What are two digit natural numbers which when divided by 7 yield 1as remainder ?
Well the smallest one is 15, and the largest is 99
They are 15, 22, .... 92, 99
They form an arithmetic sequence.
where a = 15, d = 7, term(n) = 99
Find how many terms there are, then find Sum(n) using your sum formula
Let me know what answer you get.
Find the sum of all two digit natural numbers which when divided by 7 yield 1as remainder .
2 answers
we can simply show that 8 / 7 = 1 and the remainder is 1
and to keep the remainder equals to 1 we will add multiple of 7 to the number 8 which will always gives a remainder equals to 1
we need only the 2 digit numbers but 8 is one digit number so we won't take it into account
hence we will start from 8+7 =15 and 15 is two digit number
so sum = 15+22 +29+36+43+50+57+64741+71+78+85+92+99=741
and to keep the remainder equals to 1 we will add multiple of 7 to the number 8 which will always gives a remainder equals to 1
we need only the 2 digit numbers but 8 is one digit number so we won't take it into account
hence we will start from 8+7 =15 and 15 is two digit number
so sum = 15+22 +29+36+43+50+57+64741+71+78+85+92+99=741
5 answers