The sum of all the digits of the numbers from 1 to 100 is?

(a) 5050 (b) 903

(c) 901 (d) 900

6 Answers

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  • KevinM
    Lv 7
    8 years ago
    Favorite Answer

    Note that this question asks for the sum of the DIGITS of these numbers - not the sum of the numbers themselves.

    The last digit will go from 1-0 ten times. So that's (1+2+3...+9+0) * 10 = 450

    The second digit will go from 1-9 for ten numbers each, and 0 once for 100: (1+...+9)*10 = 450

    The first digit is only 1 in the number "100", so the total is:

    450 + 450 + 1 = 901

    The answer is (c) 901.

  • 8 years ago

    You are asking for the sum of the digits and not the sum of the counting numbers (right?).

    Are you including 1 and 100? I am assuming so.

    There are

    11 zeros

    21 ones

    All the rest have 20.

    21 ones = 21

    20 twos = 40

    20 threes = 60

    20 fours = 80

    20 fives = 100

    etc.

    The answer should be 901. If you take out 1 and 100, you reduce the total to 899 which is not given so the answer is 901.

  • Anonymous
    8 years ago

    Let:

    a₁ - first term,

    an - n-th term,

    d - common difference of successive terms

    n - number of terms

    ------------------------------

    Sum of digits in an arithmetic progression (each number is previous number plus some constant 'd') is given by formulae:

    Sn = (n / 2)(a₁ + an)

    and

    Sn = (n / 2)[2a₁ + (n - 1)d]

    Which formula you use, depends on what things are given.

    I'll use 1st one:

    Sn = (n / 2)(a₁ + an)

    Sn = (100 / 2)(1 + 100)

    Sn = 50(101)

    Sn = 5050 <~~~~~~ answer (a)

  • 8 years ago

    (100+1)*100/2=5050

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  • 8 years ago

    1 + 2 + 3...... + 100

    100+99+98.....+1

    -------------------------

    101+101.....100times

    =101*100

    but as done

    we added the sum twice so

    101*100/2= sum

    10100/2

    =5050

  • 8 years ago

    a

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