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How many integers n greater than 10 and less than 100 a

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How many integers n greater than 10 and less than 100 a  [#permalink]

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New post 23 Nov 2018, 08:41
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Question Stats:

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How many integers n greater than 10 and less than 100 are such that, if the digits of n are reversed, the resulting integer is n+9 ?

(A) 5
(B) 6
(C) 7
(D) 8
(E) 9

Project PS Butler : Question #36


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Re: How many integers n greater than 10 and less than 100 a  [#permalink]

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New post 23 Nov 2018, 08:55
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HKD1710 wrote:
How many integers n greater than 10 and less than 100 are such that, if the digits of n are reversed, the resulting integer is n+9 ?

(A) 5
(B) 6
(C) 7
(D) 8
(E) 9

Project PS Butler : Question #36


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Let n=ab(as it is 2 digit number bcz >10 and less than 100)
so reverse will be =ba

now as per qsn
reverse-orgn=9--------(1)

we know ab can be written as 10*a+b(54=10*5+4)
ba=10*b+a

now substituting in 1

10b+a-10a-b=9
9b-9a=9
b-a=1
so this condition will hold true only if diff of two digits is 1

so from 11 to 20
this is true only for 12(2-1=1)

then from 21-30 true for 23(3-2=1) make sure 21 is not valid as(1-2=-1 not 1) b-a=1
so this way
each set of 10 numbers has 1 number for which b-a=1
total we have 9 such sets (11-20,21-30,---91-99)
so 8 such numbers bcz in 91-99 we cant have this condition

i.e 12,23,34,45,56,67,78,89,
so 8 such numbers
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How many integers n greater than 10 and less than 100 a  [#permalink]

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New post 23 Nov 2018, 11:51
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Let A be tens digit & B be units digit

\(X = 10A + B\)

After reversing digits, \(X _{new}\) \(= 10B + A\)

Difference between \(X\) and \(X_{new}\) is \(9\), so I get following equation

\((10A + B) - (10B + A) = 9\)

\(9A – 9B = 9\)

\(9(A-B) = 9\)

\(A-B = 1\)

\(A = B+1\)

Since units digit X is +1 more than tens digit, I get following numbers

12, 23, 34, 45, 56, 67, 78, 89

IMO: D :)
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Re: How many integers n greater than 10 and less than 100 a  [#permalink]

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New post 09 Feb 2019, 20:44
HKD1710 wrote:
How many integers n greater than 10 and less than 100 are such that, if the digits of n are reversed, the resulting integer is n+9 ?

(A) 5
(B) 6
(C) 7
(D) 8
(E) 9



First of all i didn't understand the question and when i did understood it, lot of time had passed, so marked it incorrect.

Lets see, 10<n<100, digits of n are reversed, resulting number will be n+9

10x +y = 10y+x +9
9 (x-y) = 9
x-y = 1

This means that the difference between the tens digit and units digit is 1
I messed up where ??, i had to consider these as well
12
23
34
45
56
67
78
89

Now if i took 98 it wont give 108 after reversing, so out
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Re: How many integers n greater than 10 and less than 100 a  [#permalink]

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New post 09 Feb 2019, 21:26
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The two digit number (assume) = 10a+b
then by the given condition
10a+b = 10b+a+9 (digits are reversed and 9 added)
=> 9a - 9b = 9
=> a - b =1

we have following number between 10 and 100 which satisfy above condition
12, 23 , 34, 45, 56,67,78,89
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How many integers n greater than 10 and less than 100 a  [#permalink]

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New post Updated on: 10 Feb 2019, 11:39
HKD1710 wrote:
How many integers n greater than 10 and less than 100 are such that, if the digits of n are reversed, the resulting integer is n+9 ?

(A) 5
(B) 6
(C) 7
(D) 8
(E) 9


note pattern: 12/21, 23/32, 34/43...
12+11(n-1)<100
11n<99
n<9
n=8
D

Originally posted by gracie on 09 Feb 2019, 21:39.
Last edited by gracie on 10 Feb 2019, 11:39, edited 2 times in total.
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How many integers n greater than 10 and less than 100 a  [#permalink]

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New post 09 Feb 2019, 23:46
10A+B=n
10B+A=n+9

10B+A=10A+B+9
9B-9A=9
B-A=1
Integers between 11 and 99, that have unit digit 1 greater than tenths digit-12, 23, 34, 45 ..... = total 8

Ans - D
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How many integers n greater than 10 and less than 100 a   [#permalink] 09 Feb 2019, 23:46
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