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# M01 #10

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Manager
Status: livin on a prayer!!
Joined: 12 May 2011
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Re: M01 #10 [#permalink]

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04 Nov 2011, 19:56
Answer is 300, D.

Units Digit: 007, 017....097 = 10
Tens Digit: 070, 071...079 = 10

So there are 10 + 10 = 20 instances of 7 in each hundred numbers. So in 1000 numbers, there are 10 x 100 numbers = 20 x 10 = 200 instances of 7 in 1000 for units and tens digits.

Hundreds digit: 700, 701...799 = 100 instances of 7 for the hundreds digit.

So from 1-1000, there are 200 + 100 = 300 instances of 7.

Hope that helps!!!
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Re: M01 #10 [#permalink]

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18 Nov 2011, 02:42
single digit number 7 will appear once.
2 digit number 7 will appear = 9 time at unit digit and 10 times at 10th digit. = 19 times
3 digit number 7 will appear = 100 times for 100th digit + 90 times fo 10th times +90 times for 90 times for unit digit = total 280

so total number of times 7 appears between 1 to 1000 = 280+19+1 = 300
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Re: M01 #10 [#permalink]

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14 Jan 2012, 21:57
IMO - D
Best way to solve this .

100's place = 1x10x10 = 100
10's place = 10x1x10 = 100
1's place = 10x10x1 = 100
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18 Jul 2014, 11:00
Surely the easiest way to resolve this is to use the fact that the 7 appears one tenth of the time. So out of the thousand numbers, it appears in the 100's slot 1/10(1000)=100 times, in the 10's slot 1/10(1000)=100 times, and in the units slot 1/10(1000)=100 times. Is my thinking incorrect?
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Re: M01 #10 [#permalink]

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18 Jul 2014, 12:25
Chakolate wrote:
Surely the easiest way to resolve this is to use the fact that the 7 appears one tenth of the time. So out of the thousand numbers, it appears in the 100's slot 1/10(1000)=100 times, in the 10's slot 1/10(1000)=100 times, and in the units slot 1/10(1000)=100 times. Is my thinking incorrect?

Yes. Check here: m01-70917.html#p826207
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Re: M01 #10   [#permalink] 18 Jul 2014, 12:25

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# M01 #10

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