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# How many times will the digit 7 be written when listing the

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Manager
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How many times will the digit 7 be written when listing the [#permalink]

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26 Apr 2008, 06:18
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

How many times will the digit 7 be written when listing the integers from 1 to 1000?

The solution offered explains that:

For any two-digit number the 7 could be the first digit. In that case, there are 10 ways to place the second digit, i.e. 0-9. In case, 7 is the unit's digit there are 9 ways to place the first digit. Since, 0 cannot take the tens digit. Thus, 10+9 = 19 numbers that will have 7.

Now here while explaing, GMAT club forgets that 77 has been counted twice in this calculation. i.e. the solution says that when 7 is the unit digit there are 9 ways to place the first digit. But according to me when 7 is the unit digit there are 8 ways to place the first digit since 77 has already been accounted while calculating when 7 was the first digit and there are 10 ways to place the second digit.

If we count the no. of two digit numbers containing 7 , we find that there are 18 such numbers and not 19 as expalined in the solution.
{17, 27, 37, 47, 57, 67, 70 to 79, 87, 97}= 18 two digit numbers containing 7.

Similary mistake seem to have been done in calculating 3 digit combination of 7.

Can some body explain if i am on the right track or the answer offered is faulty?
_________________

Neelabh Mahesh

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Re: Test m01, question 10; Faulty solution [#permalink]

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26 Apr 2008, 09:12
I think from the question it says how many times 7 is written.. 7 is written twice in 77
for two digited numbers it 17 27 37 47 57 67 77 87 97seven is written in units place
70 71 72 73 74 75 76 77 78 79 i.e total is 19...

I hope this explains the solution
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Re: Test m01, question 10; Faulty solution [#permalink]

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26 Apr 2008, 22:50
neelabhmahesh wrote:
How many times will the digit 7 be written when listing the integers from 1 to 1000?

The solution offered explains that:

For any two-digit number the 7 could be the first digit. In that case, there are 10 ways to place the second digit, i.e. 0-9. In case, 7 is the unit's digit there are 9 ways to place the first digit. Since, 0 cannot take the tens digit. Thus, 10+9 = 19 numbers that will have 7.

Now here while explaing, GMAT club forgets that 77 has been counted twice in this calculation. i.e. the solution says that when 7 is the unit digit there are 9 ways to place the first digit. But according to me when 7 is the unit digit there are 8 ways to place the first digit since 77 has already been accounted while calculating when 7 was the first digit and there are 10 ways to place the second digit.

If we count the no. of two digit numbers containing 7 , we find that there are 18 such numbers and not 19 as expalined in the solution.
{17, 27, 37, 47, 57, 67, 70 to 79, 87, 97}= 18 two digit numbers containing 7.

Similary mistake seem to have been done in calculating 3 digit combination of 7.

Can some body explain if i am on the right track or the answer offered is faulty?

unit digit = 10x10 = 100
tens digit = 10 x 10 =100
hundred = 100 x 1 = 100
total = 300
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Re: Test m01, question 10; Faulty solution [#permalink]

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27 Apr 2008, 11:45
i got same answer as fistail but had different approach. (mine isn't as elegant as fistail's approach)
this is how i did.
1-99
100-199
...
...
700-799
..
900-1000
there are 20 7's in each hundreds except 700's. so 20*9=180.
for 700-799, there is 7 in every number(100). also there are 20 7's like other hundreds (20) so, 100+20=120.
adding them together, i got 300.
Re: Test m01, question 10; Faulty solution   [#permalink] 27 Apr 2008, 11:45
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