
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
Answer
499.2k+ views
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 \[{{S}_{n}}=\dfrac{n}{2}\left( a+l \right)\], 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 = \[{{2}^{nd}}\] term - \[{{1}^{st}}\] term = 104 – 101 = 3.
We know the formula for sum of AP as,
= (first term + last term) \[\times \] ${\dfrac{\text{number of terms}}{2}}$.
Sum of n terms, \[{{S}_{n}}=\left( \dfrac{a+l}{2} \right)n\].
We know that the last term of an AP is given as,
\[{{a}_{n}}={{a}_{1}}+\left( n-1 \right)d\]
We know last term, \[l={{a}_{n}}=998,{{a}_{1}}=101,d=3\]
\[\begin{align}
& \therefore 998=101+\left( n-1 \right)\times 3 \\
& 998-101=3\left( n-1 \right) \\
\end{align}\]
Let us find the value of n.
\[n=299+1=300\]
Thus we got the number of terms in the series, n = 300.
Sum of an AP, \[{{S}_{n}}=\left( \dfrac{a+l}{2} \right)n\]
\[\therefore {{S}_{n}}=\left( \dfrac{101+998}{2} \right)\times 300=\dfrac{1099\times 300}{2}=164850\]
Thus we got the sum of all 3 digit numbers that leave a remainder of 2 when divided by 3 is 164,850.
\[\therefore \] Option (d) is the correct answer.
Note: For solving questions like these, the basic concept of Arithmetic progression and its formula should be known. If you know the formulas then they are the direct application of the value and solving it. Since we had already obtained common difference d = 3 and n = 300, we could also have used formula for sum as \[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\].
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 = \[{{2}^{nd}}\] term - \[{{1}^{st}}\] term = 104 – 101 = 3.
We know the formula for sum of AP as,
= (first term + last term) \[\times \] ${\dfrac{\text{number of terms}}{2}}$.
Sum of n terms, \[{{S}_{n}}=\left( \dfrac{a+l}{2} \right)n\].
We know that the last term of an AP is given as,
\[{{a}_{n}}={{a}_{1}}+\left( n-1 \right)d\]
We know last term, \[l={{a}_{n}}=998,{{a}_{1}}=101,d=3\]
\[\begin{align}
& \therefore 998=101+\left( n-1 \right)\times 3 \\
& 998-101=3\left( n-1 \right) \\
\end{align}\]
Let us find the value of n.
\[n=299+1=300\]
Thus we got the number of terms in the series, n = 300.
Sum of an AP, \[{{S}_{n}}=\left( \dfrac{a+l}{2} \right)n\]
\[\therefore {{S}_{n}}=\left( \dfrac{101+998}{2} \right)\times 300=\dfrac{1099\times 300}{2}=164850\]
Thus we got the sum of all 3 digit numbers that leave a remainder of 2 when divided by 3 is 164,850.
\[\therefore \] Option (d) is the correct answer.
Note: For solving questions like these, the basic concept of Arithmetic progression and its formula should be known. If you know the formulas then they are the direct application of the value and solving it. Since we had already obtained common difference d = 3 and n = 300, we could also have used formula for sum as \[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\].
Recently Updated Pages
The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

A certain household has consumed 250 units of energy class 11 physics CBSE

The lightest metal known is A beryllium B lithium C class 11 chemistry CBSE

What is the formula mass of the iodine molecule class 11 chemistry CBSE

Trending doubts
State the laws of reflection of light

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

How do I convert ms to kmh Give an example class 11 physics CBSE

Describe the effects of the Second World War class 11 social science CBSE

Which of the following methods is suitable for preventing class 11 chemistry CBSE
