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Nesterboy [21]
2 years ago
15

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

Mathematics
2 answers:
slamgirl [31]2 years ago
7 0

Step-by-step explanation:

Let the two-digit number is xy

<u>This can be written as:</u>

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

<u>The difference between the number and product of its digits is:</u>

  • d = 10x + y - xy

<u>Rewrite this as below:</u>

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

     x(10 - y) - (10 - y) + 10 =

     (x - 1)(10 - y) + 10

<u>We see that:</u>

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

<u>The value of d is then:</u>

  • 0 + 10 ≤ d ≤ 8*10 + 10
  • 10 ≤ d ≤ 90
<h3>Proved</h3>
Andre45 [30]2 years ago
5 0

Solution:

<em>Let's take the </em><em>greatest two-digit number</em><em> as an example.</em>

  • => 99

<u>Proving...</u>

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

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

Hence proved. ✔✔

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