CAT4MBA

1 digit – 10
2 digit – 2 x 90 = 180
3 digit – 3 x 401 = 1203
Total 1393

1.

from page no. 1-9 : 1x10=10

from page no. 10-99 : 2x90=180

from page no. 100-500 : 3x401=1203

10+180+1203= 1393

2.

0 1 2 3 4 5 6 7 8 9

10 digits are used.

what u say.....

somebody tell me where did 91-99 go !

how i find dat....lemme xplain...

each number frm 1 to 9 comes 20 times from 1 to 99...from here we can calculate no of digits i.e 20*9=180 digits
n frm 1 to 99, "9" zeroes are there..i.e in 10,20,30...90

now lets cum to 3 digits....strting from 100 to 199...we vil add zeroes of 100 at later stage...!!!1
so 1 wil cum at 100's place at 100 times
and 1 wil cum at 10s place...10 times
and 1 wil cum at unit's place=10 times

so no of ones=180

now total no of ones=100+10+10=120

now lets calculate other digits...

no of twos vil cum=20 times(10+10, same funda as we hv calculated in finding no of ones except there is no 100's place)
so...rest of the digits except 1 wil cum 20 times..

so total no of digits included 2 =9*20=180

so total no of digits from 100 to 199=120+180=400
i.e 400+2(no of zeroes in 100)=402

this funda vil go on..n vil calculate no of digits from 200-300 which is again 402
nd 300-400 again 402

nd 400-500 again 402

so 402*4=1608
nd final ansr is 1608+180=1788 no of digits..

correct mee if um wrong..

thnx..

i think the answer is 1392

Pages 1-9 : 9
Pages 10-99 : 2*90 = 180
Pages 100-500 : 3*401 = 1203
Therefore, total no. of digits used = 9+180+1203 = 1392.

hii.. wots wrong in ma aaproachh.....?????

your written solution is very complex. but if u elobrate it i can tell u wats wrong.