Question

the sum of a two digit number and the number obtained by reversing the digits is 44. if the digits of the number differ by 1, find the number. How many such number are there?​

Answer 1

◑ Given ◐

The sum of a two digit number and the number obtained by reversing the digits is 44. if the digits of the number differ by 2

◑ To find ◐

• Required numbers

◑ Solution ◐

• First condition

The sum of a two digit number and the number obtained by reversing the digits is 44.

✝ Let the tens digit be x and ones digit be y

• Original number = 10x + y
• Reversed number = 10y + x

✝ (Original number) + (Reversed number) = 44

→ 10x + y + 10y + x = 44

→ 11x + 11y = 44

→ 11(x + y) = 44

→ x + y = 4 -------(i)

• Second condition

The digits of the number differ by 2

→ x - y = 2 -----(ii)

Add both the equations

→ x + y + x - y = 4 + 2

→ 2x = 6

→ x = 6/2

→ x = 3

Substitute the value of x in eqⁿ (ii)

→ x - y = 2

→ 3 - y = 2

→ y = 3 - 2

→ y = 1

•°• Original number = 10x + y = 31

•°• Reversed number = 10y + x = 13

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Answer 2

Let’s assume the digit at unit’s place as x and ten’s place as y. Thus from the question, the number needed to be found is 10y + x.

From the question it’s told as, the two digits of the number are differing by 2. Thus, we can write

x – y = ±2………….. (i)

Now after reversing the order of the digits, the number becomes 10x + y.

Again from the question, it’s given that, the sum of the numbers obtained by reversing the digits and the original number is 66. Thus, this can be written as;

(10x+ y) + (10y+x) = 66

⇒ 10x + y + 10y + x = 66

⇒ 11x +11y = 66

⇒ 11(x + y) = 66

⇒ x + y = 66/11

⇒ x + y = 6………….. (ii)

Now, we have two sets of systems of simultaneous equations

x – y = 2 and x + y = 6

x – y = -2 and x + y = 6

Let’s first solve the first set of system of equations;

x – y = 2 …………. (iii)

x + y = 6 ………….. (iv)

On adding the equations (iii) and (iv), we get;

(x – y) + (x + y) = 2+6

⇒ x – y + x + y = 8

⇒ 2x =8

⇒ x = 8/2

⇒ x = 4

Putting the value of x in equation (iii), we get

4 – y = 2

⇒ y = 4 – 2

⇒ y = 2

Hence, the required number is 10 × 2 +4 = 24

Now, let’s solve the second set of system of equations,

x – y = -2 …………. (v)

x + y = 6 ………….. (vi)

On adding the equations (v) and (vi), we get

(x – y)+(x + y )= -2 + 6

⇒ x – y + x + y = 4

⇒ 2x = 4

⇒ x = 4/2

⇒ x = 2

Putting the value of x in equation 5, we get;

2 – y = -2

⇒ y = 2+2

⇒ y = 4

Hence, the required number is 10×4+ 2 = 42

Therefore, there are two such possible numbers i.e, 24 and 42.