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What is the sum of all the 3-digit numbers that leave a remainder of 5 when divided by 7?
- a)71079
- b)64887
- c)67890
- d)45778
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
What is the sum of all the 3-digit numbers that leave a remainder of 5...
The 3-digit numbers that leave a remainder of 5 when divided by 7 are:
103, 110, 117, ………….., 999
The given terms are in A.P., with a common difference of 7.
103, 110, 117, ………….., 999
The given terms are in A.P., with a common difference of 7.
nth term, an = a + (n – 1) d
Here, a = 103, an = 999,d = 7
So, 999 = 103 + (n – 1)7
Or (n – 1)7 = 999 – 103 = 896
Or (n – 1) = 896/7 = 128
Or n = 128 + 1 = 129
So, the total number of such terms is 129.
Here, a = 103, an = 999,d = 7
So, 999 = 103 + (n – 1)7
Or (n – 1)7 = 999 – 103 = 896
Or (n – 1) = 896/7 = 128
Or n = 128 + 1 = 129
So, the total number of such terms is 129.
Sum of an AP = [(First term + Last term)/2] × number of terms So, Sum of all the 3-digit numbers that leave a remainder of 5 when divided by 7 = [(103 + 999)/2] × 129 = 71079. Hence, option (a) is correct.
Most Upvoted Answer
What is the sum of all the 3-digit numbers that leave a remainder of 5...
Understanding the Problem
To find the sum of all 3-digit numbers that leave a remainder of 5 when divided by 7, we first need to identify the range of these numbers.
Identifying the Range
- The smallest 3-digit number is 100.
- The largest 3-digit number is 999.
Finding the Smallest and Largest Valid Numbers
- Smallest 3-digit number leaving a remainder of 5 when divided by 7:
- We start from 100. The next number that meets our condition can be found using:
- 100 mod 7 = 2
- To adjust to a remainder of 5, we add 3, resulting in:
- 100 + 3 = 103.
- Largest 3-digit number leaving a remainder of 5 when divided by 7:
- We start from 999:
- 999 mod 7 = 4
- To adjust to a remainder of 5, we subtract 1, resulting in:
- 999 - 1 = 998.
Identifying the Sequence of Numbers
The valid numbers form an arithmetic series where:
- First term (a) = 103
- Last term (l) = 998
- Common difference (d) = 7
Finding the Number of Terms (n)
- The formula for the nth term of an arithmetic sequence is:
- l = a + (n-1)d
- Setting up the equation:
- 998 = 103 + (n-1) * 7
- This gives us n = 129.
Calculating the Sum
- The sum of an arithmetic series can be calculated using the formula:
- Sum = n/2 * (a + l)
- Plugging in the values:
- Sum = 129/2 * (103 + 998) = 129/2 * 1101 = 71079.
Conclusion
The sum of all 3-digit numbers that leave a remainder of 5 when divided by 7 is 71079, confirming that option 'A' is correct.
To find the sum of all 3-digit numbers that leave a remainder of 5 when divided by 7, we first need to identify the range of these numbers.
Identifying the Range
- The smallest 3-digit number is 100.
- The largest 3-digit number is 999.
Finding the Smallest and Largest Valid Numbers
- Smallest 3-digit number leaving a remainder of 5 when divided by 7:
- We start from 100. The next number that meets our condition can be found using:
- 100 mod 7 = 2
- To adjust to a remainder of 5, we add 3, resulting in:
- 100 + 3 = 103.
- Largest 3-digit number leaving a remainder of 5 when divided by 7:
- We start from 999:
- 999 mod 7 = 4
- To adjust to a remainder of 5, we subtract 1, resulting in:
- 999 - 1 = 998.
Identifying the Sequence of Numbers
The valid numbers form an arithmetic series where:
- First term (a) = 103
- Last term (l) = 998
- Common difference (d) = 7
Finding the Number of Terms (n)
- The formula for the nth term of an arithmetic sequence is:
- l = a + (n-1)d
- Setting up the equation:
- 998 = 103 + (n-1) * 7
- This gives us n = 129.
Calculating the Sum
- The sum of an arithmetic series can be calculated using the formula:
- Sum = n/2 * (a + l)
- Plugging in the values:
- Sum = 129/2 * (103 + 998) = 129/2 * 1101 = 71079.
Conclusion
The sum of all 3-digit numbers that leave a remainder of 5 when divided by 7 is 71079, confirming that option 'A' is correct.
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