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08 May 2020, 08:39
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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:
95%
(hard)
Question Stats:
46% (03:03) correct
54%
(02:47)
wrong
based on 74
sessions
History
Date
Time
Result
Not Attempted Yet
Alex has forgotten his six-digit ID number. He remembers the following: the first two digits are either 15 or 26, the number is even and 6 appears twice. If Alex uses a trial and error process to find his ID number at the most, how many trials does he need to succeed?
(A) 729
(B) 972
(C) 2051
(D) 2052
(E) 2072
Are You Up For the Challenge: 700 Level Questions
(A) 729
(B) 972
(C) 2051
(D) 2052
(E) 2072
Are You Up For the Challenge: 700 Level Questions
08 May 2020, 22:39
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Case 1- If the number starts with 15
i) last digit is 6
he needs to select 1 more 6; he can choose any of the 10 digits(except 6) for remaining 2 places
Total possible numbers \(= 9*9*\frac{3!}{2!} = 243\)
ii) last digit is not 6
Number of ways to choose last digit = 4 {0,2,4 or 8}
Total possible numbers \(= 9*4*\frac{3!}{2!} = 108\)
Case 2- If the number starts with 26
i) last digit is 6
he can choose any of the 10 digits(except 6) for remaining 3 places
Total possible numbers \(= 9*9*9 = 729\)
ii) last digit is not 6
Number of ways to choose last digit = 4 {0,2,4 or 8}
Total possible numbers \(= 9*9*4*\frac{3!}{2!} = 1200\)
In the worst scenario, he'll choose the correct one in the last.
Total number of trials = 243+108+729+972 = 2052
i) last digit is 6
he needs to select 1 more 6; he can choose any of the 10 digits(except 6) for remaining 2 places
Total possible numbers \(= 9*9*\frac{3!}{2!} = 243\)
ii) last digit is not 6
Number of ways to choose last digit = 4 {0,2,4 or 8}
Total possible numbers \(= 9*4*\frac{3!}{2!} = 108\)
Case 2- If the number starts with 26
i) last digit is 6
he can choose any of the 10 digits(except 6) for remaining 3 places
Total possible numbers \(= 9*9*9 = 729\)
ii) last digit is not 6
Number of ways to choose last digit = 4 {0,2,4 or 8}
Total possible numbers \(= 9*9*4*\frac{3!}{2!} = 1200\)
In the worst scenario, he'll choose the correct one in the last.
Total number of trials = 243+108+729+972 = 2052
Bunuel
08 May 2020, 23:13
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Hey nick
why did u multiplied by 3!/2!
Posted from my mobile device
why did u multiplied by 3!/2!
Posted from my mobile device
