find the sum of all two digit natural numbers which when divided by 7 yield 1 as remainder..
741
15, 22, 29, 36
Take it from there.
741
740
2,3,5,7....
To find the sum of all two-digit natural numbers that, when divided by 7, yield a remainder of 1, we need to follow these steps:
Step 1: Determine the range of numbers to consider. In this case, we are looking for two-digit numbers, which means numbers from 10 to 99.
Step 2: Find the first number within the range that satisfies the given condition. In this case, we need to find the smallest two-digit number that, when divided by 7, yields a remainder of 1.
The smallest two-digit number that satisfies this condition is 15. We calculate this by adding the remainder (1) to the divisor (7).
Step 3: Find the last number within the range that satisfies the given condition. In this case, we need to find the largest two-digit number that, when divided by 7, yields a remainder of 1.
The largest two-digit number that satisfies this condition is 92. We calculate this by subtracting the remainder (1) from the largest possible number within the range (99), then subtracting the next divisor (7).
Step 4: Calculate the sum of the numbers within the given range:
To find the sum of an arithmetic series (numbers in a sequence with a common difference), we can use the formula:
Sum = (n/2) * (first term + last term)
In this case, the first term is 15, the last term is 92, and the common difference is 7.
Sum = (n/2) * (first term + last term)
= (78/2) * (15 + 92)
= 39 * 107
= 4167
Therefore, the sum of all two-digit natural numbers that, when divided by 7, yield a remainder of 1 is 4167.