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The sum of all 2-digited numbers which leave remainder 1 when divided by 3 is
- a)1605
- b)1616
- c)1604
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
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The sum of all 2-digited numbers which leave remainder 1 when divided ...
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The sum of all 2-digited numbers which leave remainder 1 when divided ...
Problem:
The sum of all 2-digit numbers which leave a remainder of 1 when divided by 3 is:
a) 1605
b) 1616
c) 1604
d) none of these
Solution:
To find the sum of all 2-digit numbers that leave a remainder of 1 when divided by 3, we can follow these steps:
Step 1: Determine the range of 2-digit numbers:
The range of 2-digit numbers is from 10 to 99.
Step 2: Identify the numbers that satisfy the given condition:
To find the numbers that leave a remainder of 1 when divided by 3, we can use the modulo operation. If a number leaves a remainder of 1 when divided by 3, then it can be written as 3k + 1, where k is an integer.
We need to find all values of k such that 3k + 1 is a 2-digit number. This condition is satisfied when k = 3, 4, ..., 33.
Step 3: Calculate the sum:
To find the sum of these numbers, we can use the formula for the sum of an arithmetic series:
Sum = (n/2)(first term + last term)
In this case, the first term is 3 * 3 + 1 = 10 and the last term is 3 * 33 + 1 = 100. The number of terms, n, is 33 - 3 + 1 = 31.
Using the formula, the sum of these numbers is:
Sum = (31/2)(10 + 100) = 31 * 55 = 1705
Conclusion:
Therefore, the sum of all 2-digit numbers that leave a remainder of 1 when divided by 3 is 1705. However, none of the given answer choices match this result, so the correct answer is none of these.
The sum of all 2-digit numbers which leave a remainder of 1 when divided by 3 is:
a) 1605
b) 1616
c) 1604
d) none of these
Solution:
To find the sum of all 2-digit numbers that leave a remainder of 1 when divided by 3, we can follow these steps:
Step 1: Determine the range of 2-digit numbers:
The range of 2-digit numbers is from 10 to 99.
Step 2: Identify the numbers that satisfy the given condition:
To find the numbers that leave a remainder of 1 when divided by 3, we can use the modulo operation. If a number leaves a remainder of 1 when divided by 3, then it can be written as 3k + 1, where k is an integer.
We need to find all values of k such that 3k + 1 is a 2-digit number. This condition is satisfied when k = 3, 4, ..., 33.
Step 3: Calculate the sum:
To find the sum of these numbers, we can use the formula for the sum of an arithmetic series:
Sum = (n/2)(first term + last term)
In this case, the first term is 3 * 3 + 1 = 10 and the last term is 3 * 33 + 1 = 100. The number of terms, n, is 33 - 3 + 1 = 31.
Using the formula, the sum of these numbers is:
Sum = (31/2)(10 + 100) = 31 * 55 = 1705
Conclusion:
Therefore, the sum of all 2-digit numbers that leave a remainder of 1 when divided by 3 is 1705. However, none of the given answer choices match this result, so the correct answer is none of these.
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Community Answer
The sum of all 2-digited numbers which leave remainder 1 when divided ...
Series is :- { 10 , 13 , 16 , ............. , 97}last term (L)= 97first term (A)=10common difference (D)= 3no. of terms(N) = ?L = A+(N-1)Don putting values of A , L and D in above equation we get N = 30.
sum of N terms = (N/2)(A+L) = 15 × 107 = 1605 ans.
sum of N terms = (N/2)(A+L) = 15 × 107 = 1605 ans.
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