The 1089 Trick, math that surprises you lesson

Choose a three-digit number whose first and last digits are different, such as $731$ , reverse its digits to get $137$ , and subtract the smaller number from the larger to get $594$ . Reverse that answer to $495$ and add, and the total is $1089$ , the same total that every other allowed starting number gives. Why should the starting number make no difference? This page answers with nothing more than the value of each digit place, and explains the two cases in which the trick seems to fail.

Key terms

1089 trick

Definition The 1089 trick is a number trick in base ten. Start with a three-digit number, from $100$ to $999$ , whose first digit and last digit are different. Write its digits in reverse order, keeping three digit places, so that a last digit $0$ becomes a leading zero, as when $910$ reverses to $019$ . Subtract the smaller of the two numbers from the larger and write the difference with three digit places, again keeping a leading zero, as in $099$ . Reverse the three digits of the difference and add the result to the difference. The final sum is always $1089$ . The reason is that the difference equals $99$ times the gap between the first and last digits, so it is one of only nine numbers, and each of those nine adds to its own reverse to make $1089$ . Both conditions matter. When the first and last digits are equal, the number equals its reverse, the difference is $0$ and the final sum is $0$ . When a difference such as $099$ loses its leading zero and is reversed as $99$ , the final sum is $99 + 99 = 198$ .

In plain words Reversing a three-digit number leaves its middle digit in place, so the middle digit cancels in the subtraction and only the gap between the outside digits survives. That gap can only be one of the numbers $1$ to $9$ , which leaves nine possible differences, and each of them finishes at the same total.

Example Start with $368$ . Its reverse $863$ is the larger number, so the subtraction is $863 - 368$ , and then the difference is added to its reverse,

\begin{aligned} 863 - 368 &= 495, \\[0.4em] 495 + 594 &= 1089. \end{aligned}

The outside digits $3$ and $8$ differ by $5$ , and $99 \cdot 5 = 495$ , so the difference could have been predicted before subtracting. Of the $900$ three-digit numbers, $810$ have different first and last digits, and every one of them ends at $1089$ . The other $90$ , such as $525$ , equal their own reverses and end at $0$ .

Place value

Definition In base ten, the value of a digit depends on its place in the number. The rightmost place counts units, the next counts tens, the next hundreds, and each place is worth ten times the place to its right. Three digits $a$ , $b$ and $c$ , each a whole number from $0$ to $9$ , written in that order, stand for

100a + 10b + c.

This is a three-digit number in the usual sense when $a$ is not $0$ . A zero adds nothing to the value, but it still holds its place, so $099$ is the number $99$ written with three places.

In plain words A digit is worth more the further left it stands. Moving a digit from the units place to the hundreds place multiplies its contribution by $100$ .

Example The number $368$ is $300 + 60 + 8$ , and its reverse $863$ is $800 + 60 + 3$ . The $6$ contributes $60$ to both, so it drops out of the difference. The hundreds places differ by $800 - 300 = 500$ and the units places by $3 - 8 = -5$ , so

\begin{aligned} 863 - 368 &= 500 - 5 \\[0.4em] &= 495. \end{aligned}

Digit reversal

Definition Reversing a number written with a fixed number of digit places means writing the same digits in the opposite order, in the same number of places. Leading zeros count as digits, so with three places $099$ reverses to $990$ , and $910$ reverses to $019$ , which is $19$ . The number of places has to be fixed first, because $99$ written with two places reverses to $99$ , while $099$ reverses to $990$ . With three places, reversal sends $100a + 10b + c$ to $100c + 10b + a$ and leaves the middle digit in the tens place, so a number minus its reverse is

\begin{aligned} &(100a + 10b + c) \\[0.4em] &\quad - (100c + 10b + a) \\[0.4em] &\quad = 99(a - c). \end{aligned}

This is a multiple of $99$ that does not depend on the middle digit $b$ . Reversing twice in the same number of places gives back the original digits.

In plain words Reversal swaps the first and last digits and leaves the middle digit alone. Only the swap changes the value, and it always changes it by a multiple of $99$ .

Example The numbers $501$ , $581$ and $591$ have the same outside digits and different middle digits. Subtracting their reverses gives the same difference every time,

\begin{aligned} 501 - 105 &= 396, \\[0.4em] 581 - 185 &= 396, \\[0.4em] 591 - 195 &= 396, \end{aligned}

and $396 = 99 \cdot 4$ , where $4 = 5 - 1$ is the gap between the outside digits. All three runs of the trick then end with $396 + 693 = 1089$ .

Multiples of 99

Definition A multiple of $99$ is a number $99k$ for a whole number $k$ . In the 1089 trick the difference is $99d$ , where $d$ is the gap between the first and last digits, a whole number from $1$ to $9$ . So the difference is one of the nine multiples

\begin{gathered} 099, \ 198, \ 297, \\[0.4em] 396, \ 495, \ 594, \\[0.4em] 693, \ 792, \ 891. \end{gathered}

Written with three places, the number $99d$ has the digits $d - 1$ , $9$ and $10 - d$ , because $99d = 100(d - 1) + 90 + (10 - d)$ . Its reverse is $100(10 - d) + 90 + (d - 1)$ , which equals $99(11 - d)$ , so the final sum of the trick is

\begin{aligned} &99d + 99(11 - d) \\[0.4em] &\quad = 99 \cdot 11 = 1089. \end{aligned}

In plain words Every possible difference has a $9$ in the middle and outside digits that add to $9$ . Its reverse is another multiple of $99$ , and the two multipliers always add to $11$ .

Example A gap of $d = 3$ gives the difference $297$ . Its reverse is $792 = 99 \cdot 8$ , and $3 + 8 = 11$ . Column by column, the units give $7 + 2 = 9$ , the tens give $9 + 9 = 18$ , so $8$ is written and $1$ is carried, and the hundreds give $2 + 7 + 1 = 10$ . That makes $10$ hundreds, $8$ tens and $9$ units, which is $1089$ . The columns work the same way for every gap. The units column and the hundreds column both add the two outside digits, which sum to $9$ , the tens column adds $9 + 9 = 18$ , and its carry makes the hundreds total $10$ .

Check yourself

Predict the difference for the starting number $903$ from its outside digits, then finish the trick.

Show the answer

The outside digits are $9$ and $3$ , with gap $6$ , so the difference should be $99 \cdot 6 = 594$ . Indeed $903 - 309 = 594$ . Reversing gives $495$ , and

594 + 495 = 1089.

Two people try the trick. One starts from $545$ and ends at $0$ . The other starts from $554$ , gets the difference $99$ , reverses it to $99$ and ends at $198$ . Which rule did each of them miss, and what should the second person get?

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The number $545$ has equal first and last digits, so it is not an allowed start. It equals its own reverse, the difference is $0$ , and no later step can turn $0$ into $1089$ .

The number $554$ is allowed, since its outside digits $5$ and $4$ are different. The difference is $554 - 455 = 99$ , which must be written with three places as $099$ . Its three-place reverse is $990$ , and

099 + 990 = 1089.

Reversing $99$ as a two-digit number changed the rule of the trick, and that is why the sum came out as $198$ .

Extend the trick to four digits. Start with a four-digit number whose first and last digits are different, reverse it with four places, subtract the smaller number from the larger, reverse the difference with four places and add. Try $5132$ and $4321$ . Is the final sum always the same?

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No. The reverse of $5132$ is $2315$ , and the reverse of $4321$ is $1234$ . For $5132$

\begin{aligned} 5132 - 2315 &= 2817, \\[0.4em] 2817 + 7182 &= 9999, \end{aligned}

while for $4321$

\begin{aligned} 4321 - 1234 &= 3087, \\[0.4em] 3087 + 7803 &= 10890. \end{aligned}

With four places the two middle digits swap places too, so they no longer cancel. If the larger of the two numbers has the digits $p$ , $q$ , $r$ and $s$ , the difference is

999(p - s) + 90(q - r),

which depends on the inner gap $q - r$ as well as on the outer gap $p - s$ . Checking all $8100$ four-digit numbers with different first and last digits shows exactly three final sums. The sum is $10890$ when $q$ is larger than $r$ , it is $9999$ when $q$ is smaller than $r$ , and it is $10989$ when $q = r$ , as for $1000$ , whose difference is $0999$ .