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If two positive integers a and b are chosen at random between 1 and 50 inclusive, what is the approximate probability that a number of the form 7α+7b is divisible by 5?
- a)1/5
- b)1/4
- c)1/2
- d)2/3
- e)3/4
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
If two positive integers a and b are chosen at random between 1 and 50...
There are four possible combinations (7 & 3, 9 & 1, 3 & 7, and 1 & 9) where the sum (7a + 7b) will be divisible by 5.
The periodicity of the repetition of the power of 7 is 4. This means that every 1st, 5th, 9th, and so on time, the unit digit will be 7. The 2nd, 6th, and subsequent times will have a unit digit of 9, while the 3rd, 7th, and subsequent times will have a unit digit of 3.
The probability of obtaining each of these unit digits is 12 (approximated as 50/4) out of 50.
Therefore, the probability for 7a is 12/50, and the probability for 7b is also 12/50.
Since there are a total of 4 combinations mentioned, the combined probability is calculated as (12/50 * 12/50) * 4 (approximated).
Simplifying this expression, we get (1/4) * (1/4) * 4 = 1/4.
Hence, the approximate probability of the sum (7a + 7b) being divisible by 5 is 1/4.
Most Upvoted Answer
If two positive integers a and b are chosen at random between 1 and 50...
Understanding the Problem
We want to find the probability that a number of the form \(7a + 7b\) is divisible by 5, where \(a\) and \(b\) are chosen randomly from the integers between 1 and 50.
Factor Out the Common Term
The expression can be simplified:
- \(7a + 7b = 7(a + b)\)
Now, we need to determine when \(7(a + b)\) is divisible by 5.
Divisibility by 5
For \(7(a + b)\) to be divisible by 5, \(a + b\) must be congruent to 0 modulo 5 because 7 is not divisible by 5. Thus, we require:
- \(a + b \equiv 0 \mod 5\)
Possible Values of \(a + b\)
The possible sums \(a + b\) can range from:
- Minimum: \(1 + 1 = 2\)
- Maximum: \(50 + 50 = 100\)
The values of \(a + b\) modulo 5 can be 0, 1, 2, 3, or 4.
Counting Valid Outcomes
Since \(a\) and \(b\) can take any integer value from 1 to 50:
- There are \(50 \times 50 = 2500\) total combinations of \(a\) and \(b\).
Now, we can find how many of these combinations result in \(a + b \equiv 0 \mod 5\).
Distribution of Remainders
The sums \(a + b\) can yield remainders that are uniformly distributed. Therefore, each of the five possible remainders (0 through 4) will occur with equal frequency.
- Thus, the probability that \(a + b \equiv 0 \mod 5\) is:
\[
\frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{1}{5} \text{ (for 0 mod 5)}
\]
However, we need to account for all combinations yielding the same modulo result.
Final Probability Calculation
Since we only need \(a + b\) divisible by 5, we find:
- Combinations yielding \(0 \mod 5\) are 1 out of 5.
Hence, the approximate probability that \(7a + 7b\) is divisible by 5 is \( \frac{1}{4} \).
Thus, the correct answer is option B.
We want to find the probability that a number of the form \(7a + 7b\) is divisible by 5, where \(a\) and \(b\) are chosen randomly from the integers between 1 and 50.
Factor Out the Common Term
The expression can be simplified:
- \(7a + 7b = 7(a + b)\)
Now, we need to determine when \(7(a + b)\) is divisible by 5.
Divisibility by 5
For \(7(a + b)\) to be divisible by 5, \(a + b\) must be congruent to 0 modulo 5 because 7 is not divisible by 5. Thus, we require:
- \(a + b \equiv 0 \mod 5\)
Possible Values of \(a + b\)
The possible sums \(a + b\) can range from:
- Minimum: \(1 + 1 = 2\)
- Maximum: \(50 + 50 = 100\)
The values of \(a + b\) modulo 5 can be 0, 1, 2, 3, or 4.
Counting Valid Outcomes
Since \(a\) and \(b\) can take any integer value from 1 to 50:
- There are \(50 \times 50 = 2500\) total combinations of \(a\) and \(b\).
Now, we can find how many of these combinations result in \(a + b \equiv 0 \mod 5\).
Distribution of Remainders
The sums \(a + b\) can yield remainders that are uniformly distributed. Therefore, each of the five possible remainders (0 through 4) will occur with equal frequency.
- Thus, the probability that \(a + b \equiv 0 \mod 5\) is:
\[
\frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{1}{5} \text{ (for 0 mod 5)}
\]
However, we need to account for all combinations yielding the same modulo result.
Final Probability Calculation
Since we only need \(a + b\) divisible by 5, we find:
- Combinations yielding \(0 \mod 5\) are 1 out of 5.
Hence, the approximate probability that \(7a + 7b\) is divisible by 5 is \( \frac{1}{4} \).
Thus, the correct answer is option B.
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Community Answer
If two positive integers a and b are chosen at random between 1 and 50...
There are four possible combinations (7 & 3, 9 & 1, 3 & 7, and 1 & 9) where the sum (7a + 7b) will be divisible by 5.
The periodicity of the repetition of the power of 7 is 4. This means that every 1st, 5th, 9th, and so on time, the unit digit will be 7. The 2nd, 6th, and subsequent times will have a unit digit of 9, while the 3rd, 7th, and subsequent times will have a unit digit of 3.
The probability of obtaining each of these unit digits is 12 (approximated as 50/4) out of 50.
Therefore, the probability for 7a is 12/50, and the probability for 7b is also 12/50.
Since there are a total of 4 combinations mentioned, the combined probability is calculated as (12/50 * 12/50) * 4 (approximated).
Simplifying this expression, we get (1/4) * (1/4) * 4 = 1/4.
Hence, the approximate probability of the sum (7a + 7b) being divisible by 5 is 1/4.
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