Question

find the sum of all 3 digit numbers divisible by 11?

Answer 1

The sum of all 3 digits divisible by 11 is 44,550.

As we know that, the below numbers which are being divisible by 11, such that

110, 121, 132, 990

The above series looks like the number are in A.P

To find ‘the last term of a series’ is = a + (n-1) d.

Where a = 110, d = 11.

Such that,

$\begin{array}{l}{990=110+(n-1) 11} \\ {880=(n-1) 11} \\ {80=n-1} \\ {n=81}\end{array}$

Sum of all the digits $=\frac{n}{2}(2 a+(n-1) d)$

$\begin{aligned} &=\frac{81}{2}(2(110)+80(11)) \\ &=\frac{81}{2}(220+880) \\ &=\frac{81}{2}(1100) \\ &=81 \times 550 \\ &=44,550 \end{aligned}$

Answer 2

Answer:

44540

Step-by-step explanation:

The sum of all 3 digits divisible by 11 is 44,550.

As we know that, the below numbers which are being divisible by 11, such that

110, 121, 132, 990

The above series looks like the number are in A.P

To find ‘the last term of a series’ is = a + (n-1) d.

Where a = 110, d = 11.

Such that,

\begin{gathered}\begin{array}{l}{990=110+(n-1) 11} \\ {880=(n-1) 11} \\ {80=n-1} \\ {n=81}\end{array}\end{gathered}

990=110+(n−1)11

880=(n−1)11

80=n−1

n=81

Sum of all the digits =\frac{n}{2}(2 a+(n-1) d)=

2

n

(2a+(n−1)d)

\begin{gathered}\begin{aligned} &=\frac{81}{2}(2(110)+80(11)) \\ &=\frac{81}{2}(220+880) \\ &=\frac{81}{2}(1100) \\ &=81 \times 550 \\ &=44,550 \end{aligned}\end{gathered}

=

2

81

(2(110)+80(11))

=

2

81

(220+880)

=

2

81

(1100)

=81×550

=44,550