3*3=9 (we have 3 options for the both digits)
2. Number that contains 2 1s = 1
number of legitimate numbers = 10
Eliminate A
Eliminate B, C and E
If n=2,
1. Numbers that don't contain 1 = 3*3=9 (we have 3 options for the both digits)
2. Number that contains 2 1s = 1
number of legitimate numbers = 10
Eliminate A
Bunuel
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Joined: 06 Dec 2020
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Schools: Stanford '24 (S)
14 Aug 2021, 07:14
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Hello All,
We can use manual brute force.
Let M(n) be a set of numbers where n is the number of digits in each number in the set and each digit of each number is selected from {1, 2, 3, 4}. Each digit can be used an unlimited number of times.
Let Legit(M(n)) be a subset of M(n) where each element in Legit(M(n)) is a number that has digit 1 an even number of times or a number that has no digit 1
M(1) = {1,2,3,4}
Legit(M(1)) = {2,3,4}
Number of elements in Legit(M(1)) = 3
M(2) = {11, 12, 13, 14, 21, 22, 23, 24, 31, 32, 33, 34, 41, 42, 43, 44}
Legit(M(2)) = {11, 22, 23, 24, 32, 33, 34, 42, 43, 44}
Number of elements in Legit(M(2)) = 10
We can continue for n = 3 but not necessary plus will burn too much time. Stop at n=2 if you want to use less than 2 minutes on the question.
M(3) = {111, 112, 113,...,434, 444} and
Legit(M(3)) ={112, 112,...,434, 444}
M(3) has 64 elements and Legit(M(3)) has 36 elements.
Only option D satisfies
Number of elements in Legit(M(n))
Number of elements in Legit(M(1)) = 3 and
Number of elements in Legit(M(2)) = 10
Cheers!
S
We can use manual brute force.
Let M(n) be a set of numbers where n is the number of digits in each number in the set and each digit of each number is selected from {1, 2, 3, 4}. Each digit can be used an unlimited number of times.
Let Legit(M(n)) be a subset of M(n) where each element in Legit(M(n)) is a number that has digit 1 an even number of times or a number that has no digit 1
M(1) = {1,2,3,4}
Legit(M(1)) = {2,3,4}
Number of elements in Legit(M(1)) = 3
M(2) = {11, 12, 13, 14, 21, 22, 23, 24, 31, 32, 33, 34, 41, 42, 43, 44}
Legit(M(2)) = {11, 22, 23, 24, 32, 33, 34, 42, 43, 44}
Number of elements in Legit(M(2)) = 10
We can continue for n = 3 but not necessary plus will burn too much time. Stop at n=2 if you want to use less than 2 minutes on the question.
M(3) = {111, 112, 113,...,434, 444} and
Legit(M(3)) ={112, 112,...,434, 444}
M(3) has 64 elements and Legit(M(3)) has 36 elements.
Only option D satisfies
Number of elements in Legit(M(n))
Number of elements in Legit(M(1)) = 3 and
Number of elements in Legit(M(2)) = 10
Cheers!
S
