Question

A number of two digits is increased by 54 when the digits are reversed. The ten digits is three times the unit digit. Find the number​

Answer 1

Let n be the unknown number. We can write it as

n = 10a + b

with a and b integers between 1 and 9 (either with positive or negative sign).

Reversing the digits gives another number

m = 10b + a

The first number is increased by 54 when the digits are reversed, which means

m = n + 54   →   10b + a = 10a + b + 54   →   9b - 9a = 54   →   b - a = 6

The digit in the tens place of n is 3 times the digit in the ones place, so

a = 3b

Substitute this into the previous equation and solve for b :

b - a = b - 3b = -2b = 6   →   b = -3

Solve for a :

a = 3b = 3(-3) = -9

Then the original number is n = 10a + b = 10(-9) + (-3) = -93

Answer 2

ANSWER:

Let the digit in the unit's place be x and the digit in the ten's place be y. Then,

Number = 10y + x

According to the given condition, we have

y = 3x ...[i]

Number obtained by reversing the digits = 10x + y

If the number is decreased by 54, the digits are reversed.

∴ Number − 54 = Number obtained by reversing the digits.

⇒10y + x − 54 = 10x + y

⇒9x − 9y = − 54

⇒x − y = − 6 ...[ii]

Putting y = 3x in equation [ii], we get

x − 3x = − 6

⇒x = 3.

Putting x = 3 in y = 3x, we get y = 9.

Hence, number = 10y + x = 10 × 9 + 3 = 93.

Reversible number = 10x + y = 10 × 3 + 9 = 39.

Hence, the numbers are 39 & 93.