A two – digit number decreased by 27 when its digits are reversed and difference between square of digit at unit place and square of the digit at tens place of the original number is same as difference between original number and the number obtained after reversing of digits. Find the original number?

Explanation

Let the digit at unit place & tens place of original number be y and x respectively. So, original number = 10 × 𝑥+ 𝑦 And, number obtained after reversing of digits = 10 × 𝑦+ 𝑥 ATQ, (10 × 𝑥+ 𝑦) −(10 × 𝑦+ 𝑥) = 27 9𝑥−9𝑦= 27 𝑥−𝑦= 3 …(i) Now, 𝑥2 −𝑦2 = 27 (𝑥+ 𝑦)(𝑥−𝑦) = 27 …(ii) On solving (i) & (ii), we get: 𝑥+ 𝑦= 9 …(iii) On solving (i) & (iii), we get: 𝑥= 6, 𝑦= 3 So, original number = 10 × 𝑥+ 𝑦 = 63