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Multiplying by 9, 99 and 999 Without Doing the Multiplication

Nine is ten minus one, so multiply by the round number and take one lot away. The same move runs at 99, 999 and 9,999, with a digit-sum check built in.

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Key takeaways

  • Nine is ten minus one, so multiplying by nine is multiplying by ten and taking one lot of the original away.
  • The same identity runs at 99, 999 and 9,999: shift the number left by the right number of places, then subtract the original once.
  • The common error is subtracting 1 rather than one lot of the number being multiplied, which the magnitude of the answer makes obvious if you estimate first.
  • Every multiple of 9, 99, 999 and 9,999 has a digit sum that reduces to nine, because ten leaves a remainder of one when divided by nine.
  • The digit-sum check catches many slips but not all of them, since swapping two digits leaves the sum unchanged.

Nine is ten minus one. So nine lots of a number are ten lots with one lot taken away, and 47 × 9 = 470 − 47 = 423. That single observation covers 9, 99, 999 and every other number sitting one below a power of ten, and it replaces a multiplication most people find awkward with a shift and a subtraction.

The move at nine

Shift the number one place left, then subtract the original once.

  • 63 × 9 = 630 − 63 = 567
  • 28 × 9 = 280 − 28 = 252
  • 154 × 9 = 1,540 − 154 = 1,386

The subtraction is the only real work, and it is the kind that counting up as a subtraction strategy handles cleanly if the borrowing bothers you: from 154 to 1,540 is 46 to reach 200, then 1,340 to reach 1,540, giving 1,386.

The error worth naming is subtracting 1 instead of one lot. Writing 470 − 1 = 469 for 47 × 9 is a slip that a rough estimate kills instantly: nine lots of 47 must be a little under 470, not one below it.

The same move at 99, 999 and beyond

The identity generalises: n × (10ᵏ − 1) = n × 10ᵏ − n. Shift by however many places the power of ten requires, then subtract the original once.

  • 38 × 99 = 3,800 − 38 = 3,762
  • 84 × 99 = 8,400 − 84 = 8,316
  • 26 × 999 = 26,000 − 26 = 25,974
  • 125 × 9,999 = 1,250,000 − 125 = 1,249,875

This is where the method genuinely beats the alternatives. Nobody wants to multiply 125 by 9,999 any other way, and the two steps here are a shift you cannot get wrong and a subtraction of a small number from a round one.

The related family runs the other direction — numbers just above a power of ten. For 101 you shift and add rather than subtract: 47 × 101 = 4,700 + 47 = 4,747. Same identity, opposite sign.

Numbers near a power of ten, generally

Nothing about the method requires the gap to be one. Any number close to a round figure can be handled as the round figure minus the difference.

  • 47 × 8 = 470 − 94 = 376, since 8 = 10 − 2 and two lots of 47 is 94.
  • 45 × 98 = 4,500 − 90 = 4,410, since 98 = 100 − 2.
  • 72 × 97 = 7,200 − 216 = 6,984, since 97 = 100 − 3 and three lots of 72 is 216.

The gap has to stay small enough that multiplying by it is trivial. At 10 − 4 you are multiplying by four to save multiplying by six, which is no saving at all. In practice the method pays for gaps of one, two and three, and the further shortcuts for numbers sitting near 100 on either side are covered in multiplication shortcuts and why they work.

Why every multiple of nine has digits that add to nine

The pattern in the nine times table — 9, 18, 27, 36, 45, and so on, each with digits summing to 9 — is not a coincidence of small numbers. It follows from place value.

Ten is nine plus one. A hundred is 99 plus one. A thousand is 999 plus one. Every power of ten is one more than a multiple of nine. So a number like 3,762 can be rewritten as

3 × (999 + 1) + 7 × (99 + 1) + 6 × (9 + 1) + 2

which separates into a pile of multiples of nine — 3 × 999, 7 × 99, 6 × 9 — plus 3 + 7 + 6 + 2, the digit sum. The multiples of nine are divisible by nine no matter what the digits are, so the whole number is divisible by nine exactly when its digit sum is.

NRICH's article on divisibility tests, by Tim Rowland at the University of Cambridge, sets out this argument directly, observing that ten is nine plus one and that every power of ten is therefore one more than a multiple of nine, which is what makes the digit-sum tests for three and nine work. A companion NRICH article on digital roots notes that the pattern is usually first met by children spotting it in the nine times table, and that repeatedly adding digits until a single digit remains is the same operation as working modulo nine. The NRICH problem Adding All Nine makes the connection concrete from the other side: the digits 1 to 9 total 45, so any number built from all of them once is divisible by nine.

Using the digit sum as a check

Any product with 9, 99, 999 or 9,999 as a factor is a multiple of nine, because 99 = 9 × 11, 999 = 9 × 111 and 9,999 = 9 × 1,111. So the answer's digits must reduce to nine. Check the results above:

  • 423 → 4 + 2 + 3 = 9
  • 3,762 → 3 + 7 + 6 + 2 = 18 → 1 + 8 = 9
  • 25,974 → 2 + 5 + 9 + 7 + 4 = 27 → 2 + 7 = 9
  • 1,249,875 → 1 + 2 + 4 + 9 + 8 + 7 + 5 = 36 → 3 + 6 = 9

The check costs about a second and it catches the most common failure, which is losing or mis-shifting a digit.

It is not a proof. Swapping two digits leaves the digit sum untouched, so 3,762 written as 3,726 passes the test and is still wrong. Treat the digit sum as a filter that removes most errors, not as a verification, and keep a rough estimate alongside it.

The finger trick, and what it is not

Hold up both hands and fold down the n-th finger from the left. For 7 × 9, fold the seventh: six fingers stand to its left and three to its right, giving 63.

It is correct, and its scope is exactly 1 to 10 times 9. The reason is that the fingers left standing count n − 1 tens, and those to the right count 10 − n units, which sums to 10(n − 1) + (10 − n) = 9n. The construction needs ten fingers, so it cannot reach 11 × 9 or 12 × 9, and it says nothing about 99.

That makes it a lookup device for one row of the times table rather than a method. Worth showing a child; not worth carrying as an adult, because the shift-and-subtract move covers the same row and everything beyond it.

Practising it

The bottleneck is not the shift, which is free. It is the subtraction, and specifically the subtraction of an awkward number from a round one.

IQ Test & Brain Training Games, our own app at Reign Creative, drills both halves. Multiplication Rush runs timed rounds across difficulties from the 1–9 tables up to two-digit by two-digit products, Subtraction Sprint covers the take-away step including a borrowing-practice difficulty, and the Accuracy Trainer runs ten untimed questions with a hint and an explanation available. The app's technique library carries the nine and ninety-nine move as a named method with its own worked example and its own named common mistake — which is, as above, subtracting one instead of one lot. The app itself is described on the IQ Test & Brain Training Games app page.

A sensible sequence: ten products by 9, then ten by 99, then a mixed set where some factors are 8 or 98 so the gap is no longer always one. Adding the digit-sum check to every answer from the start makes it a habit rather than something you remember to do when an answer looks odd.

The companion move — turning a multiplication into a division, as with 5, 25 and 50 — is covered in multiplying by 5, 25 and 50 by dividing instead, and the full ordering of methods is in the mental math practice guide. More of this material sits in the mental math topic hub and under education and brain training articles.

Good to know

Frequently asked questions

What is the quickest way to multiply by 9?
Multiply by 10 and subtract the original number once. For 47 times 9, take 470 and subtract 47 to get 423. It works because 9 equals 10 minus 1, so nine lots of a number are ten lots with one lot removed.
How do you multiply by 99 in your head?
Multiply by 100 and subtract the original once. For 38 times 99, take 3,800 and subtract 38 to get 3,762. The same move handles 999 with a shift of three places and 9,999 with a shift of four.
Why do the digits of multiples of nine add up to nine?
Because ten is one more than nine, and so is every power of ten. That makes any number equal to a multiple of nine plus the sum of its digits, so a number is divisible by nine exactly when its digit sum is, and repeatedly adding the digits of a multiple of nine always lands on nine.
Does the finger trick work for larger numbers?
No. Folding a finger works only for 1 through 10 multiplied by 9, because it relies on having ten fingers to split. It is a lookup device for one row of the times table, not a method, and it does not extend to 12 times 9 or to 99.

Sources

  1. Divisibility Tests (opens in a new tab)Rowland, T., NRICH, Millennium Mathematics Project, University of Cambridge — accessed
  2. Digital Roots (opens in a new tab)Bagnall, B., NRICH, Millennium Mathematics Project, University of Cambridge — accessed
  3. Adding All Nine (opens in a new tab)NRICH, Millennium Mathematics Project, University of Cambridge — accessed

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