How many positive integers less than or equal to 2017 contain the digi

Question

How many positive integers less than or equal to 2017 contain the digit 0?

(A) 469
(B) 471
(C) 475
(D) 478
(E) 481

AOBang · Accepted Answer

Approach this problem from opposite direction.

Do not calculate number of numbers with digit zero in them but let us find number of numbers which do not have any digit as zeros in them. So starting from - single digit number, 1 to 9 => 9 numbers Then - all 2 digit number, Unit place can be any number from 1 to 9 and also tense place number can be any number from 1 to 9. ^9 C_{1} * ^9 C_{1} = 9 * 9 => 81 numbers Then - all 3 digit number, Unit place, Tense place & hundreds place can be any number from 1 to 9. ^9 C_{1} * ^9 C_{1} * ^9 C_{1} = 9 * 9 * 9 => 729 numbers Then - all 4 digit number (<2000), Thousands place number is now fixed as 1. Unit place, Tense place & hundreds place can be any number from 1 to 9. ^1 C_{1} * ^9 C_{1} * ^9 C_{1} * ^9 C_{1} = 1* 9 * 9 * 9 => 729 numbers Now out of our total number 2017 subtract these numbers to get required ans. = 2017 - 9 - 81 - 729 - 729 = 469 option A To encourage hit a 'Fearless Kudos' !!

Archit3110 · Answer

total integers 2017
unit digit except 0 ; 9
2 diigt no except 0 ; 9*9
and 3 digit no except 0 ; 9^3 
and 4 digit no starting with 1 ; 1 * 9^3 ; 729
2017-9-81-729-729 ; 469
IMO A

stne · Answer

What is the logic of subtracting the 729 second time ?
Can you elaborate ?

Archit3110 · Answer

stne 3 and 4 digit no ; would both by 729

stne · Answer

Indeed my question is worthy of being laughed at, thanks man.
Careless me!

RusskiyLev · Answer

It is a tricky question! And requires a lot of attention! We are given that 0 < n =< 2017 and we need to find the number of "n"s which contain 0s. 
First of all, let us count the numbers where last digit is   0: 1 0 , 2 0 , 3 0 , ... 8 0 , 9 0 , 10 0    There are 10 such numbers in every hundred of numbers. As our maximum value is 2017 (n =< 2017), we have 20 hundreds, thus   10 * 20 = 200  
Let us now count numbers where tens digit is 0:   1 0 1, 1 0 2, 1 0 3, ... 1 0 8, 1 0 9  . We do not include here the number 100, since it was already captured in the previous calculation. The number of 0s here is 9, and we again multiply it by 20 using the same principle described above:   20 * 9 = 180  
Now let us count numbers where hundreds digit is 0:   1 0 00, 1 0 01, 1 0 02, 1 0 03, ... 1 0 08, 1 0 09  . Now, we know that these numbers extend till number 1099, and the next occasion is number 2000. Till number 1099, there are 100 numbers, but let us not forget about 10 numbers calculated already in the 1st step, and 9 numbers in the 2nd step.   100 - 10 - 9 = 81  . In addition, we have 8 numbers in 2000s:   2 0 10, 2 0 11, 2 0 12, ... 2 0 16, 2 0 17  . 
Now let us sum all these numbers: 
  200 + 180 + 81 + 8 = 469

Answer:  A

vs224 · Answer

If 4 digit number starting with 1 is 729. How 2000, 2001, .. 2017 are included in it.

Vigasimrair · Answer

RusskiyLev  may you please clarify: why numbers which were counted on the 2nd step (0 - tens digit) do not include 2001...2009 (which you've counted on the last step)
thanks a lot!

Vigasimrair · Answer

fantastic and elegant solution, thank you

RusskiyLev · Answer

Vigasimrair , You are correct, I made a mistake in the explanation, and I have already edited it. These numbers were actually counted in the second step, and in the last step we should have counted the numbers 2010 - 2017. Thanks for noting!

stne · Answer

Hi  vs224 ,

Actually 2001 ........2017 are NOT included. Which means ,all numbers NOT including 0 have been subtracted and then what we are left with are the numbers that include zero.

One digit numbers EXCLUDING zero - 1,2,3,4,5,6,7,8,9 Total nine such numbers.
Two digit numbers EXCLUDING zero- 12,12,13...........99 Total 81 such numbers ( 9 options for the unit digit and nine options for the tens digit , hence 9*9 =81)
Three digit numbers EXCLUDING zero-111......999 ( 9*9*9=729)
For Four digit numbers EXCLUDING zero , there are two categories , 
1) Those starting with 1 , of the form 1000 - 1999 to remove all that,which do not contain zero, we do 1*9*9*9 here for the left most digit we have only one option,i.e. 1
for hundreds digit we have nine options , because we cannot include 0, hence 1 to 9 similarly for the tens and units digit.
Hence total 4 digit numbers not including 0 and beginning with one are 1*9*9*9=729

2) Those starting with 2, of the form 2000-2999, but note, we are only asked till 2017, hence from 2000-2017 all numbers actually include zero. So nothing to subtract here.

So total positive numbers that do  NOT  include the digit zero and are less than or equal to 2017 are 
9+81+729+729=1578 
Hence total positive numbers that include the digit zero in them, and are less than or equal to 2017 are 2017-1578 =469
Hope it's clear.

Please let me know if anything is still unclear, thanks.

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

How many positive integers less than or equal to 2017 contain the digi

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

How many positive integers less than or equal to 2017 contain the digi

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards