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

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    Question

    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

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    Find the sum of three

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

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

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    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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    Chapter 3: Arithmetic Progression - Q.3 (B)

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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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    8 Answers

    Hungenahalli Sitaramarao Badarinath

    , 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

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    ∑ 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

    583 viewsView upvotes

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    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

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    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

    657 viewsView upvotes

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    Frank Abbing

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    Mohammed 7 month ago
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