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

User Aghd
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2 Answers

3 votes

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. ✔✔

User Vikzilla
by
8.1k points
12 votes

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

User Wellie
by
9.0k points

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