A faulty car odometer proceeds from digit 3 to digit 5, always skippin

No of times 4 appears in a hundred=20
In 2005, 4 has appeared=20*20=400
Therefore car has actually travelled=2005-400=1605

Posted from my mobile device

I am not sure what concept is this question really checking and is it really relevant to gmat..

my take below
odometer works where in 1 unit position keeps moving with distance and its total gets accumulated over the range
so here suppose inital reading is 000000
so first reading 000001,000002,00003,000005 now meter is skipping after every3rd unit so it skips total 10 times in range from 0-100
given total final reading 002005 ; meter must have skipped 2005/100; 200.5 times
so actual reading 2005-200.5 = ~ 1804
IMO E

Total miles actually travelled = Odometer reading - No. of times 4 appear between 0000 to 2005

No. of times 4 appear between 0000 to 2005 = No. of times 4 appear between 000 to 999 + No. of times 4 appear between 1000 to 2005

now,

between 000 to 999 , there are 1000 different numbers and each number is a 3 digit number so , in and all there 1000*3 digits = 3000 digits

by symmetry, each digit between 0 to 9 appears equal no. of times so every digit appears 3000/10 no. of times .... hence 4 appears 300 no. of times

similary 4 appears 300 no. of times between 1000 to 1999
and 4 appears 1 no. of times between 2000 to 2005

so in an all 4 appears 300+300+1 = 601 no. of times
so, Total miles actually travelled = 2005 - 601 = 1404

Inyour calculation you have accounted only one 4 in 44, the other 4 is not accounted for... so it would mean 20 nos. of 4s in every 100

Moreover you have not considered these 20 nos. of 4s which will appear in 400 and 1400 series..... so that would mean there will be total 542 +18 + 20 + 20 + 1 4s in all

Consider it as 4digit numbers

_ _ _ _ in how many ways we can fill this up to 2005
Considering first blank to be 0, 1 it is 2 ways rest is 10
So total number ways this can be filled for digits 1-1999
2*10*10*10-1(considering one number where all are zeroes)=1999
Now number starting with 2 is 2000,2001,2002,20032004,2005 so total is 1999+6=2005
same way if 4 is not considered 4 digit number up to 1999 can be filled in 2*9*9*9-1(considering all as zero)=1457
Rest digit with from 2000-2005 with excluding 2004 is 5digirs
So total is 1457+5=1462

Posted from my mobile device

Consider it as 4digit numbers

_ _ _ _ in how many ways we can fill this up to 2005
Considering first blank to be 0, 1 it is 2 ways rest is 10
So total number ways this can be filled for digits 1-1999
2*10*10*10-1(considering one number where all are zeroes)=1999
Now number starting with 2 is 2000,2001,2002,20032004,2005 so total is 1999+6=2005
same way if 4 is not considered 4 digit number up to 1999 can be filled in 2*9*9*9-1(considering all as zero)=1457
Rest digit with from 2000-2005 with excluding 2004 is 5digirs
So total is 1457+5=1462

Posted from my mobile device

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.