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How many integers k greater than 100 and less than 1000 are [#permalink]
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17 Sep 2010, 05:39
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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 reversed, the resulting integer is k + 99? A. 50 B. 60 C. 70 D. 80 E. 90
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Re: Jow meny intergers [#permalink]
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17 Sep 2010, 06:13
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 3digit integer. Any 3 digit integer can be expressed as \(100a+10b+c\). Now, we are told that \((100a+10b+c)+99=100c+10b+a\) > \(99c99a=99\) > \(ca=1\) > 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 3digit 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. Answer: D.
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Re: Jow meny intergers [#permalink]
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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]
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17 Sep 2010, 06:18
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]
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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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How many integers K greater that 100 and less than 1000 are [#permalink]
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15 May 2014, 13:45
How many integers K greater that 100 and less than 1000 are there such that if the hundreds and the units digits of K are reversed, the resulting integer is K+99?
a) 50 b) 60 c) 70 d) 80 e) 90



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Re: How many integers K greater that 100 and less than 1000 are [#permalink]
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15 May 2014, 18:22
Ans is D. The hundredth digit is always smaller from units digit by 1 as obtained by solving the equation 100a+10b+c+99=100c+10b+a. So 'a' can have values from 1 to 8 and for every value of a, 'b' varies from 0 to 9. So total integers are 8×10.
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Re: How many integers K greater that 100 and less than 1000 are [#permalink]
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15 May 2014, 18:22
tinku21rahu wrote: How many integers K greater that 100 and less than 1000 are there such that if the hundreds and the units digits of K are reversed, the resulting integer is K+99?
a) 50 b) 60 c) 70 d) 80 e) 90 I think the answer is D  80 integers. My logic is as follows: Since we are adding 99, the hundreds digit will need to be less than the ones digit if the corresponding number when the two digits flipped is to be larger. Since the change in magnitude is just shy of 100, the different between the two can only be one. Look at an example case of 100 < X < 200  102 > 201
 112 > 211
 ...
 192 > 291
What about for 800 < X < 900?  809 > 908
 819 > 918
 ...
 899 > 998
For each of these samples, there are 10 numbers between the two stated bounds (the tens digit ranges from 0 to 9). Since we can't do this for 900 < X < 1,000, there are 8 sets of 10, or 80 total integers. Hopefully that makes sense (and correct!). Let me know if I'm missing something here! EDIT: Dang! Looks like I was beat to the punch...



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Re: How many integers K greater that 100 and less than 1000 are [#permalink]
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16 May 2014, 08:15
GordonFreeman: What you said said is 100% correct and that was the way i solved it, but that took me lot of time, Do you feel there is any easier was like PuneetSood Said.
PuneetSood: What you said seems to be an Easy Way. Thank You!



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Re: How many integers K greater that 100 and less than 1000 are [#permalink]
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16 May 2014, 08:43
tinku21rahu wrote: GordonFreeman: What you said said is 100% correct and that was the way i solved it, but that took me lot of time, Do you feel there is any easier was like PuneetSood Said.
PuneetSood: What you said seems to be an Easy Way. Thank You! PuneetSood's method is definitely slicker than mine. I only started my GMAT journey this week so when I approach problems like this I like to see all the moving parts. After I am more comfortable with a problem type I would opt for an approach like PuneetSood's.



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Re: How many integers K greater that 100 and less than 1000 are [#permalink]
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16 May 2014, 19:13
tinku21rahu wrote: GordonFreeman: What you said said is 100% correct and that was the way i solved it, but that took me lot of time, Do you feel there is any easier was like PuneetSood Said.
PuneetSood: What you said seems to be an Easy Way. Thank You! Really appreciate, GordonFreeman



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Re: How many integers K greater that 100 and less than 1000 are [#permalink]
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25 May 2014, 09:58
great approach, this is exactly the kind of help I was looking to find when I joined the gmat club community, thank you!



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Re: How many integers K greater that 100 and less than 1000 are [#permalink]
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26 May 2014, 21:17
Let a=Hundreds Digit b=Tens Digit c=Units Digit
Since we know that the hundreds digit and units digit (when reversed) is k+99,
100a+10b+c=a+10b+10c+99 99a99c=99 99(ac)=99 ac=1
From here, we can assume that the hundreds digit and units digit must be 1 digit apart in order to satisfy this condition.
Only 9 options are possible: 9x8,8x7,7x6,6x5,5x4,4x3,3x2,2x1,1x0
We are also told in the question stem that we are restricted among 3 digits only which eliminates the 1x0 option. Now, there are only 8 possible options left. Since there is no restriction in the tens digits,
8 x 10 (tens digit can be from 09) = 80



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Re: How many integers K greater that 100 and less than 1000 are [#permalink]
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27 May 2014, 00:59



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Re: How many integers k greater than 100 and less than 1000 are [#permalink]
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27 May 2014, 17:04
Let the 3 digit number be xyz If the units and hundred's digit is reversed then the number becomes zyx
Given: 100z + 10y + x = 100x + 10y + z + 99
100z + 10y + x = 100x + 10y + z + 99
or, 99z  99x = 99 or, 99(zx)=99 or, z=x+1
which means z can take only 1 value i.e 1 more than x
Now in the original number xyz, y can take 10 values 0 through 9 x can take 18(8 values) and for each x, z can take exactly 1 value (from 2 through 9)
so 8 (values for x and z) X 10 (values of y) = 80. Hence D



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Re: How many integers k greater than 100 and less than 1000 are [#permalink]
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25 Aug 2017, 11:24
Answer is Dsee the pic..
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