A two-digit positive number is such that the product of its digits is 24. When 18 is added to the number, the digits interchange their places. Which smallest positive number should be subtracted from the given number to make it a perfect square?
A two-digit positive number is such that the product of its digits is 24. When 18 is added to the number, the digits interchange their places. Which smallest positive number should be subtracted from the given number to make it a perfect square? Correct Answer 10
Given:
A two-digit positive number is such that the product of its digits is 24.
When 18 is added to the number, the digits interchange their places.
Concept used:
(A + B)2 = (A - B)2 + 4AB
Calculation:
Let the unit and ten's place digits of the number be P and Q respectively.
The number = 10 × Q + P = 10Q + P
When the digits exchange = 10 × P + Q =10P + Q
According to the question,
PQ = 24 ....(1)
10Q + P + 18 = 10P + Q
⇒ 9P - 9Q = 18
⇒ P - Q = 18/9
⇒ P - Q = 2 ...(2)
⇒ (P - Q)2 = 22
⇒ (P - Q)2 + 4PQ = 22 + 4PQ
⇒ (P + Q)2= 22 + 4 × 24 (From 1)
⇒ (P + Q)2 = 100
⇒ (P + Q) = 10 ....(3)
Now solving (2) and (3),
P = 6, Q = 4
So, the number = 10 × 4 + 6 = 46
Closest perfect square integer less than 46 = 36
Now, the smallest positive number that should be subtracted from 46 to make it a perfect square
⇒ 46 - 36
⇒ 10
∴ 10 should be subtracted from the given number to make it a perfect square.