Question
Question: What is the sum of all 3 digit numbers that leave a remainder of ‘2’ when divided by 3? (a) 530,82...
What is the sum of all 3 digit numbers that leave a remainder of ‘2’ when divided by 3?
(a) 530,820
(b) 179,587
(c) 560,758
(d) 164,850
Solution
Hint: Find the smallest and largest 3 digit number when divided by 3 leaves as 2. Then form the arithmetic progression of such numbers. Find first term, last term common difference and number of terms. Substitute these values in the formula to find sum of n terms which is given by Sn=2n(a+l), where n is number of terms, a is first term and l is last term.
Complete step by step solution:
The smallest 3 digit number we know is 100. But the smallest 3 digit number that will give us a remainder 2, when divided by 3 is 101.
Similarly the largest 3 digit number is 999. But the largest 3 digit number that will give us a remainder 2, when divided by 3 is 998.
Thus we can say that the lowest 3 digit number is 101 and the highest number is 998. The next number that will leave a remainder of 2 when divided by 3 after 101 is 104 and the next is 107 etc. Thus we can form an AP with it,
101, 104, 107,……. 998
Thus in the Arithmetic progression (AP) we know the first term, a = 101 and the last term, l = 998.
The common difference = 2nd term - 1st term = 104 – 101 = 3.
We know the formula for sum of AP as,
= (first term + last term) × 2number of terms.
Sum of n terms, Sn=(2a+l)n.
We know that the last term of an AP is given as,
an=a1+(n−1)d
We know last term, l=an=998,a1=101,d=3