KVPY SA Exam 2014 Question Paper

Question : 4 of 80

Marks: +1, -0

Let x and y be two 2-digit numbers such that y is obtained by reversing the digits of x. Suppose they also satisfy

x ^2-y ^2 = m ^2

for some positive integer m. The value of

x + y + m

Solution:

x = ab = 10a + b

y = ba = 10b + a

(1 ≤ a, b ≤ 9)

x^ 2 - y^ 2 = m^ 2

100a^2 + b^2 + 20ab - (100b^2 + a^2 + 20ab)

= m^2

99(a^2 - b^2 ) = m^2

9 × 11 × (a - b) (a + b) = m^2

m is integer so
a = 6 {Note: a – b ≠ 9 (think why ?)}
b = 5

m^2 = 9 × 11 × 1 × 11

⇒ m = 3 × 11

x + y + m = 10a + b + 10b

+ a + 33

= 11(a + b) + 33 = 11(11) + 33

= 121 + 33 = 154

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