Division does not have a remainder. The remainder is an artifact (do not use this word on a six year old) of an incomplete division.
When we do long division, a remainder can take place at any stage of the division process. In fact, a remainder keeps it going from one step to the next.
When we divide 529 by 3, what do we do first? We note that 3 "goes into" 5 one time, and from this, there is a remainder of 2. If we didn't care about accuracy down to the unit, we could just stop right there. We could pad our partial quotient with enough zeros to make $100$, and then append the remaining digits $29$ to the remainder. And thus $529\div 3 = 100, R\ 229$. Surely enough, checking our result, if we multiply $3\times 100$ we get $300$, and if we add $229$ to it, we get $529$.
Of course we know that $529\div 3$ isn't $100$. This is just an approximation which is good to within $100$. The actual number is in fact $100$-something. It's not less than $100$, and it's not $200$ or more.
Now usually we do not do this. We don't stop division at the hundreds or tens to take a funny remainder. We usually stop division at the units, and take the remainder there.
This is because many objects that we work with in the real world cannot be divided beyond the unit. If we want to distribute 13 toys among 4 children, everyone gets three toys, and there is a toy left over. We don't want to divide that toy into four, because it will be destroyed. Thus, we use a form of inexact mathematical division which gives us a model for this real-world constraint: integer division with a remainder.
But non-unit remainders can be useful too. Suppose we are distributing a large number of toys at the wholesale level, and there are 24 toys in a box. Customers must get a whole box; we do not divide boxes. Boxes are only opened in the retail store to sell individual toys.
Now suppose I have $1272$ toys in my warehouse and $5$ stores approach me, all wanting to buy $300$ toys. I don't have $1500$ toys, so I decide to give the customers an equal number of toys based on what I have. Now $1272 \div 5$ gives us $254.4$, or $254\ R\ 2$. Great, each customer can have $254$ toys, with two toys left over I can keep. But wait, the toys are in boxes of $24$, so that can't work! I cannot ship $254$ toys, because that is 10 whole boxes, plus 14 toys from an open box. The calculation has to be done with boxes, not with toys. In fact I have 53 boxes of toys, and I can give the customers 10 boxes each, and have 3 boxes left over. So the remainder is $3\times 24 = 72$ toys. In other words, in a real and useful sense, the remainder of dividing $1272 \div 5$ can be $240$ with remainder of $72$, when we require the result to be a multiple of 24.
So, on to the question: could there be a remainder in multiplication? The answer is no. A remainder is a special way of expressing the error in a calculation, peculiar to inexact division. We do not have to use a remainder to express the error in an inexact division: we can simply express division error as the difference between the approximate quotient and the exact one. For instance $17\div 8$ doesn't have to be $2\ R\ 1$; it could just be $2$ with an error of $-1/8$, since the exact quotient is $2\frac{1}{8}$ and $2 - 2\frac{1}{8} = -\frac{1}{8}$. The result is $\frac{1}{8}$ less than the exact result, and so that is the error.
Multiplications cannot have a remainder for two reasons. Firstly, multiplicationsare usually carried to completion and therefore exact. There is no reason why a multiplication of two integers would be left incomplete, in the same way that we can stop a division short and take stock of what is left. Secondly, multiplications can be inexact, when the inputs are fractions (possibly themselves inexact) and we truncate the result to a given number of significant digits. However, when multiplication is inexact, we do not express the error as a remainder.
The concept of "remaining" is peculiar to division. When we divide a number, and make the result slightly smaller so that the division "works out evenly" we have something "left over". This does not apply in multiplication.
Note also that in (ordinary) multiplication, the two operands are equivalent and it is commutative. In $5\times 4$ both values are products, and it is the same as $4\times 5$. One of the products is called a multiplicand and the other a multiplier, but they can readily switch roles. But in division $5\div 4$ is not $4\div 5$: dividend and divisor cannot switch roles, and the remainder is closely related to the divisor. So right off the bat we have a conceptual problem with a remainder in multiplication: which of the two products should be related to the remainder, like the divisor is in division? If we round off the result of a multiplication to some multiple, and choose to express the error as a remainder related to one of the products, should we choose the multiplicand? or the multiplier? And why?
There is (at least) one option - however depending on whether a solution for the problem of "fractional iteration of logarithm" is available (aka: "Tetration". Note that in the tetration-forum there is an exchange on various approaches to this problem).
Assume for the moment that a version of that fractional iteration is available, and let us denote
$ \log_b^{\circ h}(x) $ for the $h$-times iterated logarithm to some given base $b$, including the opportunity to have $h$ also fractional.
Assume also that we have this well-defined for the interval $h=-1 \cdots +1$ (so for $h=-1$ this means actually exponentiation) then
$$ f_b(x,y,0) = \exp_b^{\circ 0} ( \log_b^{\circ 0} (x)+\log_b^{\circ 0} (y)) = x+y \\
f_b(x,y,1) = \exp_b^{\circ 1} ( \log_b^{\circ 1} (x)+\log_b^{\circ 1} (y)) = x \cdot y
$$
and for $0<h<1$ we have then some operation in between well defined:
$$ f_b(x,y,h) = \exp_b^{\circ h} ( \log_b^{\circ h} (x)+\log_b^{\circ h} (y)) = x \{\circ _h \} y
$$
where the operation-symbol of the circle and the index $h$ means that interpolated operation between "+" (for $h=0$) and "*" (for $h=1$).
Now let's see, whether our assumption of the existence of such fractional iterations is justified/can be realized:
- If the base $b$ is $\exp(1)$ then one can resort to Hellmuth Kneser's solution for the fractional iteration of $\exp(x)$, which provides an analytic expression for this, real for real arguments, as far as I understood things correctly.
- But one can also use the base $b$ from the interval $1<b< e^{1/e}$ and might use the easier and better understandable Schröder-mechanism for a solution for the iterative loagrithm/exponentiation. Especially the procedere for the base $b=\sqrt{2}$ is discussed and described in broadth & width in the tetration-forum.
- In the wikipedia-entry "tetration" there is some introductory information for that basic problem.
(Note: this is just one ad-hoc-approach; it does not, for instance, provide a solution comparable with one which might be expected, when we aproach the problem via the interpolation of the Ackermann-function as indicated by Daniel Geisler)
Addition, multiplication and "half multiplication" tables for
$x,y=1 \ldots 10$ . I used the base
$b=t^{1/t} =\sqrt2 \approx 1.414$ with
$t_l=2$ resp
$t_u=4$ for the logarithm and the exponentiation. For the half-exponentials/half-logarithms the formulae relate on lower fixpoint
$t_l=2$ for
$x \le 3$ and on upper fixpoint
$t_u=4$ for
$x>3$ and use the well known Schröder-mechanism for the half-iterate.
[update]: In the comment MphLee provided a link into wikipedia, where this "hyperoperation"-scheme is mentioned as due to A. Bennet (but not then generalized to fractional orders / fractional iteration heights of the $\log()$ and the $\exp()$ )
addition $h=0$ ($\text{"+"}=\circ _0$)
o_0="+"| 1 2 3 4 5 6 7 8 9 10 |
- + - - - - - - - - - - +
1 | 2.00000 3.00000 4.00000 5.00000 6.00000 7.00000 8.00000 9.00000 10.0000 11.0000 |
2 | 3.00000 4.00000 5.00000 6.00000 7.00000 8.00000 9.00000 10.0000 11.0000 12.0000 |
3 | 4.00000 5.00000 6.00000 7.00000 8.00000 9.00000 10.0000 11.0000 12.0000 13.0000 |
4 | 5.00000 6.00000 7.00000 8.00000 9.00000 10.0000 11.0000 12.0000 13.0000 14.0000 |
5 | 6.00000 7.00000 8.00000 9.00000 10.0000 11.0000 12.0000 13.0000 14.0000 15.0000 |
6 | 7.00000 8.00000 9.00000 10.0000 11.0000 12.0000 13.0000 14.0000 15.0000 16.0000 |
7 | 8.00000 9.00000 10.0000 11.0000 12.0000 13.0000 14.0000 15.0000 16.0000 17.0000 |
8 | 9.00000 10.0000 11.0000 12.0000 13.0000 14.0000 15.0000 16.0000 17.0000 18.0000 |
9 | 10.0000 11.0000 12.0000 13.0000 14.0000 15.0000 16.0000 17.0000 18.0000 19.0000 |
10 | 11.0000 12.0000 13.0000 14.0000 15.0000 16.0000 17.0000 18.0000 19.0000 20.0000 |
- + - - - - - - - - - - +
"half multiplication" $h=0.5$ ($\circ _{0.5}$) with base $b=\sqrt2$ using Schröder-mechanism
o_0.5 | 1 2 3 4 5 6 7 8 9 10 |
----- + - - - - - - - - - - +
1 | 1.42469 2.55554 3.67186 4.77842 5.87792 6.97204 8.06187 9.14818 10.2315 11.3123 |
2 | 2.55554 4 5.39214 6.75005 8.08354 9.39847 10.6987 11.9868 13.2648 14.5340 |
3 | 3.67186 5.39214 7.02490 8.60114 10.1372 11.6429 13.1245 14.5863 16.0315 17.4625 |
4 | 4.77842 6.75005 8.60114 10.3749 12.0939 13.7715 15.4163 17.0342 18.6296 20.2057 |
5 | 5.87792 8.08354 10.1372 12.0939 13.9820 15.8183 17.6136 19.3754 21.1090 22.8185 |
6 | 6.97204 9.39847 11.6429 13.7715 15.8183 17.8034 19.7397 21.6361 23.4990 25.3332 |
7 | 8.06187 10.6987 13.1245 15.4163 17.6136 19.7397 21.8094 23.8332 25.8183 27.7704 |
8 | 9.14818 11.9868 14.5863 17.0342 19.3754 21.6361 23.8332 25.9783 28.0799 30.1443 |
9 | 10.2315 13.2648 16.0315 18.6296 21.1090 23.4990 25.8183 28.0799 30.2932 32.4652 |
10 | 11.3123 14.5340 17.4625 20.2057 22.8185 25.3332 27.7704 30.1443 32.4652 34.7408 |
- + - - - - - - - - - - +
multiplication $h=1$ ($\text{"*"}=\circ _{1}$)
o_1="*"| 1 2 3 4 5 6 7 8 9 10 |
- + - - - - - - - - - - +
1 | 1.00000 2.00000 3.00000 4.00000 5.00000 6.00000 7.00000 8.00000 9.00000 10.0000 |
2 | 2.00000 4.00000 6.00000 8.00000 10.0000 12.0000 14.0000 16.0000 18.0000 20.0000 |
3 | 3.00000 6.00000 9.00000 12.0000 15.0000 18.0000 21.0000 24.0000 27.0000 30.0000 |
4 | 4.00000 8.00000 12.0000 16.0000 20.0000 24.0000 28.0000 32.0000 36.0000 40.0000 |
5 | 5.00000 10.0000 15.0000 20.0000 25.0000 30.0000 35.0000 40.0000 45.0000 50.0000 |
6 | 6.00000 12.0000 18.0000 24.0000 30.0000 36.0000 42.0000 48.0000 54.0000 60.0000 |
7 | 7.00000 14.0000 21.0000 28.0000 35.0000 42.0000 49.0000 56.0000 63.0000 70.0000 |
8 | 8.00000 16.0000 24.0000 32.0000 40.0000 48.0000 56.0000 64.0000 72.0000 80.0000 |
9 | 9.00000 18.0000 27.0000 36.0000 45.0000 54.0000 63.0000 72.0000 81.0000 90.0000 |
10 | 10.0000 20.0000 30.0000 40.0000 50.0000 60.0000 70.0000 80.0000 90.0000 100.000 |
- + - - - - - - - - - - +
halfway between multiplication and commutative(!) exponent $h=1.5$($\text{"??"}=\circ _{1.5}$)
3/2 | 1 2 3 4 5 6 7 8 9 10 |
- + - - - - - - - - - - +
1 | 0.856374 1.42469 1.95136 2.45125 2.93192 3.39780 3.85179 4.29587 4.73152 5.15984 |
2 | 1.42469 4.00000 7.02490 10.3749 13.9820 17.8034 21.8094 25.9783 30.2932 34.7408 |
3 | 1.95136 7.02490 13.8017 21.9113 31.1374 41.3357 52.4022 64.2572 76.8377 90.0920 |
4 | 2.45125 10.3749 21.9113 36.4401 53.5757 73.0498 94.6618 118.254 143.699 170.889 |
5 | 2.93192 13.9820 31.1374 53.5757 80.7547 112.281 147.853 187.226 230.200 276.603 |
6 | 3.39780 17.8034 41.3357 73.0498 112.281 158.529 211.396 270.556 335.735 406.696 |
7 | 3.85179 21.8094 52.4022 94.6618 147.853 211.396 284.818 367.723 459.769 560.660 |
8 | 4.29587 25.9783 64.2572 118.254 187.226 270.556 367.723 478.278 601.830 738.030 |
9 | 4.73152 30.2932 76.8377 143.699 230.200 335.735 459.769 601.830 761.492 938.375 |
10 | 5.15984 34.7408 90.0920 170.889 276.603 406.696 560.660 738.030 938.375 1161.30 |
- + - - - - - - - - - - +
Commutative(!) expon $h=2$($\text{"??"}=\circ _{2}$)
2 | 1 2 3 4 5 6 7 8 9 10 |
- + - - - - - - - - - - +
1 | 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 |
2 | 1.00000 4.00000 9.00000 16.0000 25.0000 36.0000 49.0000 64.0000 81.0000 100.000 |
3 | 1.00000 9.00000 32.5416 81.0000 164.317 292.874 477.416 729.000 1058.95 1478.85 |
4 | 1.00000 16.0000 81.0000 256.000 625.000 1296.00 2401.00 4096.00 6561.00 10000.0 |
5 | 1.00000 25.0000 164.317 625.000 1761.64 4107.92 8404.54 15625.0 27000.1 44040.9 |
6 | 1.00000 36.0000 292.874 1296.00 4107.92 10543.5 23393.4 46656.0 85775.3 147885. |
7 | 1.00000 49.0000 477.416 2401.00 8404.54 23393.4 55587.9 117649. 227926. 411822. |
8 | 1.00000 64.0000 729.000 4096.00 15625.0 46656.0 117649. 262144. 531441. 1.00000E6 |
9 | 1.00000 81.0000 1058.95 6561.00 27000.1 85775.3 227926. 531441. 1.12138E6 2.18701E6 |
10 | 1.00000 100.000 1478.85 10000.0 44040.9 147885. 411822. 1.00000E6 2.18701E6 4.40409E6 |
- + - - - - - - - - - - +
(Remark: because double-log of number $1$ would be negative infinity, I replaced this by approximation and insertion of $1+1e-20$ instead of $1$)
We can even define negative-half index of operation:
-1/2 | 1 2 3 4 5 6 7 8 9 10 |
- + - - - - - - - - - - +
1 | 2.55554 3.32163 4.13229 4.97809 5.85143 6.74644 7.65862 8.58454 9.52153 10.4676 |
2 | 3.32163 4.00000 4.73255 5.50980 6.32348 7.16675 8.03408 8.92107 9.82421 10.7407 |
3 | 4.13229 4.73255 5.39214 6.10270 6.85632 7.64601 8.46581 9.31075 10.1767 11.0604 |
4 | 4.97809 5.50980 6.10270 6.75005 7.44491 8.18072 8.95160 9.75241 10.5787 11.4268 |
5 | 5.85143 6.32348 6.85632 7.44491 8.08354 8.76647 9.48824 10.2439 11.0289 11.8394 |
6 | 6.74644 7.16675 7.64601 8.18072 8.76647 9.39847 10.0719 10.7823 11.5252 12.2968 |
7 | 7.65862 8.03408 8.46581 8.95160 9.48824 10.0719 10.6987 11.3644 12.0652 12.7973 |
8 | 8.58454 8.92107 9.31075 9.75241 10.2439 10.7823 11.3644 11.9868 12.6460 13.3386 |
9 | 9.52153 9.82421 10.1767 10.5787 11.0289 11.5252 12.0652 12.6460 13.2648 13.9184 |
10 | 10.4676 10.7407 11.0604 11.4268 11.8394 12.2968 12.7973 13.3386 13.9184 14.5340 |
- + - - - - - - - - - - +
and the negative $-1$- operation
-1 | 1 2 3 4 5 6 7 8 9 10 |
- + - - - - - - - - - - +
1 | 3.00000 3.54311 4.16993 4.87350 5.64386 6.46968 7.33985 8.24439 9.17493 10.1248 |
2 | 3.54311 4.00000 4.54311 5.16993 5.87350 6.64386 7.46968 8.33985 9.24439 10.1749 |
3 | 4.16993 4.54311 5.00000 5.54311 6.16993 6.87350 7.64386 8.46968 9.33985 10.2444 |
4 | 4.87350 5.16993 5.54311 6.00000 6.54311 7.16993 7.87350 8.64386 9.46968 10.3399 |
5 | 5.64386 5.87350 6.16993 6.54311 7.00000 7.54311 8.16993 8.87350 9.64386 10.4697 |
6 | 6.46968 6.64386 6.87350 7.16993 7.54311 8.00000 8.54311 9.16993 9.87350 10.6439 |
7 | 7.33985 7.46968 7.64386 7.87350 8.16993 8.54311 9.00000 9.54311 10.1699 10.8735 |
8 | 8.24439 8.33985 8.46968 8.64386 8.87350 9.16993 9.54311 10.0000 10.5431 11.1699 |
9 | 9.17493 9.24439 9.33985 9.46968 9.64386 9.87350 10.1699 10.5431 11.0000 11.5431 |
10 | 10.1248 10.1749 10.2444 10.3399 10.4697 10.6439 10.8735 11.1699 11.5431 12.0000 |
- + - - - - - - - - - - +
And to see that different ways to define the fractional iterate of exponentiation lead to different multiplication-tables I show here the "half-multiplication" taken by base $b=2$ and the implementation via something which I called "polynomial tetration" (which is based on eigendecomposition of the truncated carlemanmatrices, and seems to approximate the Kneser-solution when the truncation allows larger matrix-sizes)
"half-multiplication" (h=0.5) by "polynomial tetration", base b=2, matrixsize=16x16
1.55799 2.75402 3.92565 5.08186 6.22771 7.36568 8.49732 9.62384 10.7460 11.8646
2.75402 4.31898 5.81047 7.25673 8.67206 10.0641 11.4377 12.7963 14.1424 15.4777
3.92565 5.81047 7.57736 9.27212 10.9176 12.5260 14.1053 15.6608 17.1964 18.7150
5.08186 7.25673 9.27212 11.1905 13.0426 14.8449 16.6080 18.3393 20.0440 21.7259
6.22771 8.67206 10.9176 13.0426 15.0851 17.0659 18.9982 20.8910 22.7508 24.5824
7.36568 10.0641 12.5260 14.8449 17.0659 19.2138 21.3042 23.3478 25.3524 27.3236
8.49732 11.4377 14.1053 16.6080 18.9982 21.3042 23.5440 25.7302 27.8714 29.9745
9.62384 12.7963 15.6608 18.3393 20.8910 23.3478 25.7302 28.0521 30.3235 32.5519
10.7460 14.1424 17.1964 20.0440 22.7508 25.3524 27.8714 30.3235 32.7197 35.0681
11.8646 15.4777 18.7150 21.7259 24.5824 27.3236 29.9745 32.5519 35.0681 37.5322
[update]: Appendix.
In the (in WP) cited article of A Bennett (1915) for the interpolation to fractional orders (there $n$ here $h$) A.Bennet misses the possibility to take another base $b$ for iterated logarithm and exponentiation, where $b \ne e$ and actually $1<b<e^{1/e}$ where we can iterate infinitely and also have a "regular" method for interpolation to fractional orders/iterates using E. Schröder's method which I've used here. See the screenshot of some paragraph in A. Bennett's short paper:
Best Answer
$(A-B)(A-C)\;=\;A^2-AB-AC+BC\;=\;A(A-(B+C))+BC.$
Therefore, with $A=10^2$ and $B=3$ and $C=4$, we have $$(97)(96)=(10^2-3)(10^2-4)=10^2(10^2-(3+4))+3\cdot 4= (100)(93) +12.$$
Another example: $(91)(92)=10^2(10^2-(9+8))+9\cdot8=8300 +72.$