The number of times 2 occurs in 1000 to 9999 I came across two methods Total 4 digit numbers that

@karthika- when considering at the unit place we get 9*10*10*1=900
for tenths position- 9*10*1*10=900
for 100th position - 9*1*10*10=900
for 1000th position- 1*10*10*10=1000
so total no of 2's=(900+900+900+1000)=3700
11 years agoHelpfull: Yes(24) No(1)

2 (without repition at units place) --- 8*9*9*1 = 648
2 (without repition at tens place) --- 8*9*1*9 = 648
2 ( without repition at 100s place ) --- 8*1*9*9 = 648
2 ( without repition at 1000s place ) --- 1*9*9*9 = 729
2 ( at units nd 10s place ) -- (8*9*1*1) * 2 = 144
2 ( at units nd 100s place ) -- ( 8*1*9*1 ) * 2 = 144
2 ( at units and 1000s place ) -- ( 1*9*9*1 ) * 2 = 162
2 ( at 10s nd 100s place ) -- ( 8*1*9*1 )*2 = 144
2 ( at 10s and 1000s place ) -- (1*9*9*1)*2 = 162
2 ( at 100s and 1000s place ) -- (1*1*9*9)*2 = 162
2 ( at 1s 10s 100s place ) -- 3*8 -- 24
2 ( at 10s 100s 1000s place ) -- 3*9 -- 27
2 ( at 1s 10s 100s 1000s place ) -- 1(2222) as it contains 4 no. of 2's -- 4
total numbr of 2's from 1000 to 9999 = 3646
hope answer is correct !!
11 years agoHelpfull: Yes(7) No(4)
3700 is the correct method...
11 years agoHelpfull: Yes(5) No(0)
Solution 1 is correct because in Second method calculation is done by fixing 2 at unit,tenths,hundreds,thousands place and then 2 is allowed to repeat.
thus in this method we have counted many nos like 2222; 4 times.
Thus according to me 1st method is correct as it applies simple concept of arrangement
11 years agoHelpfull: Yes(3) No(4)
Ok guys..but please explain y the first method is not ryt?
11 years agoHelpfull: Yes(2) No(0)
method 2 gives the correct answer
11 years agoHelpfull: Yes(2) No(0)
2 Solution is Correct, in First solution you have actually calculated the no. of 4 digit no. in which 2 is there in any place.. While in 2 solution, total no. of occurance of Digit 2 is calculated, which is desired solution.
11 years agoHelpfull: Yes(2) No(0)
i have the strong reason of it
2nd one is correct it's because in first method we can count only once either 2222 occurs four times in a no. thats why in method one counting of 2 became less
11 years agoHelpfull: Yes(1) No(0)
Second one
11 years agoHelpfull: Yes(0) No(0)
0-99: 1 recurrence of digit 7 (single digit) + 9 (from "ones" in double digit) + 10 (from "tens" in double digit). (1+9+10)=20

100-999: 10 (from "ones" in triple digit numbers) + 90 (from "tens" in triple digit numbers) + 100 (from "hundreds" in triple digit numbers) = 10*(1+9+10)=200

1,000-9,999: again we see: 100*(1+9+10)=2,000
11 years agoHelpfull: Yes(0) No(3)
praksh singh plz explaain hy frst option is wrong!!!!!!!!!
11 years agoHelpfull: Yes(0) No(0)
1st method is correct
as for example take no of time 2 occurs between 10 and 99
using first method we will get 2digit no that we can form as(9*10-8*9)=18
ie 12,22,32,42,52,62,72,82,92,20,21,23,24,25,26,27,28,29 total of 18
however using second method we should have answer as
unit place 9*1=9
tens place 1*10=10
total of 19 which is wrong .because 22 we are counting twice
11 years agoHelpfull: Yes(0) No(3)
2 occurs in 4-digit num once=((9*9*9)+(8*9*9)*3)*1=2673
2 occurs in 4-digit num 2 times=(3*(9*9)+3*(8*9))*2=459
2 occurs in 4-digit num 3 times=(3*9+8*1)*3=105
2 occurs in 4-digit num 4 times=4*1,ans=3700 adding all above
11 years agoHelpfull: Yes(0) No(0)
@praveen-In the first method we will get the total no of 4 digit nos that contain 2 at any place, but as we have to count the total no of 2's so we consider the second method..
@Roshan kumar- The question is to count the total no of 2's that occours not the numbers.... and yes for the number 22 we count twice as we have two 2's occuring
11 years agoHelpfull: Yes(0) No(0)
Method number 1 is right, and the reason why I say is because of this link. "http://mathforum.org/library/drmath/view/55936.html"
6 years agoHelpfull: Yes(0) No(0)

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