Question

find the sum of all 3 digit number which when divided by 11 leave a remainder 5

Answer 1

Step-by-step explanation:

Hey guy..!!

Gud evening..

115,126,137,148,159,170,181,192,203,214 etc..

Brainliest please..

Answer 2

The required sum is 45059.

Step-by-step explanation:

Consider the provided information.

99 is the last two digit number which is divisible by 11.

Add 5 to 99 that will gives us the first 3 digit number which when divided by 11 leave a remainder 5.

99+5=104

104 is the first 3 digit number which when divided by 11 leave a remainder 5.

Now add 11 in 104 that will gives you the next number.

104+11=115

990 is the last 3 digit number which is divisible by 11.

Add 5 in 990 that will gives us the last 3 digit number which when divided by 11 leave a remainder 5.

990+5=995

Hence, the required sequence is 104, 115, .........995

Calculate the total number of terms using the formula: $a_n=a+(n-1)d$

Substitute a=104, d=11 and aₙ = 995 in above formula.

$995=104+(n-1)11\\891=11(n-1)\\n=82$

To find the sum use the formula: $S_n=\frac{n}{2}(a+l)$

Where l is the last term.

Substitute respective values in above formula.

$S_n=\dfrac{82}{2}(104+995)$

$S_n=41(1099)$

$S_n=45059$

Hence, the required sum is 45059.

#Learn more

Find the sum of all 3 digit numbers which leave remainder 3 when divided by 5

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