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# find the sum of three-digit natural numbers, which are divisible by 4.

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## How many three digit natural numbers are divisible by 4 ?

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## How many three digit natural numbers are divisible by 4?

Hard Open in App Solution Verified by Toppr

Thethreedigitalnumberwhichisdivisibleby4is

100,104,108,..........996

firstterm=100 d=104−100=4 lastterm=996 Then, Tn=a=(n−1)×4 996=100+(n−1)×4 996−100=(n−1)4 4 896 ​ =(n−1) 224=(n−1) ∴n=225

225threedigitsnumberwhichisdivisibleby4

142 5

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## Find the sum of three

Find the sum of three-digit natural numbers, which are divisible by 4 Find the sum of three-digit natural numbers, which are divisible by 4

### SOLUTION

The three-digit natural numbers divisible by 4 are

100, 104, 108, ......, 996

The above sequence is an A.P.

∴ a = 100, d = 104 – 100 = 4

Let the number of terms in the A.P. be n.

Then, tn = 996

Since tn = a + (n – 1)d,

996 = 100 + (n – 1)(4)

∴ 996 = 100 + 4n – 4

∴ 996 = 96 + 4n ∴ 996 – 96 = 4n ∴ 4n = 900 ∴ n = 9004 = 225 Now, Sn = n ttn n2(t1+tn) ∴ S225 = 2252(100+996) = 2252(1096) = 225 × 548 = 123300

∴ The sum of three digit natural numbers, which are divisible by 4 is 123300.

Concept: Sum of First n Terms of an AP

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## How to find the sum of all 3

Answer (1 of 8): The smallest 3-digit number divisible by 4 is 100 while the largest is 996. The sum of all 3-digit number divisible by 4 is an exercise in an AP. a = 100, d = 4, 996 = a+(n-1)d = 100 + (n-1)*4, or 249–25 = 224 = (n-1) n = 225 S225 = (225/2)[2*100 + (225–1)*4] = (225/2)[200... How do I find the sum of all 3-digit numbers which are divisible by 4?

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, Civil engineer

Answered 3 years ago · Author has 43.8K answers and 28.6M answer views

The smallest 3-digit number divisible by 4 is 100 while the largest is 996.

The sum of all 3-digit number divisible by 4 is an exercise in an AP.

a = 100, d = 4,

996 = a+(n-1)d = 100 + (n-1)*4, or

249–25 = 224 = (n-1)

n = 225

S225 = (225/2)[2*100 + (225–1)*4]

= (225/2)[200+ 896] = (225/2)*1096 = 12330o. Answer :123300.

6.2K viewsView upvotesView 1 share

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Alex Moon

, BS (in progress) Mathematics, Michigan State University

Answered Mar 17, 2022 · Author has 2.6K answers and 960.4K answer views

∑ k=1 n k= n 2 +n 2 ∑k=1nk=n2+n2

The above is the well known arithmetic series,

let’s play around with it!

∑ k=m+1 n k= ∑ k=1 n k− ∑ k=1 m k= n 2 +n− m 2 −m 2

∑k=m+1nk=∑k=1nk−∑k=1mk=n2+n−m2−m2

But we need to scale this by

4 4

to deal with multiples of

4 4 : 2( n 2 +n− m 2 −m) 2(n2+n−m2−m)

This is the sum of the

(m+1 ) th (m+1)th thru n th nth multiples of 4 4 , we see 100=4(m+1),996=4n 100=4(m+1),996=4n implies that m=24,n=249 m=24,n=249

, those are the upper and lowerbounds on

3− 3− digit multiples of 4 4 . Our answer is

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(Continue reading) Soufiane Erraji

, B.S Mathematics, Joseph Fourier University (1997)

Answered 3 years ago · Author has 101 answers and 64.6K answer views

S = 100 + 104 + 108 + … + 996

S = 4 * (25 + 26 + 27 + … + 249)

S = 4 * [ (1 + 2 + 3… + 249) - (1 + 2 + 3…+24)]

S = 4 * [ (249 * 250 / 2) - (24 * 25 / 2)]

S = (249 * 500) - (24 * 50) = 123 300

1.2K views Vincent Herriau

, Principal (2009-present)

Answered 3 years ago

Let us divide it by 4 : the sum of all integers from 1 to 249

It is equal to 249x250/2

Now multiply by 4 : 249 x 250 x 2 = 124500

QED

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Timothy Chang

, Lecturer at Dongguan University of Technology (2017-present)

Answered 3 years ago · Author has 379 answers and 416.4K answer views

If N N is a 3 3 -digit number, then 100≤N<1000 100≤N<1000

. This problem asks for the sum of all such

N N

that is divisible by

4 4 . Let N=4k N=4k where k k

is another positive integer.

100≤N=4k<1000⟹25≤k<250

100≤N=4k<1000⟹25≤k<250

.

It is convenient to use Python to find the sum of all multiples of

4 4 between 100 100 and 1,000 1,000

(inclusive on the left side):

sum(4*k for k in range(25, 250))

The answer is 123,300 123,300 . 626 views Guillermo Owen , former Professor

Answered 3 years ago · Author has 174 answers and 83.9K answer views

These go from 100 to 996. Thy form an arithmetic progression with common difference 4. There are (996–100)/4 + 1 = 225 of them The usual formula for such a progression is

Sum = (first + last)*number of terms / 2 = 1096*225/2 = 123,300

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Frank Abbing

, former Pensioner at Philips (1965-1990)

Answered 2 years ago · Author has 504 answers and 46.5K answer views

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Mohammed 7 month ago

Guys, does anyone know the answer?