vs224 wrote:
stne wrote:
(A) 469
(B) 471
(C) 475
(D) 478
(E) 481
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
What is the logic of subtracting the 729 second time ?
Can you elaborate ?
If 4 digit number starting with 1 is 729. How 2000, 2001, .. 2017 are included in it.
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.
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