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Hint: Here we get the three digit numbers which are multiple of 6 in the form of AP series .To find the sum of digits use the formula

${S_n} = \dfrac{n}{2}({\text{first term}} + {\text{last term}})$

We know that least three digit number is $100$

If we divided $100$ by $6$ we get remainder as $4$

We know that greatest two digit number that is multiple of $6$=$100 - 4 = 96$

Now the least three digit number which is multiple of $6$ =$96 + 6 = 102$

From this we can say that the least three digit number which is multiple of $6$ is $102$

To find the sum of all three digit numbers that are multiple of $6$

Let us add $6$to the first which mean least three digit number $102$ and lets continue the processing adding ‘$6$’ to the resultant number to get next numbers.

Then the next number will be $108,114,120.....$

Then series is $102,108,114,120....$

The above series is of AP where the first term a=$102$, d=$6$

To find the sum of the numbers we have to find the n value

We know that ${n^{th}}$of AP is ${a_n} = a + (n - 1)d$

And again here we need the ${n^{th}}$term value nothing but maximum value that is multiple of $6$

We know that greatest three digit number =$999$

So here if we divide $999$ with $6$ the remainder will be $3$

So to get the maximum three digit number which is multiple of $6$ let us subtract $3$ from $999$ which gives the $3$-digit number that is multiple of$6$.

$ \Rightarrow 999 - 3 = 996$

So here the maximum $3$-digit number that is multiple of $6$ is $996$.

Then here ${a_n} = 996$

Now let us find n value

$

\Rightarrow {a_n} = a + (n - 1)d \\

\Rightarrow 996 = 102 + (n - 1)6 \\

\Rightarrow (n - 1)6 = 894 \\

\Rightarrow n - 1 = 149 \\

\Rightarrow n = 150 \\

\therefore n = 150 \\

$

From this we can say there are total $150$ numbers in the series that are multiple by $6$

Sum of the terms$ \Rightarrow $${S_n} = \dfrac{n}{2}({\text{first term}} + {\text{last term}})$

Let us substitute the value

$

\Rightarrow {S_n} = \dfrac{{150}}{2}(102 + 996) \\

\Rightarrow {S_n} = 75 \times 1098 \\

\Rightarrow {S_n} = 82350 \\

$

Therefore sum of all three digit numbers which are multiple of $6$=$82350$

Note: Make note that to find the sum of all three digit numbers multiple of $6$ it’s important to find the n value. So in this problem we have to find least number that is multiple of 6 and to find sum of number 3 digit numbers that are multiple of 6 we have the how many 3digit numbers are present that are divisible by 6.So we have used nth term of AP to get n value, as the numbers that are multiple of 6 are in AP. And finally we have used the sum of n terms formula to get a sum of numbers that are multiple of 6.

${S_n} = \dfrac{n}{2}({\text{first term}} + {\text{last term}})$

We know that least three digit number is $100$

If we divided $100$ by $6$ we get remainder as $4$

We know that greatest two digit number that is multiple of $6$=$100 - 4 = 96$

Now the least three digit number which is multiple of $6$ =$96 + 6 = 102$

From this we can say that the least three digit number which is multiple of $6$ is $102$

To find the sum of all three digit numbers that are multiple of $6$

Let us add $6$to the first which mean least three digit number $102$ and lets continue the processing adding ‘$6$’ to the resultant number to get next numbers.

Then the next number will be $108,114,120.....$

Then series is $102,108,114,120....$

The above series is of AP where the first term a=$102$, d=$6$

To find the sum of the numbers we have to find the n value

We know that ${n^{th}}$of AP is ${a_n} = a + (n - 1)d$

And again here we need the ${n^{th}}$term value nothing but maximum value that is multiple of $6$

We know that greatest three digit number =$999$

So here if we divide $999$ with $6$ the remainder will be $3$

So to get the maximum three digit number which is multiple of $6$ let us subtract $3$ from $999$ which gives the $3$-digit number that is multiple of$6$.

$ \Rightarrow 999 - 3 = 996$

So here the maximum $3$-digit number that is multiple of $6$ is $996$.

Then here ${a_n} = 996$

Now let us find n value

$

\Rightarrow {a_n} = a + (n - 1)d \\

\Rightarrow 996 = 102 + (n - 1)6 \\

\Rightarrow (n - 1)6 = 894 \\

\Rightarrow n - 1 = 149 \\

\Rightarrow n = 150 \\

\therefore n = 150 \\

$

From this we can say there are total $150$ numbers in the series that are multiple by $6$

Sum of the terms$ \Rightarrow $${S_n} = \dfrac{n}{2}({\text{first term}} + {\text{last term}})$

Let us substitute the value

$

\Rightarrow {S_n} = \dfrac{{150}}{2}(102 + 996) \\

\Rightarrow {S_n} = 75 \times 1098 \\

\Rightarrow {S_n} = 82350 \\

$

Therefore sum of all three digit numbers which are multiple of $6$=$82350$

Note: Make note that to find the sum of all three digit numbers multiple of $6$ it’s important to find the n value. So in this problem we have to find least number that is multiple of 6 and to find sum of number 3 digit numbers that are multiple of 6 we have the how many 3digit numbers are present that are divisible by 6.So we have used nth term of AP to get n value, as the numbers that are multiple of 6 are in AP. And finally we have used the sum of n terms formula to get a sum of numbers that are multiple of 6.

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