Finding 0's

• From $1$ to $100$ there are $\color{#20A900}10$ numbers with $0$ as a digit :

$\quad \color{#E81990}10,20,30,40,50,60,70,80,90,100$

• From $101$ to $200$ there are $\color{#20A900}19$ numbers with $0$ as a digit :

$\quad \color{#3D99F6}101,102,103,104,105,106,107,108,109,110,120,130,140,150,160,170,180,190,200$

• From $201$ to $1000$ every set of $100$ numbers follow the same sequence. So, a total of $\color{#20A900}9$ sets follow this sequence.

$\text{Total number of numbers with 0 as a digit = } {\color{#E81990}10} + {\color{#3D99F6}9(19)}$

$\implies {\color{#E81990}10} + {\color{#3D99F6}171} = {\color{#20A900}181}$

Since the number can't begin with a zero, it is either a multiple of $10$ (ending in a zero), or of the form $\overline{a0b}$ where $a,b\neq0$ to avoid double counting numbers that fall into the first situation.

There are $\dfrac{1000}{10} = \color{#D61F06}{100}$ multiples of $10$ for the first situation (this goes up to and including $1000$ ).

For the second situation, there are $9$ possibilities independently for $a$ and $b$ so there are $9 \times 9 = \color{#D61F06}{81}$ integers.

$\color{#D61F06}{100} \color{#333333}{ + } \color{#D61F06}{81} \color{#333333}{= \boxed{181}}$

9 two-digit numbers contain zero:10,20,30,40,50,60,70,80,90. 1000 also contains zero. There are 900 three-digit numbers. Those that do not contain zero are with digits between 1 to 9. Therefore we have $9^3$ =729 three-digit numbers with no zero or 900-729=171 three-digit numbers containing zero. The total of all numbers from 1 to 1000, containing zero is 9+1+171=181

It's easier to count the complement: all the numbers that don't contain a 0.

• Of all the single-digit numbers there are 9 that don't have a 0
• Of all the two-digit numbers there are 9 * 9 = 81 that don't have a 0
• Of all the three-digit numbers there are 9 * 9 * 9 = 729 that don't have a 0

That means there are a total of 9 + 81 + 729 = 819 numbers that don't contain a 0. Therefore there are 1000 - 819 = $\boxed{181}$ numbers that do contain a 0.

One tenth of the integers from 1 to 1000 have a 0 as their last digit.

One tenth of the integers from 100 to 1000 have a 0 in their second digit.

The only number having a 0 in its third digit is already part of both other groups.

One tenth other the second group (0 in the second digit) is also part of the first group.

Thus there are: 1000/10 from the first group; 900/10 for the second group; (900/10)/10 for the overlap of the groups.

This makes 100 + 90 - 9 = 181

Numbers containing zeros have the following distinct forms and counts (x, y can be any non-zero digit):

X0..........9 possibilities

X0y........9*9=81 possibilities

Xy0........9*9=81 possibilities

X00........9 possibilities

1000......1 possibility

Since x and y are non-zero, there are no numbers that are double-counted.

9+81+81+9+1=181

Supposed we want to find the number doesn't contain $0$ , three digits number will be $729$ , two digits number will be $81$ , and single-digit will be $9$ . Thus, there are $1000-729-81-9=181$ numbers contain $0$ .

There are 10 from one to 100. 9 from 200-300 9 from 300-400 9 from 400 to 500 and 9 for everything else. Ad them up, you get 181

If you look at 0s in the 1s place that happens, in a set of numbers from 1-100 or 901-1000 (example 950), 10 times. That happens 10 times between 1-1000 so 10 times 10 equals 100 numbers.

Then there is 0s in the 10s place. That happens 9 times in a set of numbers from 101-200 (not including ones with 0s in the 1s place) and that only happens 9 times between 1-1000 because it doesn't happen between 1-99 as it needs 3 digits to have a 0 in the 10s place so 9 times 9 equals 81.

100+81=181 =>QED

From 1 to 100 only multiples of 10 has zeros (10, 20, 30, ...). From 1 to 1000 there are 10 times such series (10, 20, ..., 110, 120, ..., 210, 220, ...): $10 * 10 = 100$ .

Plus from 100 on, the first 9 numbers will also get a zero (101, 102, 103, ..., 109). From 100 to 1000 there are 9 such series (101, 102, ..., 201, 202, ..., 301, 302, ...): $9 * 9 =81$ .

The total number of zeros is the sum of these: $100 + 81 = 181$

In the first 100 there is 10, 20, 30 etc up to 100 which makes 10. From 100-200, there is 101, 102, 103 etc up to 110. Then there is 120, 130 etc up to 200. That makes 19. Each hundred after this follows the exact same number since the only thing that changes is the first digit. So that makes 9 sets of 19 + the 10 from 1-100. That makes 181. It does say the first 1000 positive integers so I thought it would include 1000 but it doesn't otherwise the answer would be 182.

The required number can be of the form(here [RN] means a random number that can be 1,2,3,4,5,6,7,8,9)

(1)[RN]0 In this [RN] can have 9 values hence their are 9 such numbers (2)[RN][RN]0 in this each [RN] can have 9 values hence total=9×9=81 (3)[RN]0[RN] in this again each [RN] can have 9 values total=9×9=81 (4)[RN]00 in this[RN] can have 9 values hence their are 9 such numbers (5)1000 their is only 1 such number TOTAL=9+81+9+81+1=181

Sorry if this is a repeat of any other previous solution, I just want to share how I came to a solution :-)

the above implies that the number of integers containing at least one zero could be generalised to

$\frac{n}{10}$ + 9 × ( $\frac{n}{100}$ -1)

So for n = 1000

$\frac{1000}{10}$ + 9 × ( $\frac{1000}{100}$ - 1) = 100 + 9 × (10 - 1) = 100 + 81 = 181

I'm having a little trouble understanding this although I arrived to the same conclusion, only with a slightly different method. Now, from positive integers to a 1000 exclude zero itself and include 1000 itself, so you have 1 digit numbers, 2 digit numbers, 3 digit numbers, and a 4 digit number.

_ - 1 digit numbers: 9 of them don't have a zero: 9^1 = 9 numbers

_ _ - 2 digit numbers: on "ones", 9 numbers don't have a zero, and on "tens" 9 numbers don't have a zero: 9^2 = 81 numbers

_ _ _ - 3 digit numbers: on "ones", 9 numbers don't have a zero, on "tens" 9 numbers don't have a zero, and on "hundreds" 9 numbers don't have a zero: 9^3 = 729

Now 9+81+729 = 819 numbers don't have a zero, meaning that 999-819 = 180 do have a zero. And if we add the "1000", that makes it 181 numbers that have a zero. BUT wouldn't this mean that that includes zeros in places of "hundreds" in three digit numbers (9 of them), and zeros in places of "tens" in two digit numbers (also 9 of them)? That is: shouldn't it be 181-18 = 163?

Between 1 - 100 } 10 no.s

Between 100 - 200 } 10 + 9 no.s (Numbers in the form of $\overline{x0y}$ )

Between 200 - 300 } 19 no.s (( Same rule applies ))

Between 300 - 1000 } 19 $\times$ 7

$\therefore$ between 1 - 1000 = 10 + 19 + 19 + (19 $\times$ 7) = $\large \color{#20A900} 181$

Any multiple of 10 has a 0. There are 100 "10s" in a 1000.

101 has 0 in it too. So we need to include all solutions like 101.

The number 101 is of the form a0b where a and b are digits.

We know b can't be 0 because we already did that and a can't be 0 because that doesn't make sense.

Thus a is 1-9 and b is 1-9. We know 9x9 = 81.

100 + 81 = 181.

Its 100 // coz 10,20,30,.... //+81 // a01a02a03...a09 for all a belongs N 1-9 //