Division by 9
(Note: The first part of this article was taken from page 37 of our book 25 Math Short Cuts which can be purchased through any of our MATHInic affiliates)
Most of us want to avoid 9 in almost all calculations. But we can make calculations easier by thinking of 9 as the difference of 10 and 1, (10 – 1). This fact is particularly useful in division by 9. Every 10 contains a 9 and a remainder of 1. So every multiple of 10 less than 90 will have a quotient and remainder equal to its tens digit. So 20 ÷ 9 = 2 r. 2 40 ÷ 9 = 4 r. 4 And 70 ÷ 9 = 7 r. 7 Extending this observation, we can readily obtain the quotient when small numbers are divided by 9. Take the case of 34. When divided by 9, the quotient is equal to the tens digit 3 and the remainder is equal to the sum of the tens and units digits, 3 + 4 = 7 42 ÷ 9 = 4 r. 6 71 ÷ 9 = 7 r. 8 26 ÷ 9 = 2 r. 8 69 ÷ 9 = 6 r. 15 In 15, we can still get a 9 and a remainder of 6 so 69 ÷ 9 is 7 remainder 6. At this point we would like to stress that the following results are equivalent: 69 ÷ 9 = 6 r. 15 = 7 r. 6 = 8 r. – 3 but 7 r. 6 is the best form. This technique can also be used for longer number. To divide longer numbers by 9, get the running sum of the digits of the number. We can mentally answer 102003000 ÷ 9, for example, by
 noting that the dividend has 9 digits, then the answer must have 8 digits
 get the running total of the digits of the dividend: the first digit of the answer is 1 while the second is 1 + 0 = 1. So the answer begins with 11 million…
 the next digits are 1 + 2 = 3, 3 + 0 = 3 and 3 + 0 = 3: 333 thousand …
 the last three digits of the answer are 3 + 3 = 6, 6 + 0 = 6 and 6 + 0 = 6: 666
 And the remainder is 6 + 0 = 6.

 52 ÷ 9 =
 71 ÷ 9 =
 123 ÷ 9 =
 142 ÷ 9 =
 2121 ÷ 9 =
 3113 ÷ 9 =
 11221 ÷ 9 =
 22111 ÷ 9 =
 300211 ÷ 9 =
 321001 ÷ 9 =
Answers to exercises in previous article (see https://www.mathinic.com/blog/multiplyingbyxa/) :
 \((x + 3)(x – 1) = x^2 + 2x – 3\)
 \((x + 3)(x – 2) = x^2 + x – 6\)
 \((2x + 5) (x – 1) = 2x^2 + 3x – 5\)
 \((2x + 5) (x – 2) = 2x^2 + x – 10\)
 \((2x^2 – 3x – 4) (x – 1) = 2x^3 – 5x^2 – x + 4\)
 \((2x^2 – 3x – 4) (x – 2) = 2x^2 – 7x^2 + 2x + 8\)
 \((2x^2 – 3x – 4) (x – 3) = 2x^3 – 9x^2 + 5x + 12\)
 \((x^3 + 4x^2 – 2x – 5) (x – 1) = x^4 + 3x^3 – 6x^2 – 3x + 5\)
 \((x^3 + 4x^2 – 2x – 5) (x – 2) = x^4 + 2x^3 – 10x^2 – x + 10\)
 \((x^3 + 4x^2 – 2x – 5) (x – 3) = x^4 + x^3 – 14x^2 + x + 15\)