Question

Prove that the difference between a two-digit number and the product of its digits will always be a 2 digit number.

Answer 1

Step-by-step explanation:

Let the two-digit number is $xy$

This can be written as:

• 10x + y, where 1 ≤ x ≤ 9 and 0 ≤ y ≤ 9

The difference between the number and product of its digits is:

• d = 10x + y - xy

Rewrite this as below:

d = 10x - xy + y - 10 + 10 =

     x(10 - y) - (10 - y) + 10 =

     (x - 1)(10 - y) + 10

We see that:

• 0 ≤ x - 1 ≤ 8 according to the condition given above
• 1 ≤ 10 - y ≤ 10 again according to the condition given above

The value of d is then:

• 0 + 10 ≤ d ≤ 8*10 + 10
• 10 ≤ d ≤ 90

Proved

Answer 2

Solution:

Let's take the greatest two-digit number as an example.

• => 99

Proving...

• 99 - (9 x 9) = Two-digit number
• => 99 - 81 = Two-digit number
• => 18 = Two-digit number

Since the difference between the greatest two-digit number and the product of its digits is a two-digit number, the difference between the other two-digit numbers '10, 11, 12.... 97, 98' and the product of its digits will always be a two-digit number, as the product of its digit is less than the two-digit number.

Hence proved. ✔✔