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13 Jan 2022, 06:30
Kudos
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
76% (02:14) correct
24%
(02:33)
wrong
based on 38
sessions
History
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Time
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Not Attempted Yet
If m is a natural number less than or equal to 100, what is the probability that \((7^m + 11^m)\) is divisible by 5?
A. \(\frac{1}{4}\)
B. \(\frac{3}{7}\)
C. \(\frac{8}{25}\)
D. \(\frac{13}{50}\)
E. \(\frac{3}{8}\)
A. \(\frac{1}{4}\)
B. \(\frac{3}{7}\)
C. \(\frac{8}{25}\)
D. \(\frac{13}{50}\)
E. \(\frac{3}{8}\)
18 Jan 2022, 23:17
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Bunuel
11^m will always give us the unit digit as 1.
we need to make sure the sum is divided by 5 which is possible only when the last digit is either 0 or 5.
7^m will give us unit digits as 7, 9, 3 and 1.
Only 9 will give us the unit digit as 0.
Which we will get when m is = 2, 6, 10, 14 and so on.
So the first term is 2 and the last term will be 98. |How.? 100 is completely divisible by 4 we are looking for the power of 2 so the last digit can be either 102 or 98. 102 is more than 100. so 98 it is.
Number of terms between 2 and 98 = ?
Using AP formula for nth term = a+(n-1)d =98
a =2 ; d =4
n will be = 25.
So there will be 25 terms which will be divisible by 5.
Total number of terms = 100.
Probablity = 25/100 = 1/4
A is in Alpha is the answer.
18 Jan 2022, 23:25
Kudos
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Bunuel
We know m≤100 and since m is a natural number m can take value from 1-100 only. For \(7^m+11^m\) to be divisible by 5, the unit digit of \(7^m+11^m\) has to be either 0 or 5.
Now, we know the cyclicity of 7 is 4 and that unit digits of 7 repeat in the pattern of 7,9,3,1, and the unit digit of 11 raised to any power remains 1.
This implies unit digit of \(7^m+11^m\) can be 0 only when the unit digit of \(7^m\) is 9. The units digit can never be 5.
Which happens when m = 2, 6, 10 14, 18 ....98. This makes 25 such cases.
Hence the probability that the given equation is divisible by 5 is \(\frac{25}{100}\) = \(\frac{1}{4}\)
IMO, the answer is A
