Question

3. Find the sum of all three digits number divisible by 7 leaves a remainder 2​

Answer 1

Answer:

The lowest 3 digit number is 101 and the highest number is 998.

Therefore, the AP is 101,104,107,.......,998 where, the first term is a=101, the common difference is d=104−101=3 and the nth term is a

n

=998

Now, let us find the number of terms

We know that the nth term of AP(Arithmetic Progression) is given by

a

n

=a+(n−1)d

Now, substituting the values, we get:

a

n

=a+(n−1)d

⇒998=101+(n−1)3

⇒(n−1)3=998−101

⇒(n−1)3=897

⇒n−1=

3

897

⇒n−1=299

⇒n=299+1

⇒n=300

We also know that the sum to nth term in an AP is given by,

S=

2

n

(1st term + last term)

Therefore, we have:

S=

2

300

(101+998)=150×1099=164850

Hence, the sum of all 3 digit numbers that leave a remainder of 2 when divided by 3 is 164,850.