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

Question

A faulty car odometer proceeds from digit 3 to digit 5, always skipping the digit 4, regardless of position. If the odometer now reads 002005, how many miles has the car actually traveled?

(A) 1404
(B) 1462
(C) 1604
(D) 1605
(E) 1804

baru · Accepted Answer

When the odometer says that it traveled for 5miles then it means that it actually only travelled 4 miles

So +1 added to the odometer with out travelling.

Similarly 4,14,24,34,54,64,74,84,94 -(40-is skipped ) . Here a Total of 9*1 + 10 = 19 miles are added for a hundred
Similarly 400 ,1400 are also skipped - 200 more added

Total extra miles added before 2000 = 19(18) +200 = 342 + 200 = 542
for 2005 , 1 mile is for five miles travelled after 2000 , So Grand Total = 542+1 = 543

miles travelled = 2005-543 = 1462- Option B

If its clear - Kudos Please

deushyant · Answer

Thought of this as a combinations problem and that made it quite easy.
Single digit number without 4 = 9
Double digit numbers without 4 = 8x9 = 72
Triple = 8x9x9 = 648
4 digit with first fixed as 1 = 1x9x9x9 = 729
Add all plus 4 miles for 2000-2005 gives:
9+72+648+729+4 = 1462

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Ypsychotic · Answer

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

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Archit3110 · Answer

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

Chikazo · Answer

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

Chikazo · Answer

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

sameeruce08 · Answer

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

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Dereno · Answer

Usual odometer reading moves from digits O (zero) to 9(nine) I.e.,  DECIMAL NOTATION  . But, here let's see a pattern going on. Let us have two scenario's - actual vs faulty.

ACTUAL. FAULTY
0.              0. 
1.              1
2.              2
3.              3
4.              5
5.              6
6.              7
7.              8
8.              9.

This decimal notation is now converted to a system of base 9. The faulty odometer have skipped a number to show values in base 9.

Faulty 002005 ————-> the corresponding value of 5 in actual is 4, and two is 2. ( see the table above)

Therefore , 002004 in base 9 equals (moving left to right)

4 *9^0 = 4
0 *9^1 = 0
0 *9^2 = 0

2 *9^3 = 729*2 = 1458

Adding all 1458+4= 1462.

Rohanx9 · Answer

Quickest way imo is using combinations.

2005 = Four digits A B C D

Numbers 0-1999:

A can be 2 numbers (0, 1)
B, C and D can be 9 numbers (0, 1, 2, 3, 5, 6, 7, 8, 9)

Combinations = 2 * 9 * 9 * 9 =1458

Adding the numbers 2000, 2001, 2003 and 2005 (4 in total) we get:

1458 + 4 = 1,462 (Answer)

agrotechpepper · Answer

Really nice, but you did a small mistake. Here, when you write " Adding the numbers 2000, 2001, 2003 and 2005 (4 in total) we get: ", you missed  2002,  but the answer comes out to be correct since you added  0000  in your this calculation - 
" Numbers 0-1999:

A can be 2 numbers (0, 1)
B, C and D can be 9 numbers (0, 1, 2, 3, 5, 6, 7, 8, 9)

Combinations = 2 * 9 * 9 * 9 =1458 "

So actually, its 1458 - 1 (for 0000) + 5 = 1462.

But overall good method.

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

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

Prep Toolkit

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