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

(a) 5050 (b) 903

(c) 901 (d) 900

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• KevinM
Lv 7
8 years ago

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

• 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

• 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