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15 Aug 2022, 01:30
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E
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Difficulty:
85%
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Question Stats:
54% (03:03) correct
46%
(02:46)
wrong
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A number m is randomly selected from the set {11,13,15,17,19}, and a number n is randomly selected from {1999,2000,2001,...,2018}. What is the probability that \(m^n\) has a units digit of 1?
(A) \(\frac{1}{5}\)
(B) \(\frac{1}{4}\)
(C) \(\frac{3}{10}\)
(D) \(\frac{7}{20}\)
(E) \(\frac{2}{5}\)
Are You Up For the Challenge: 700 Level Questions
(A) \(\frac{1}{5}\)
(B) \(\frac{1}{4}\)
(C) \(\frac{3}{10}\)
(D) \(\frac{7}{20}\)
(E) \(\frac{2}{5}\)
Are You Up For the Challenge: 700 Level Questions
15 Aug 2022, 02:37
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A number m is randomly selected from the set {11,13,15,17,19}, and a number n is randomly selected from {1999,2000,2001,...,2018}. What is the probability that \(m^n\) has a units digit of 1?
Units digit will be 1 of only odd numbers, so m is odd.
The powers given are 20 in consecutive numbers, so 5 each of the form 4k, 4K+1, 4k+2, and 4K+3.
Now, digit 1 will always give units digit 1….n can be any of h the 20 numbers and m can be only one, that is 11 => 20*1 or 20 ways
Digit 3 has the cyclicity 3,9,7,1, so n has to be of the form 4k. Similarly for 7 too, 7,9,3,1….Thus m can take two values, 13 & 17, while n can take any of 5 values of form 4k => 2*5 or 10 ways.
Digit 9 has the cyclicity 9,1,9,1, so n has to be of the form 4k or 4k+2. Thus, m can take one value, 19, while n can take any of 5 values each of form 4k and 4k+2 => 1*10 or 10 ways.
Total possibilities are 20+10+10=40
Total ways = m*n = 20*5 = 100
P = \(\frac{40}{100}=\frac{2}{5}\)
Units digit will be 1 of only odd numbers, so m is odd.
The powers given are 20 in consecutive numbers, so 5 each of the form 4k, 4K+1, 4k+2, and 4K+3.
Now, digit 1 will always give units digit 1….n can be any of h the 20 numbers and m can be only one, that is 11 => 20*1 or 20 ways
Digit 3 has the cyclicity 3,9,7,1, so n has to be of the form 4k. Similarly for 7 too, 7,9,3,1….Thus m can take two values, 13 & 17, while n can take any of 5 values of form 4k => 2*5 or 10 ways.
Digit 9 has the cyclicity 9,1,9,1, so n has to be of the form 4k or 4k+2. Thus, m can take one value, 19, while n can take any of 5 values each of form 4k and 4k+2 => 1*10 or 10 ways.
Total possibilities are 20+10+10=40
Total ways = m*n = 20*5 = 100
P = \(\frac{40}{100}=\frac{2}{5}\)
12 Oct 2023, 13:04
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Bunuel
Total possibilities \(= 5*20 =100\)
For 11, raised to any number will give unit digit 1 = Thus 20
For 13, raised to any multiple of 4 from the second set will always give unit digit 1 = Thus 5
For 17, raised to any multiple of 4 from the second set will always give unit digit 1 = Thus 5
For 15, raised to any among the second set will NEVER give unit of 1 = Thus 0
For 19, raised to any even number from second set will always give unit of 1 = Thus 10
Therefore total favorable \(= 20+5+5+10 = 40\)
Overall total cases \(= 5*20 =100\)
Prob: \(\frac{40}{100} =\frac{2}{5} \)
Ans E
Hope it's clear.
