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Director
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Starting with 1, positive integers are written one after the [#permalink]
 02 Nov 2005, 23:55
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Starting with 1, positive integers are written one after the other. What is the 40,000th digit that will be written?
A) 2
B) -2
C) 3
D) 6
E) None of the above (   )
Need proof that you worked all the 40,000!
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SVP
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Re: Patience is a virtue! - PS [#permalink]
03 Nov 2005, 00:00
gsr wrote: Starting with 1, positive integers are written one after the other. What is the 40,000th digit that will be written? A) 2 B) -2 C) 3 D) 6 E) None of the above ( ) Need proof that you worked all the 40,000! oh my , one of the typically advanced arithmetic problems which I did try before but ................ I TOTALLY forget how to solve it LOL ...ahem, let me try to wake my subconsciouness up first 
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Director
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Re: Patience is a virtue! - PS [#permalink]
03 Nov 2005, 00:02
laxieqv wrote: oh my , one of the typically advanced arithmetic problems which I did try before but ................ I TOTALLY forget how to solve it LOL ...ahem, let me try to wake my subconsciouness up first While you try to wake your subconsciousness, let me put mine to 'sleep' coz i got work tomorrow! Happy counting!!
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Director
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Re: Patience is a virtue! - PS [#permalink]
03 Nov 2005, 00:03
gsr wrote: Starting with 1, positive integers are written one after the other. What is the 40,000th digit that will be written? A) 2 B) -2 C) 3 D) 6 E) None of the above ( ) Need proof that you worked all the 40,000! my my my gsr you are a busy bee tonight! you must had a tough day at work and are intent on distributing some real punishment on the 2 other people on this board. 
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Director
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Re: Patience is a virtue! - PS [#permalink]
03 Nov 2005, 00:10
Titleist wrote: my my my gsr you are a busy bee tonight! you must had a tough day at work and are intent on distributing some real punishment on the 2 other people on this board. LOL!
Good Night! See ya tomorrow!
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VP
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E. 1
numbers total number of digits
1-9 9*1 = 9
10-99 90*2 = 180
100-999 900*3 = 2700
1000-9999 9000*4 = 36000
10000-99999 90000 * 5 = 450000
first 4 if written one after other will have : 9+180+2700+36000 = 38889 digits
rest of the 40000 - 38889 = 1111 digit comes from numbers 10000-99999 range
each of these numbers have 5 digits so we will need ceil(1111/5) = 223 numbers
last number: 99999 + 223 = 10222
40000 digit: 1
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SVP
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duttsit wrote: E. 1
numbers total number of digits
1-9 9*1 = 9
10-99 90*2 = 180
100-999 900*3 = 2700
1000-9999 9000*4 = 36000
10000-99999 90000 * 5 = 450000
first 4 if written one after other will have : 9+180+2700+36000 = 38889 digits
rest of the 40000 - 38889 = 1111 digit comes from numbers 10000-99999 range
each of these numbers have 5 digits so we will need ceil(1111/5) = 223 numbers
last number: 99999 + 223 = 10222
40000 digit: 1
again duttsit, you're too keen .....but , it's 9999 instead of 99999 
Still do I have an intuition that my teacher solved it a very different, short and sweet way ....I'll try to recall 
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Director
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I got it using brute force.. You guys correct me if I am wrong..
123456789123... -- the 10th digit is 1.. the 20th digit wil be 2.. the 30th digit will be 3.. and so on.. till we get to 100th digit which will again be 1..
40,000th digit is a multiple of 100.. 100+100+100 *400 times.. gets me 40,000.. so the 40,000th digit should be 1.
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Director
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Good job guys!
OA is E ('1')
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