|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
firas92
Current Student
Joined: 16 Jan 2019
Last visit: 02 Dec 2024
Posts: 616
Given Kudos: 142
Location: India
Concentration: General Management
Schools: Tepper '23 (M$) Foster '23 (D)
GMAT 1: 740 Q50 V40
WE:Sales (Other)
Products:
19 Aug 2020, 07:01
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
53% (03:10) correct
47%
(03:02)
wrong
based on 87
sessions
History
Date
Time
Result
Not Attempted Yet
Let \(a\) and \(b\) be two 2 digit numbers such that \(b\) is obtained by reversing the digits of \(a\). It is also known that they satisfy the relation \(a^2-b^2=k^2\) where \(k\) is some positive integer. What is the value of \(a+b+k\)?
A. 128
B. 154
C. 186
D. 220
E. 256
A. 128
B. 154
C. 186
D. 220
E. 256
19 Aug 2020, 10:14
Kudos
Bookmarks
Since a and b are 2 digit numbers and b is the reverse of a
Let a = 10x + y and b = 10y + x
\(a^2 - b^2 = k^2\)
\((10x + y)^2 - (10y - x)^2 = k^2\)
Expanding: \(100x^2 + 20xy + y^2 - 100y^2 - 20xy - y^2 = k^2\)
\(99x^2 - 99y^2 = k^2\)
\(99[x^2 - y^2] = k^2\)
99 [x + y][x - y] = \(k^2\)
Since k is an integer \(k^2\) is a perfect square. We get 2 cases here.
Case 1: \(99 * 99 = k^2\). Therefore [x + y][x - y] = 99, which gives us that x + y = 11 and x - y = 9
The values that satisfies this would be x = 10 and y = 1. This would not satisfy \(a\) and \(b\).
Case 2: 99 can be written as \(11 * 3^2\), which means to make the expression into a perfect square, [x + y][x - y] = 11
This gives us x + y = 11 and x - y = 1
The values that satisfies these equations would be x = 6 and y = 5. These satisfy satisfy \(a\) and \(b\) (65 and 56).
So k = 11 * 3 = 33
a + b + k = 10x + y + 10y + x + k = 11(x + y) + 33 = 11(11) + 33 = 154
Option B
Arun Kumar
Let a = 10x + y and b = 10y + x
\(a^2 - b^2 = k^2\)
\((10x + y)^2 - (10y - x)^2 = k^2\)
Expanding: \(100x^2 + 20xy + y^2 - 100y^2 - 20xy - y^2 = k^2\)
\(99x^2 - 99y^2 = k^2\)
\(99[x^2 - y^2] = k^2\)
99 [x + y][x - y] = \(k^2\)
Since k is an integer \(k^2\) is a perfect square. We get 2 cases here.
Case 1: \(99 * 99 = k^2\). Therefore [x + y][x - y] = 99, which gives us that x + y = 11 and x - y = 9
The values that satisfies this would be x = 10 and y = 1. This would not satisfy \(a\) and \(b\).
Case 2: 99 can be written as \(11 * 3^2\), which means to make the expression into a perfect square, [x + y][x - y] = 11
This gives us x + y = 11 and x - y = 1
The values that satisfies these equations would be x = 6 and y = 5. These satisfy satisfy \(a\) and \(b\) (65 and 56).
So k = 11 * 3 = 33
a + b + k = 10x + y + 10y + x + k = 11(x + y) + 33 = 11(11) + 33 = 154
Option B
Arun Kumar
19 Aug 2020, 10:25
Kudos
Bookmarks
Given: Let \(a\) and \(b\) be two 2 digit numbers such that \(b\) is obtained by reversing the digits of \(a\). It is also known that they satisfy the relation \(a^2-b^2=k^2\) where \(k\) is some positive integer.
Asked: What is the value of \(a+b+k\)?
Let a = 10x + y
b = 10y + x
a^2 - b^2 = (10x+y)^2 - (10y+x)^2 = 11(x+y)9(x-y) = 99(x^2 - y^2) = k^2
x^2 - y^2 = 11m^2 = {11,44,99,...}
If x^2 - y^2 = 11; k = 33; x+ y = 11; x - y = 1; x = 6; y=5
a = 65; b=56; a^2 - b^2 = 65^2 - 56^2 = 121*9 = 11^2*3^2 = 33^2; k=33
a + b + k = 65 + 56 + 33 = 154
IMO B
Asked: What is the value of \(a+b+k\)?
Let a = 10x + y
b = 10y + x
a^2 - b^2 = (10x+y)^2 - (10y+x)^2 = 11(x+y)9(x-y) = 99(x^2 - y^2) = k^2
x^2 - y^2 = 11m^2 = {11,44,99,...}
If x^2 - y^2 = 11; k = 33; x+ y = 11; x - y = 1; x = 6; y=5
a = 65; b=56; a^2 - b^2 = 65^2 - 56^2 = 121*9 = 11^2*3^2 = 33^2; k=33
a + b + k = 65 + 56 + 33 = 154
IMO B
