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Set A consists of three consecutive positive multiples of 3, and set B consists of five consecutive positive multiples of 5. If the sum of the integers in set A is equal to the sum of the integers in set B, what is the least number that could be a member of set A?
- a)69
- b)72
- c)75
- d)78
- e)81
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Set A consists of three consecutive positive multiples of 3, and set B...
A = {3k, 3k+3, 3k+6}; k>0; k is a positive integer
B = {5m, 5m+5, 5m + 10, 5m+15, 5m+20}; m>0; m is a positive integer
B = {5m, 5m+5, 5m + 10, 5m+15, 5m+20}; m>0; m is a positive integer
3k + 3k + 3 + 3k + 6 = 5m + 5m+5 + 5m + 10 + 5m+15 + 5m+20
9k + 9 = 25m + 50
9k = 25m + 41
kmin = 24
3kmin = 72
9k + 9 = 25m + 50
9k = 25m + 41
kmin = 24
3kmin = 72
Most Upvoted Answer
Set A consists of three consecutive positive multiples of 3, and set B...
A = {3k, 3k+3, 3k+6}; k>0; k is a positive integer
B = {5m, 5m+5, 5m + 10, 5m+15, 5m+20}; m>0; m is a positive integer
B = {5m, 5m+5, 5m + 10, 5m+15, 5m+20}; m>0; m is a positive integer
3k + 3k + 3 + 3k + 6 = 5m + 5m+5 + 5m + 10 + 5m+15 + 5m+20
9k + 9 = 25m + 50
9k = 25m + 41
kmin = 24
3kmin = 72
9k + 9 = 25m + 50
9k = 25m + 41
kmin = 24
3kmin = 72
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Community Answer
Set A consists of three consecutive positive multiples of 3, and set B...
Understanding Sets A and B
Set A consists of three consecutive positive multiples of 3. Let the first multiple in set A be represented as \(3n\). Therefore, the integers in set A can be expressed as:
- \(3n\)
- \(3(n+1)\)
- \(3(n+2)\)
The sum of the integers in set A is:
- \(S_A = 3n + 3(n+1) + 3(n+2) = 3n + 3n + 3 + 3n + 6 = 9n + 9 = 9(n + 1)\)
Set B Analysis
Set B consists of five consecutive positive multiples of 5. Let the first multiple in set B be \(5m\). Therefore, the integers in set B can be expressed as:
- \(5m\)
- \(5(m+1)\)
- \(5(m+2)\)
- \(5(m+3)\)
- \(5(m+4)\)
The sum of the integers in set B is:
- \(S_B = 5m + 5(m+1) + 5(m+2) + 5(m+3) + 5(m+4) = 5m + 5m + 5 + 5m + 10 + 5m + 15 + 5m + 20 = 25m + 50\)
Setting the Sums Equal
Given that \(S_A = S_B\), we equate the two sums:
- \(9(n + 1) = 25m + 50\)
Rearranging gives:
- \(9n + 9 = 25m + 50\)
- \(9n = 25m + 41\)
Finding the Values of n and m
To find integer solutions, we need \(25m + 41\) to be divisible by 9. Checking for values of \(m\):
- \(m = 0\): \(41 \mod 9 = 5\) (not divisible)
- \(m = 1\): \(66 \mod 9 = 3\) (not divisible)
- \(m = 2\): \(91 \mod 9 = 1\) (not divisible)
- \(m = 3\): \(116 \mod 9 = 0\) (divisible)
For \(m = 3\):
- \(9n = 25(3) + 41 = 116\)
- \(n = \frac{116}{9} = 12.888\) (not an integer)
Continuing this process, we find:
- \(m = 4\) yields \(141\), \(n = 15\).
- Thus, the first integer in set A: \(3n = 3(15) = 45\).
Finding valid values continues until:
- For \(m = 5\): \(166\) gives \(n = 18.44\) (not an integer).
Eventually, checking the provided options:
- The lowest \(n\) yielding \(3n\) as a valid answer is \(72\) (as \(n + 1 = 25\)).
Thus, the least
Set A consists of three consecutive positive multiples of 3. Let the first multiple in set A be represented as \(3n\). Therefore, the integers in set A can be expressed as:
- \(3n\)
- \(3(n+1)\)
- \(3(n+2)\)
The sum of the integers in set A is:
- \(S_A = 3n + 3(n+1) + 3(n+2) = 3n + 3n + 3 + 3n + 6 = 9n + 9 = 9(n + 1)\)
Set B Analysis
Set B consists of five consecutive positive multiples of 5. Let the first multiple in set B be \(5m\). Therefore, the integers in set B can be expressed as:
- \(5m\)
- \(5(m+1)\)
- \(5(m+2)\)
- \(5(m+3)\)
- \(5(m+4)\)
The sum of the integers in set B is:
- \(S_B = 5m + 5(m+1) + 5(m+2) + 5(m+3) + 5(m+4) = 5m + 5m + 5 + 5m + 10 + 5m + 15 + 5m + 20 = 25m + 50\)
Setting the Sums Equal
Given that \(S_A = S_B\), we equate the two sums:
- \(9(n + 1) = 25m + 50\)
Rearranging gives:
- \(9n + 9 = 25m + 50\)
- \(9n = 25m + 41\)
Finding the Values of n and m
To find integer solutions, we need \(25m + 41\) to be divisible by 9. Checking for values of \(m\):
- \(m = 0\): \(41 \mod 9 = 5\) (not divisible)
- \(m = 1\): \(66 \mod 9 = 3\) (not divisible)
- \(m = 2\): \(91 \mod 9 = 1\) (not divisible)
- \(m = 3\): \(116 \mod 9 = 0\) (divisible)
For \(m = 3\):
- \(9n = 25(3) + 41 = 116\)
- \(n = \frac{116}{9} = 12.888\) (not an integer)
Continuing this process, we find:
- \(m = 4\) yields \(141\), \(n = 15\).
- Thus, the first integer in set A: \(3n = 3(15) = 45\).
Finding valid values continues until:
- For \(m = 5\): \(166\) gives \(n = 18.44\) (not an integer).
Eventually, checking the provided options:
- The lowest \(n\) yielding \(3n\) as a valid answer is \(72\) (as \(n + 1 = 25\)).
Thus, the least
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