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For each positive integer k, f(k) is defined to be the number created

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Bunuel
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For each positive integer k, f(k) is defined to be the number created [#permalink] For each positive integer k, f(k) is defined to be the number created   01 Nov 2023, 06:29
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For each positive integer k, f(k) is defined to be the number created by reversing the digits of k. For instance, f(42) = 24. How many positive two-digit integers k are there where k + f(k) is the square of a positive integer?

A. 6
B. 7
C. 8
D. 9
E. 10

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For each positive integer k, f(k) is defined to be the number created [#permalink] For each positive integer k, f(k) is defined to be the number created   01 Nov 2023, 07:07
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Bunuel wrote:
For each positive integer k, f(k) is defined to be the number created by reversing the digits of k. For instance, f(42) = 24. How many positive two-digit integers k are there where k + f(k) is the square of a positive integer?

A. 6
B. 7
C. 8
D. 9
E. 10


Let's assume that the two-digit number \(k = xy\).

We can represent the number \(xy\) as \(10x + y\). For example, \(29\) can be represented as \(2*10 + 9\)

\(k = 10x + y\)

Therefore, \(f(k) = yx\) ⇒ Similar to above, \(yx\) = \(10y + x\)

\(f(k) = 10y + x\)

Given : \(k + f(k)\) is a perfect square

\(k + f(x) = 10x + y + 10y + x = 11x + 11y = 11(x+y)\)

As \(11(x+y)\) is a perfect square, this is only possible when \(x + y = 11\)

Possible pairs \((x,y) = (2,9), (3,8), (4,7), (5,6), (6,5), (7,4), (8,3), (9,2)\)

Hence \(8\) pairs result into the sum \(x+y = 11\).

Therefore, we have \(8\) values of \(k\).

Option C
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Re: For each positive integer k, f(k) is defined to be the number created [#permalink] For each positive integer k, f(k) is defined to be the number created   01 Nov 2023, 07:11
Let K = 10a+b
f(k)=10b+a

K+f(k)=10(a+b)+(a+b)
i.e, (a+b)11

For it to be perfect square a+b needs to be 11, keeping in mind a and b are single digits

The combinations possible are (2,9)(3,8)(4,7)(5,6)(6,5)(7,4)(8,3)(9,2) = 8

Answer is C imo

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For each positive integer k, f(k) is defined to be the number created [#permalink] For each positive integer k, f(k) is defined to be the number created   Updated on: 01 Nov 2023, 09:30
\(10a+b+10b+a=11(a+b)=n^2\\
11*11\\
11*44\\
11*99\\
..\\
..\\
\)
since \(0<=a,b<=9, a+b=11\)

\(\\
ab=\\
{29,38,47,56} & {92,83,74,65}\\
\)

8 values

Originally posted by Oppenheimer1945 on 01 Nov 2023, 07:40.
Last edited by Oppenheimer1945 on 01 Nov 2023, 09:30, edited 1 time in total.
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Re: For each positive integer k, f(k) is defined to be the number created [#permalink] For each positive integer k, f(k) is defined to be the number created   01 Nov 2023, 07:46
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samagra21 wrote:
\(10a+b+10b+a=11(a+b)=n^2\\
11*11\\
11*44\\
11*99\\
\\
a+b=11,44,99\\
{29,38,47,56} & {92,83,74,65}\\
\)

8 values


a and b are digits, so how can a + b be 44 or 99?
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Re: For each positive integer k, f(k) is defined to be the number created [#permalink] For each positive integer k, f(k) is defined to be the number created   01 Nov 2023, 09:31
Bunuel wrote:
samagra21 wrote:
\(10a+b+10b+a=11(a+b)=n^2\\
11*11\\
11*44\\
11*99\\
\\
a+b=11,44,99\\
{29,38,47,56} & {92,83,74,65}\\
\)

8 values


a and b are digits, so how can a + b be 44 or 99?


edited! That was for general possibilities without constraints of digits.
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Re: For each positive integer k, f(k) is defined to be the number created [#permalink] For each positive integer k, f(k) is defined to be the number created   01 Nov 2023, 09:52
Let the two digits of k be x and y
k= 10x+y
f(k) = 10y+x
sum = 11(x+y)
it can only be a perfect sqaure if (x+y) =11
(2,9)(9,2)
(3,8)(8,3)
(4,7)(7,4)
(5,6)(6,5)
therefore, 8
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For each positive integer k, f(k) is defined to be the number created [#permalink] For each positive integer k, f(k) is defined to be the number created   23 Jan 2024, 10:24
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For each positive integer \(k\), \(f(k)\) is defined to be the number created by reversing the digits of \(k\). For instance, \(f(42) = 24\). How many positive two-digit integers k are there where \(k + f(k)\) is the square of a positive integer?

A. 6
B. 7
C. 8
D. 9
E. 10


Let's represent a two-digit number \(k\) as \(10a + b\), where \(a\) is the tens digit and \(b\) is the units digit. Then \(f(k)\) becomes \(10b + a\), and \(k + f(k) = (10a + b) + (10b + a) = 11(a + b)\).

For \(11(a + b)\) to be the square of an integer, \((a + b)\) must be 11, making \(k + f(k) \) equal to \(11^2\).

\(a + b = 11\) is possible for the following 8 sets of \((a, b)\): (2, 9), (3, 8), (4, 7), (5, 6), (6, 5), (7, 4), (8, 3), and (9, 2).

Therefore, there are a total of 8 two-digit integers \(k\) where \(k + f(k)\) is the square of an integer.


Answer: C

Similar but harder question is here: https://gmatclub.com/forum/12-days-of-c ... 22993.html
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Re: For each positive integer k, f(k) is defined to be the number created [#permalink] For each positive integer k, f(k) is defined to be the number created   10 Feb 2024, 03:01
Bunuel wrote:
For each positive integer k, f(k) is defined to be the number created by reversing the digits of k. For instance, f(42) = 24. How many positive two-digit integers k are there where k + f(k) is the square of a positive integer?

A. 6
B. 7
C. 8
D. 9
E. 10

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­We're told if k = 10a+b then f(k) = 10b+a
We need to find how many two digit integers k can be.
x^2 = k+f(k)= 10a+b+10b+a
=11a+11b
=11(a+b)
Therefore x^2 = 11(a+b)
They can only be (6,5) (7,4) (8,3) (9,2) in both orders
Hence 8
Hence C
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Re: For each positive integer k, f(k) is defined to be the number created [#permalink]
10 Feb 2024, 03:01
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