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Jow meny intergers

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Jow meny intergers [#permalink]  17 Sep 2010, 05:39
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Question Stats:

40% (06:30) correct 60% (00:00) wrong based on 3 sessions
How many integers k greater than 100 and less than 1000 are there such that if the hundreds and the unit digits of k are reverced, the resulting integer is k + 99?

50
60
70
80
90

please provide explaination .. with minimal calculations
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Re: Jow meny intergers [#permalink]  17 Sep 2010, 06:13
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vanidhar wrote:
How many integers k greater than 100 and less than 1000 are there such that if the hundreds and the unit digits of k are reverced, the resulting integer is k + 99?

50
60
70
80
90

please provide explaination .. with minimal calculations

Some wording. Anyway:

k is 3-digit integer. Any 3 digit integer can be expressed as 100a+10b+c.

Now, we are told that (100a+10b+c)+99=100c+10b+a --> 99c-99a=99 --> c-a=1 --> there are 8 pairs of c and a possible such that their difference to be 1: (9,8), (8,7), ..., and (2,1) (note that a can not be 0 as k is 3-digit integer and its hundreds digit, a, is more than or equal to 1).

Also, b, tens digit of k, could tale 10 values (0, 1, ..., 9) so there are total of 8*10=80 such numbers possible.

Example:
102 - 201;
112 - 211;
122 - 221;
...
899 - 998.

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Re: Jow meny intergers [#permalink]  17 Sep 2010, 06:16
vanidhar wrote:
How many integers k greater than 100 and less than 1000 are there such that if the hundreds and the unit digits of k are reverced, the resulting integer is k + 99?

50
60
70
80
90

please provide explaination .. with minimal calculations

Not sure if this is the shortest.. But this is how I did this

There are 8 sets of integers with hundreds and units digits exchanged that satisfies k + 99.
1. 102 | 201 (satisfies k+99, where k = 102)
2. 203 | 302 (satisfies k+99, where k = 203)
3. ...
4. ...
5. ...
6. ...
7. 708 | 807
8. 809 | 908

Each set has 10 such numbers.
1. 102 | 201 (still k+99 holds good)
2. 112 | 211
3. 122 | 221
4. 132 | 231
5. ...
6. ...
7. ...
8. ...
9. 182 | 281
10. 192 | 291

Therefore, 8 sets with 10 such number in each set will give 8 x 10 = 80 integers.
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Re: Jow meny intergers [#permalink]  17 Sep 2010, 06:18
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When you understand it right it's simple. But you need to find an example :

Example1: 102 -> 201 and 102 + 99 = 201.
It will be okay for 102, 112, 122, 132, ..., 192 -> 10 times

So now we have that for the "100", "200", the "300", the "400", ... the "800" (8 times)

So 10 * 8 = 80

ANS : D.

Hope it's clear
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Re: Jow meny intergers [#permalink]  17 Sep 2010, 19:29
u_h - h_u = 99

So reversing digits should be 1 digit different

2_1 and 1_2 (1132 and 1231)
3_2
4_3
5_4
6_5
7_6
8_7
9_8

So 10 * 8 = 80
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Re: Jow meny intergers   [#permalink] 17 Sep 2010, 19:29
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