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

Question

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

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

Bunuel · Accepted Answer

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

gmatophobia · Answer

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

YB10 · Answer

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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Oppenheimer1945 · Answer

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

Bunuel · Answer

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

Oppenheimer1945 · Answer

edited! That was for general possibilities without constraints of digits.

rey1009 · Answer

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

nivivacious · Answer

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

bumpbot · Answer

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

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