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19 Jan 2026, 22:00
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D
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Difficulty:
45%
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Question Stats:
66% (01:47) correct
34%
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wrong
based on 171
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A two digit number is randomly selected and multiplied by itself x times. What is the probability that the unit’s digit of the resulting number is 1?
(1) x = 12.
(2) x is a multiple of 4.
(1) x = 12.
(2) x is a multiple of 4.
20 Jan 2026, 04:10
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ExpertsGlobal5
Any number raised to the 4th power or a power that is a multiple of 4 has the same units digit as n
Keeping that in mind, both statements provide the same information that x is a multiple of 4 (note that x cannot be 0 in statement 2)
To get the units digit as 1 when any number is raised to the 4th power (or a power that is a multiple of 4), the units digit of the original number must be 1, 3, 7 or 9
Since we can definitely find how many 2-digit numbers are there that end in 1,3,7 or 9, we can determine the probability as well (there are 36 such numbers out of 90 such 2-digit numbers => prob = 36/90 = 0.4)
Each statement separately sufficient (D)
20 Jan 2026, 10:59
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ExpertsGlobal5
Let the two digit number be n.
When the number is multiplied x times = n*n^x = n ^(x+1)
The total number of two digit numbers = 99-10+1 = 90.
We need to find : What is the probability that the unit’s digit of the resulting number is 1?
Cyclicity of numbers:
Unit digit as 1: a1*a1 multiplied any times will yield unit digit as 1.
Unit digit as 3: cyclicity of 4 = 3*3*3*3= _ _ 1.
Unit digit as 7: Cyclicity of 7 = 7*7*7*7 = _ _ 1.
Unit digit as 9: Cyclicity of 2 = 9*9 = _ _ 1.
Statement 1:
x = 12.
Then, n^(x+1) = n^13
The only cyclicity that matches this is when unit digit is 1.
10 such numbers exists = (11,21,31,41,51,62,71,81,91)
Probabilty = 9/90 = 1/10.
Hence, Sufficient
Statement 2:
x is a multiple of 4.
Then, n ^(4k+1) .
The only cyclicity that matches this is when unit digit is 1.
10 such numbers exists = (11,21,31,41,51,62,71,81,91)
Probabilty = 9/90 = 1/10.
Hence, Sufficient
Option D
