Last updated: around August 2005

Level 0 update on Feb 13, 2007 - fixed header

Type: Challenge Reliability: - Medium-high Ease of learning: - Easy-medium Time saving: - Small gain Usefulness: - Moderately useful Difficulty: - Hard Overall: - Good

Multiplying several numbers at once might seem crazy and overwhelmingly difficult, considering the math graphic above (which seems surprisingly correct if punched into a calculator). It's actually MUCH easier than it seems and many of my multiplication shortcuts also come in handy and they further amplify the speed! Though it might not seem possible to multiply a large group of numbers at once (such as 347×82×67×15), this document explains the method behind it with ease so even beginners could understand it and even make better use of multiplying just a pair of numbers. Interesting (and difficult) as it may seem, there is little difference between multiplying a pair of numbers. The technique is well-explained with plenty of examples so that even beginners at math can understand (even an average 4th grader could understand it and an average 5th grader could process it). You should definitely read this. After you read this, you won't see multiplication in the old way again.

This awesome discovery was made when I was at Triangle YMCA camp during late July of 1997, just as I got out of the 6th grade. I was playing around with multiplication (as I was bored waiting for something) and I wanted to know if it was possible to multiply a group of numbers at once as you can with addition

Today, the method is much more enhanced and is explained much better than most of my other math shortcuts in general. It's easier than what it was around July 22, 1997 (Yeah, I was just 13.52-years old at the time and I knew how to multiply a group of numbers at once without doing them in pairs.... It seems bizarre, but I've done it!]. With the discovery of general multiplication shortcuts such as the powerful copy shortcut, the best one I have of all, and the transfer shortcut multiplying a group of numbers at once can be done faster than with pairs due to the multiplier effect.

To start you with the basics, let's consider the general rules for multiplying. First, you take the top 1's digit and multiply by the bottom 1's digit. You then move to the top 10's digit and multiply by the bottom 1's digit. Then repeat this only using the bottom ten's digit instead. This illustrates what you'd do for the question:

36 57 --

You'd take 6×7, carry the 4, then 3×7+4 all on the first row. The second row would have an extra zero placed first with 6×5 done first, a 3 carried, then 3×5+3 afterwards. It's then adding the results to get 2052. That's a simple case though.

If you were to toss in an extra digit, there isn't much change. Consider this:

357 128 ---

You'd do 7×8, 5×8, and 3×8 first (with the usual digit placements and carrying), then one extra zero followed by doing 7×2, 5×2, and 3×2 next, then two extra zeros followed by doing 7×1, 5×1 and 3×1 last. Finally, you add them up to get the answer of 45,696 (For the 1, you can use the copy trick to speed up solving it, but you don't have to.). Note a pattern? Here's how I look at multiplication. It's a lot different than what you were tought in elementary school.

3 2 1 0 1 2 3 4 5 6 7 8 -------

The notation above is like a "coordinate". I call it an "index". A group of indexes from each row in multiplication is called a "combo". The order in which combos are arranged is from the top row of the numbers being multiplied to the bottom row of the numbers being multiplied. The 3×6 step is doing the 1-2 combo as I call it. The 3 is in the top row with an index of 1 and the 6 is in the second row with an index of 2. The top is 1 and the second is 2. When put together, this is 1-2, your combo. This is how combos are named. The 1×8 step is the 3-0 combo as the 1 is in the top row with an index of 1 and the 8 is in the bottom row with an index of 0. Working with combos and indexes is like counting in a different number base

To distinguish the "indexes" from the scrap (scrap, as I refer to it, are the numbers placed on top of the multiplication question that get added on next, caused from carrying numbers), I usually circle the index notations and leave the scrap placed on the top untouched [in terms of marking indicators]. I'd consider doing this question in this manner, in order of the combos: 0-0, 1-0, 2-0, 3-0, 0-1, 1-1, 2-1, 3-1, 0-2, 1-2, 2-2, 3-2, 0-3, 1-3, 2-3, then 3-3. After doing that, I'd add the results from the four rows.

Why not start from one rather than zero? The reason is because it is based on powers of ten. 10

Also, note that the combos tell you where to place your digit. For example, if you have 5 digits in on the third row, you can easily find where you left off. The third row has an index of 2 as you place two zeros at the end for a multiple of 100. With 5 digits in and you know the index of the row you're on as 2, just find the difference and the result is the index on the top number being multiplied.

Let's consider a very basic group-of-three multiplication question.

1 0 1 1 1 1 1 1 ---

Here, instead of having a combo of 2 indexes, you'd have a combo of 3 indexes. The column numbers are still ordered the same: top to bottom in the question for left-to-right in the combos. Here's what you do. First, you should always start with all zeros, or the 0-0-0 combo in this case. The next step is 1-0-0 then 0-1-0, 1-1-0, 0-0-1, 1-0-1, 1-1-0, then 1-1-1. To find the answer, let's break it up into 2 math questions: 11×11, then this result (which is 121) times 11. That comes out to 1331. Let's do this using the following pattern with the combos:

11 11 11 ---- 11 110 110 1100 ---- 1331

You get the same answer, but the number of zeros for the third and fourth rows doesn't seem right. To see why it's like this, consider it's combo. Here, you're doing the 0-0-1 combo. When deciding how many zeros to place at the end after each line, always ignore the first digit as it'll always be zero anyway. In this case, we consider the second and third combo digits, the 0 and the 1. Add the two and you get 1, thus you place 1 zero on the end, not 2. Back when I first discovered this, I didn't know when to add 1 or 2 zeros, especially with very complicated questions (such as 3 digits per row). The 4th row has 2 zeros because it starts the 0-1-1 combo, thus 1+1 (or 0+1+1 if you include the full combo) is 2 and you have 2 zeros. Get it?

Now, what if you had digits other than 1's? The concept is still the same, except when it comes to carrying. Like multiplying 2 numbers, you multiply the two digits you're working on and add the previous carried number then write down the ones value and carry anything else. With 3 numbers being multiplied, the numbers being carried can easily pass 9 thus you have 2-digit numbers to carry and add. The numbers that are carried are what I often call "scrap". Consider this example already worked out:

8 10 2 3 2 3 4 5 7 2 ---------- 1 1 1 2 3 0 1 8 4 0 8 0 5 0 6 4 4 0 0 ---------- 7 4 5 2 0

Note how you carried a 10 from doing the 0-0-1 combo. Because 3×5×7 is 105, you write the 5 down and carry the 10. This will happen when you multiply groups of numbers. The largest number that can be in the scrap is an 80.

How does it work when you have varying numbers of digits? The concept is still the same but with a slight difference. Let's consider this example:

3 2 1 0 1 2 3 4 4 6 2 9 8 --------

The combos are arranged a bit differently. The first number in the combo is always the top row, the second is always the second row, the third is always the third row and so on. To do this question, you'd do the following combos in order: 0-0-0, 1-0-0, 2-0-0, 3-0-0, 0-1-0, 1-1-0, 2-1-0, 3-1-0, 0-2-0, 1-2-0, 2-2-0, 3-2-0, 0-0-1, 1-0-1, 2-0-1, 3-0-1, 0-1-1, 1-1-1, 2-1-1, 3-1-1, 0-2-1, 1-2-1, 2-2-1, then finally 3-2-1. Note that that is 24 combos [also note that the number of required combos equals the number of digits in each row times the number of digits in the next row and so on]! Working this question out will reveal (Comments are after // in each line to help you understand how many zeros to add and their combo (x is a variable for the first digit as it can be anything and is irrelevant to figuring out how many zeros to add.).):

81214 121821 4 6 7 71012 111619 3 5 6 1 2 3 4 4 6 2 9 8 ---------------- 1 2 2 3 2 1 9 7 4 4 // x-0-0 combo - 0+0 or no extra zeros 5 9 2 3 2 0 // x-1-0 combo - 1+0 or 1 extra zero 3 9 4 8 8 0 0 // x-2-0 combo - 2+0 or 2 extra zeros 2 2 2 1 2 0 // x-0-1 combo - 0+1 or 1 extra zero 6 6 6 3 6 0 0 // x-1-1 combo - 1+1 or 2 extra zeros 4 4 4 2 4 0 0 0 // x-2-1 combo - 2+1 or 3 extra zeros ---------------- 5 5 8 7 0 5 8 4

That's a lot o' work just multiplying that! Note that you carried a 21 in the stash of scrap above (from doing the 0-1-1 combo (which gives 216)). That's pretty high. In fact, the highest you can carry is an 80 unless you add a 4th row.

This doesn't just work with groups of 2 or 3 numbers, but it can work for groups of 4, 5, or even more. The process isn't much difference, except you have more indexes in the combos, one index for each number being multiplied. If you were multiplying 6 numbers at once (wouldn't recommend it, except when shortcuts can be used that would otherwise simplify it), an example combo for a group of 6 would be 2-0-3-1-1-0. This means that, for the top number, you are on the column marked with a 2, or the hundreds place. You are in the ones place for the second number, the thousands place in the third, the tens place in the fourth and fifth and the ones place in the sixth.

How would multiplying a group of four numbers look after finished? Here's an example. I'm using | to indicate the combo columns to distinguish from the scrap.

9938 4919 4919 24 9 12 4 12 4 8733 4316 4316 21 8 10 4 10 4 |2|1|0| 3 5 2 6 3 3 4 1 × 8 7 ------------------- 1 3 5 7 6 4 2 7 3 9 2 // x-0-0-0 combo - 0+0+0 or no extra zeros 7 3 9 2 0 // x-1-0-0 combo - 1+0+0 or 1 extra zero // copy trick can be used* 1 4 7 8 4 0 0 // x-2-0-0 combo - 2+0+0 or 2 extra zeros 2 9 5 6 8 0 // x-0-1-0 combo - 0+1+0 or 1 extra zero 2 9 5 6 8 0 0 // x-1-1-0 combo - 1+1+0 or 2 extra zeros // copy trick can be used 5 9 1 3 6 0 0 0 // x-2-1-0 combo - 2+1+0 or 3 extra zeros 8 4 4 8 0 // x-0-0-1 combo - 0+0+1 or 1 extra zero 8 4 4 8 0 0 // x-1-0-1 combo - 1+0+1 or 2 extra zeros // copy trick can be used 1 6 8 9 6 0 0 0 // x-2-0-1 combo - 2+0+1 or 3 extra zeros 3 3 7 9 2 0 0 // x-0-1-1 combo - 0+1+1 or 2 extra zeros 3 3 7 9 2 0 0 0 // x-1-1-1 combo - 1+1+1 or 3 extra zeros // copy trick can be used + 6 7 5 8 4 0 0 0 0 // x-2-1-1 combo - 2+1+1 or 4 extra zeros ------------------- 7 9 4 7 8 4 6 7 2 // the final answer

* The copy trick can be applied as, where the footnote is noted, you're doing 352×3×1×7 for the x-0-0-0 combo and 352×30×1×7 for the x-1-0-0 combo and the 0 is already dealt with (due to the combo) leaving 352×3×1×7, the exact same thing.

As you can see here, that's a lot of work. Doing this for groups of more than 4 is not recommended unless many of the shortcuts can be used, especially the end-five shortcut or the transfer shortcut to get rid of some numbers. In the above example, the copy trick could be used. We'll cover the impact of my multiplication shortcuts on multiplying groups in the next section. Note how I carried a 99 in the stash of scrap above. What's the maximum? 729 is the maximum.

You can use the 8 multiplication shortcuts that I know of from this area and group multiplication to amplify the power of the 8 multiplication shortcuts and group multiplication even allows for two additional shortcuts.

When using these shortcuts in group multiplication, their power is often amplified. Due to this, you should spend more time searching for available shortcuts to reduce the work load. The most important thing to consider when using these shortcuts is the number of combos needed. Remember that the number of combos needed is calculated by counting the digits in each row and multiplying this. Getting rid of double-digit numbers in a row is the best method as it has the strongest impact. Getting rid of a digit in a 3-digit number has less of an effect, but still considerable.

The zero and decimal shortcut has no change in it's effectiveness. All zeros and decimals are dropped with a number put to the side (positive for dropped zeros, negative for a dropped decimal). This question:

35.37 120 ×473.6 ------

would become this after applying the zero and decimal shortcut:

3537 -2 12 1 ×4736 -1 -----

The process is otherwise the same. 35.37 has two digits after the decimal, thus a -2 off the side. 120 has one zero at the end, thus a 1 off the side. 473.6 has one digit after the decimal, thus a -1 off the side. To figure out how many zeros to add at the end or how many digits need to be after the deicmal, it's trickier. I recommend adding all the negative values then adding all the positive values. From these two results, find the difference and give the sign of the larger value. From this, I'd get -2 as a result (-2 + -1 of the negatives gives -3 and 1 from the positives gives 2. 3 is bigger and it's negative so the final result is -2) and thus the decimal is added so that two digits are after the decimal. If the result was, say 3, I'd add 3 zeros at the end. No other change except that you just have more numbers to the side to mess with.

The end-five shortcut allows for more freedom of choices instead of having to only transfer to the other number. This shortcut is slightly more effective when multiplying several numbers at once, 6 stars instead of 5. Consider this question:

128 16 ×75 ---

In this case, the 75 can be transferred to 128 to get a 64 and a 15 (with a 1 to the side from the zero and decimal shortcut). Since 15 is still a multiple of 5, the end-five shortcut can be applied again. In this case, you can choose between the 64 and 16. The 16 is the best because it'll cut the number of combos by half meaning less work. Remember, the number of combos you have to do is equal to the number of digits in the first number times the number of digits in the second, times those in the third and so on. This would result in this (with the process worked out and separated by | and the number of total combos where the answer would normally be):

128 | 64 1 | 64 1 16 | 16 | 8 1 ×75 | ×15 | ×3 --- | --- | -- 12 8 2

The original would yield 12 combos. The first modification gives 8 combos. The second modification yields just two. After solving that, you just need to add 2 zeros to the end (from the zero and decimal shortcut).

The copy shortcut (aka, the "copy trick"), so powerful as it is, becomes even more effective with multiplying several numbers at once and it cannot be used for 1's unless all numbers below the top in your combo are ones. The downside is that it's less noticable, but you have more available options. It gets 7 stars. The more numbers being multiplied, the more powerful it gets. All it takes is an identical number. There's even a side effect to it as well, though tricky for beginners to work with that makes it yet more powerful.

Why does it become more effective? Consider the question above with multiplying 4 numbers at once. That may seem overwhelming. Beginners could spot an obvious case quite well. However, there just happens to be another trick in the book to amplify the effect even more, but it's more difficult to spot the amplification. We'll consider this question as a guide:

352 633 21 ×48 ---

The doublets can be applied from the 633 to cut the workload down by having to do 4 fewer combos directly (just copying is all you need to do). However, there's something special with the 21 and the 48. What do you supposed it is? I'll give you a hint. What do you get when you do the 0-0-0-0 combo and the 0-0-1-1 combo? Don't worry about having to work it out or carry anything. Still don't see it? What do you get when you multiply 2×3×1×8 and 2×3×2×4? Both results are the same at 48. This means that you can simply copy the base result of the x-0-0-0 combo for the x-0-1-1 combo. There's more to it than this that can be used.

What else do you see? Try looking at the 0-0-1-0 combo and see what you get. Now look at the 0-2-0-0 combo. 0-0-1-0 gives 2×3×2×8 and 0-2-0-0 gives 2×6×1×8. These multiply to 96 for each. In this case, you can copy the result of the 0-2-0-0 combo for the 0-0-1-0 combo. The same can be applied for the 0-0-0-0 combo against the 0-2-1-1 combo which happen to total 48 (which also comes from the 0-0-0-0 and 0-0-1-1 combo) for a really powerful effect. Worked out, with the copy trick notes after //, you'll see just how effective it can be (sum added up from Windows Calculator as that's of low importance to the copy trick):

6 2 // scrap from the x-0-0-1 combo 4919 // scrap from the x-2-1-0 combo 24 9 // scrap from the x-2-0-0 combo 12 4 // scrap from the x-0-0-0 combo 3 5 2 6 3 3 2 1 × 4 8 ------------------- 1 3 5 6 6 4 2 // scrap due to adding 8 4 4 8 // base of 24 - result of x-0-0-0 8 4 4 8 0 // base of 24 - result of x-1-0-0 - copy x-0-0-0 1 6 8 9 6 0 0 // base of 48 - result of x-2-0-0 - no copy 1 6 8 9 6 0 // base of 48 - result of x-0-1-0 - copy x-1-0-0 1 6 8 9 6 0 0 // base of 48 - result of x-1-1-0 - copy x-0-1-0 3 3 7 9 2 0 0 0 // base of 96 - result of x-2-1-0 - no copy 4 2 2 4 0 // base of 12 - result of x-0-0-1 - no copy 4 2 2 4 0 0 // base of 12 - result of x-1-0-1 - copy x-0-0-1 8 4 4 8 0 0 0 // base of 24 - result of x-2-0-1 - copy x-0-0-0 8 4 4 8 0 0 // base of 24 - result of x-0-1-1 - copy x-0-0-0 8 4 4 8 0 0 0 // base of 24 - result of x-1-1-1 - copy x-0-0-0 + 1 6 8 9 6 0 0 0 0 // base of 48 - result of x-2-1-1 - copy x-2-0-0 ------------------- 2 2 4 5 9 8 5 2 8

I did that one roughly 3 times faster than the above question of similar format mainly because of the copy trick. On paper, when multiplying a bunch of numbers at once, I put the base of each combo to the side of the results I get from multiplying as I sort of have marked here (no carrying is needed). The more digits you have and the more numbers being multiplied, the faster the results are (results caused from the copy trick anyway). 36 combos seems like a lot, but the copy trick made it seem more like 12 instead.

The transfer shortcut also becomes more effective when multiplying a bunch of numbers at once due to the fact that you have more freedoms in your choices of transfer destinations and the transfer source. As an added bonus, if you can reduce the number of combos to work out, the effect is more worth it so you should spend more time searching for transfers in group multiplication, especially to get rid of double-digit numbers as they have the biggest influence when removed. In fact, the bigger the group, the more effective it gets, just like the copy trick. However, rather than doing it only when quickly noticed, when multiplying a bunch of numbers at once, you should spend more time looking for methods to transfer numbers around with. With the increased effectiveness from more choices available, spending an extra ten seconds could actually cut the work time needed by even a whole minute for 24 combos. Sometimes, it's useful for simplifying the question by reducing the number of numbers to multiply. If you can turn multiplying 4 numbers at once into 3, then do it as it's much easier that way. For this, I'd rate it 8 stars. To explain it's uses, consider this question:

144 27 21 ×15 ---

This might not seem exciting with 24 combos to do. Transfering numbers around can simplify this. First, the end-five shortcut can be used to convert the 15 to a 3 and the 144 to 72 (with a 1 to the side). I'm now down to 8 combos to do. It, however, can be reduced even further. You can actually get rid of the 3 by transferring it to the 27 to get 81. Although this doesn't reduce the number of combos, it simplifies the question quite a bit. However, you can still reduce the number of combos. It doesn't look like it, but you can. You could transfer a 7 or a 3 from the 21 to reduce it to a single digit, but this will add another digit. Remember how you find how many combos you have to do? We have 8 right now. With transferring either a 7 or a 3, it can be reduced to 6 combos instead. I'll transfer a 3 to the 72 to get a 216 and 7. You don't neccessarily have to do this as you can use the copy trick right away from the 1's and you have simple numbers to work with as it is thus not really much of an impact on speed. If you do transfer, the 1 to the side must always remain the same. Also, try to look for ways to get a case where the zero-in-the-center shortcut can be used as well. Here's the step-by-step changes as explained here with the number of combos needed for the version at the bottom.

144 | 72 1 | 72 1 | 216 1 27 | 27 | 81 | 81 21 | 21 | ×21 | × 7 ×15 | ×3 | --- | --- --- | -- 8* 6 24 8

* I recommend stopping at this point. I went further to further demonstrate this. Although there's no change in the number of combos, it's always much easier to multiply a smaller group than a larger one which makes this a recommended stopping point. It's recommended in another way though as well.

Then, when you're done, remember to add the extra zero at the end.

Unlike the other shortcuts, this one really doesn't come in handy as much. There it does make a difference and due to the complexity of group multiplication, it's useful in this way, but not as easy otherwise. It gets 6 stars for this reason. In the above example, I recommended sticking with the 8 combo version (the third one with three numbers marked with a footnote) instead of 6 because of the baby-digit shortcut. 1's are easier to work with than a screwy 7 and with the copy trick tossed in, it's a lot more helpful. The number of combos you do takes greater priority over the baby-digits shortcut, but think when transferring. If you can't really reduce the combo count that much, and you've got baby digits, use the higher combo count as it would be a lot easier. If it can be reduced otherwise (or could get the copy trick to be used), then stick with the lower combo count. You decide in this case.

This shortcut, because of the nature of multiplying by zero, has a greater impact when multiplying a bunch of numbers at once. This shortcut gets 9 stars. When you multiply anything by 0, you'll only get zero as your result no matter what else is left to multiply. The first index of the combo doesn't count though, but other indexes afterwards do. So, when you have 36 combos and 2 zeros in the center, make use of it to simplify your work load. Unlike multiplying a pair where you can skip just 2 combos, this can increase to skipping even 8 or more combos! Consider this example:

2592 309 ×602 ----

In this case, the 0 in the center really would help. With the availability of the copy trick as well (one case), that helps even more. In the example here, you can skip the x-1-x and x-x-1 combos because of the zero in the center. Here's this question worked out to help explain:

314910 // scrap from x-0-2 3 5 1 // scrap from x-2-0 1016 3 // scrap from x-0-0 2 5 9 2 3 0 9 × 6 0 2 ------------------- 1 2 2 1 1 4 6 6 5 6 // 18 base from x-0-0 1 5 5 5 2 0 0 // 6 base from x-2-0 (no copy) - used ZICS* 1 3 9 9 6 8 0 0 // 54 base from x-0-2 (no copy) - used ZICS** + 4 6 6 5 6 0 0 0 0 // 18 base from x-2-2 (copy x-0-0) - used ZICS*** ------------------- 4 8 2 1 5 8 6 5 6

* I skipped 0-1-0, 1-1-0, 2-1-0, and 3-1-0 skipping over the x-1-0 combo entirely. ZICS stands for Zero-in-the-center shortcut, or at least when I used it. This is 4 combos skipped

** Because of the zero in the center, you're skipping x-0-1, x-1-1, and x-2-1 as the x-x-1 has a zero in it. This is where I can skip 12 combos..

*** Skipped 4 combos here as well.

In this, I skipped 20 combos from 36 leaving 16 to do. The copy trick reduced it by 4 more leaving 12 combos to do directly. You have less to add as well. Normally, you'd have 9 numbers to add. The zeros in the center reduced it to only 4 from skipping combos.

This shortcut doesn't apply to group multiplication at all, but I do suspect that there is a way to do it. This gets just 2 star because it doesn't work at all as far as I know, but theories show that it is possible, just not in the way the shortcut is for two numbers being multiplied at once. I have, however, found some interesting patterns in the numbers of cubes, but not true relationship yet.

Exactly like the same-number shortcut, this doesn't work with group multiplication and thus only gets 1 star. I also, however, suspect that something similar is available, just in a different style.

These are shortcuts that are only available when you use group multiplication.

As the name says, the split shortcut is easy. I mainly use it for mental math calculations though I sometimes use it on paper. Imagine multiplying this:

32 ×18 ---

You could use the split-same-number shortcut for this (mid-point of 25), but you could go faster than using that, though not much faster. This shortcut involves adding another number to multiply, but, in return, you could use fewer combos. This is best with splitting two-digit numbers (except when another shortcut can be applied such as the end-five shortcut). So, which do we split? The choice is up to you. Prime numbers cannot be split and I only recommend it when you can get single-digit numbers as, if you kept the double-digit number, you wouldn't really get much in the way of added speed, except the baby-digits shortcut and/or the copy trick.

In this example, the 32 could split into 8 and 4 and the 18 can split into either 6 and 3 or 9 and 2. 2 is considered a baby digit (stronger than 3 anyway), so use that meaning that the 18 will be split into a 9 and a 2. From 4 combos down to 2:

32 9 ×2 --

You can then multiply almost normally from there. Also, you don't have to add anything either. You could, if you want, split the 32 into a 8 and a 4 to get 8×4×9×2. This is much more complicated to work with, but is more worthy for mental math where you can regroup the numbers for easier processing.

When multiplying a bunch of numbers at once, there happens to be another shortcut available that is completely unavailable without using group multiplication. It basically involves the transfer shortcut and the split shortcut:

164 ×73 ---

Since 73 is prime, you can't split it leaving only 164. There doesn't seem to be any shortcuts that can be applied to this question so you'd think you'd have to work it out normally at a slower speed. You can go a somewhat faster actually. Haven't you thought of splitting the 164 into 82 and 2 or even 41 and 4? From 6 combos, you get 4. 2 is a baby digit, so use twos often. This is what you'd have:

82 | 73 73 | 41 ×2 | ×4 -- | --

This shouldn't be too hard now. 41×4 is what I'd prefer as you have somewhat easier numbers to work with, especially from the 1, the biggest baby number available. This shortcut can even work to "activate" the zero-in-the-center shortcut as well as other shortcuts. Consider this:

621 ×569 ----

This doesn't seem like you can do anything. 621 is a multiple of 3. If you wanted to, you could split the 621 into 207 and 3 and move the 569 to the top. This would yield:

569 207 × 3 ---

With the 0 in the center, you can now skip 3 combos out of 9 total. There are frequently options available to make use of these long lists of shortcuts. The chances increase significantly when you multiply a group of numbers at once. Due to this significant increase of availability and power of the shortcuts, I use a lot of group multiplication.

Footnotes:

* A number base is based on the number of digits used. We generally use base 10 because we use ten digits. Base 4 would be using only 4 digits: 0, 1, 2, and 3.

** This is dependent a lot on your mental skills. Depending on how good you are, the time you can save may be 15%. Here, some of the tips and tricks from this page can be applied to enhance the speed to even a 50% time reduction and higher. Due to the multiplier effect, almost all of those shortcuts are amplified with group multiplication, especially the copy shortcut and zero-in-the-center shortcut.

*** This challenge is explained very well, but it's the complexity that lowers it. Outside the complexity, this would be a solid 9-star rating.

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