Units Digit Pattern Analysis for \(77^n\)
The question asks for the probability that the units digit of the number \(m = 77^n\) is 1, where \(n\) is a positive integer chosen at random.
To find the units digit of \(77^n\), we only need to consider the units digit of the base, which is 7. Let's examine the pattern of the units digits of powers of 7:
- \(7^1\) has a units digit of 7.
- \(7^2 = 49\) has a units digit of 9.
- \(7^3 = 343\) has a units digit of 3.
- \(7^4 = 2401\) has a units digit of 1.
- \(7^5 = 16807\) has a units digit of 7.
We can see a repeating pattern in the units digits: (7, 9, 3, 1). This cycle has a length of 4.
Relating Exponent \(n\) to Units Digit
The pattern repeats every 4 powers. We can determine the units digit based on the remainder of the exponent \(n\) when divided by 4:
- If \(n \pmod 4 = 1\), the units digit is 7.
- If \(n \pmod 4 = 2\), the units digit is 9.
- If \(n \pmod 4 = 3\), the units digit is 3.
- If \(n \pmod 4 = 0\) (which means \(n\) is a multiple of 4), the units digit is 1.
Probability Calculation
Since \(n\) is a positive integer chosen at random, we assume that each possible remainder when \(n\) is divided by 4 (1, 2, 3, or 0) is equally likely. There are 4 possible outcomes for the remainder \(n \pmod 4\).
We are interested in the case where the units digit of \(m = 77^n\) is 1. This occurs only when \(n\) is a multiple of 4, meaning \(n \pmod 4 = 0\).
There is only 1 favorable outcome (when \(n \pmod 4 = 0\)) out of the 4 possible equally likely outcomes for the remainder.
Therefore, the probability is calculated as:
\(P(\text{units digit is 1}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{1}{4}\)
Thus, the probability that the value of \(m\) will have 1 in the units place is \(\frac{1}{4}\).
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Important Questions from Probability
Two dice are thrown simultaneously. What is the probability of getting the same number on both the dice?
The probability of being 53 Sundays in year 2020 is-
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