Maths Addition, Subtraction and Multiplication |
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If you look back at the 'Person 0 to 9' article, you see that I used letters of the alphabet to mean digits from 0 to 9.
I believe that mathematics is easier for some people if they speak maths rather than picture it. Let us say that:
B is 0
Ch is 1
Sh is 2
G is 3
D is 4
N is 5
S is 6
Z is 7
X is 8
F is 9
Let us say that one of those digits needs adding to another digit; and let that other digit be expressed as:
0. O (sounds like o in 'pot')
1. I (sounds like ee in 'peep')
2. A (sounds like a in 'pat')
3. E (sounds like e in 'pet')
4. U (sounds like oo in 'hoot')
5. W is oo (eg. the Ue sound in 'tune')
6. X is ie (eg. the Ie sound in 'pie')
7. K is i (eg. the i sound in 'pip')
8. Y is ah (eg. the sound in 'park')
9. V is u (eg. the sound in 'pun')
What I want to develop is a way of expressing 37 + 89 in a way that automatically solves the question. If the 7 and 9 have no carrier to deal with then I would use the 'o' sound between the Z of 7 and the F of 9: ZoF. But I would have a rote table where Z and F are listed as ZeeF: the 'ee' means that the sum is more than 10 and so a carrier is involved. In fact the look up table would state: ZeeFie because the 'ie' is the sound for 6 which is the right side answer of the question. I could write a 6 on paper at this point.
Now I want to look at the 3 and 8 of the tens part of the question. Normally, that would be GeeX: the G is the 3, the 'ee' is the fact there is a carrier, and the X is the 8; but I already have the 'ee' from the 'ZeeF' I just did; so GeeX is expressed as 'GaX'. A rote lookup table would say 'GaXeeSh' where the eeSh is the 12 to put at the start of the answer. So the answer is 126.
Note: GeeX of 3+7 would have a rote table answer of GeeXeeCh.
I wonder if this approach would work better for some people than ordinary maths. If so then a similar rote syste could be developed for subtractions maybe.
I have other ideas about calculations. I like the idea of counting in 50s. So, if you add 45 and 47 (=92), you would note that the answer is at least 50 or higher; and you would know from memory the remainder of 92-50: learn the 42.
You could tot up a lot of 2 column numbers, noting the 50s being passed, and then re-introduce all the 50s when presenting the final answer. Or maybe 25 is the upper limit before the reset (I am playing with ideas.)
Another idea is two digit multiplication: there would be 5000 person images numbered from 0000 to 4999.
The hair or hat and the colour has an encrypted 3 digit meaning; and the outfit and colour has encrypted 3 digit meaning too.
If you need to calculate 47 x 49, person 4749 would have 3 digits of the answer encrypted as one of the hat or hair styles; and its colour.
If you need to calculate 97 x 49, you subtract 50 from the 97 to get 47; and then look up person 4749 who would have 3 digits of the answer encrypted as an outfit and outfit colour.
Each of these people has a first name based on two syllables of the Town 100 syllables; and
Each of these people (with a 0 to 49 x 0 to 49 aspect) has a surname which is a syllable representing the remainder when a 50s addition is done. Eg. 45 + 47 has a carrier and the appropriate 42 syllable represented in the surname [see the Person 00-99 article]. The suffix 'Ch' could indicate the carrier. 45 = MO, 47 = MA. So person with number code 4547 (so the first name is MOMA) would need a surname meaning "42 and add 1 to the count of 50s": LACh.
So about 2500 results would need memorising to make use of the approach. In modern times, with the wide availability of calculators, there is probably little appeal for this facility.
The 1000 Women and 1000 Male cartoon images of people involve logarithm data. Logarithms were very important before computers made difficult calculations easy. They are very weird. For instance, multiplication is achieved by a process only slightly more complicated than a basic addition calculation. Int this way, approximate answers can be calculated that are of sufficient accuracy to be useful to navigators or engineers.
Another way of thinking about summing differently is to apply the syllables form the People 00-99 article:
Left Part of Syllable | Add This | Left Vowel | Right Vowel | Add This | |
B | 0 | O | O | 0 | |
CH | 5 | O | I | 1 | |
D | 10 | O | A | 2 | |
F | 15 | O | E | 3 | |
G | 20 | O | U | 4 | |
H | 25 | I | O | 1 | |
J | 30 | I | I | 2 | |
K | 35 | I | A | 3 | |
L | 40 | I | E | 4 | |
M | 45 | I | U | 5 | |
N | 50 | A | O | 2 | |
P | 55 | A | I | 3 | |
R | 60 | A | A | 4 | |
S | 65 | A | E | 5 | |
SH | 70 | A | U | 5+1 | |
T | 75 | E | O | 3 | |
V | 80 | E | I | 4 | |
W | 85 | E | A | 5 | |
Y | 90 | E | E | 5+1 | |
Z | 95 | E | U | 5+2 | |
U | O | 4 | |||
U | I | 5 | |||
U | A | 5+1 | |||
U | E | 5+2 | |||
U | U | 5+3 |
Let's express 37+39 as a pair of syllables from that article: 37 is KA and 39 is KU; so the word I make is KAKU. If i have memorised the table above then the K and the K mean 35 and 35 which is 70; this is easy to calculate because numbers ending in 5 are easy to add together. The A and U can be looked up in the second table above and, rote, mean 6. It is easy to add the 6 to the 70. The answer is 76.
So a small amount of rote learning can make a sum be expressed in a way that is easier to solve; and the syllable pair is also an alternative way of holding an intermediate result in your short term memory if the 37+39 is just a part of a bigger calculation.
I think there is less chance of number errors when syllables are used.
I could also develop a table to state what any two letters add up to. Eg. KN is 85 because it is 35+50 (K is 35, N is 50). If I have a two syllable word like KAKU then the KK would, by rote, give me 70 (K 35 plus K 35) and AU is +5 and +1 to give me 76. KN and NK are going to have the same table answer of 85 because of the similarity in adding 35+50 or 50+35.
How many 3 digits are there when faced with adding three digits together? 1+2+1 will be similar to 1+1+2; so that helps reduce the number of three digit numbers to learn rote the total of.
There are about 220 results that, if rote learned, would speed up the summing of three digits:
Ascending Order | Sum |
000 | 0 |
100 | 1 |
110 | 2 |
111 | 3 |
200 | 2 |
210 | 3 |
211 | 4 |
220 | 4 |
221 | 5 |
222 | 6 |
300 | 3 |
310 | 4 |
311 | 5 |
320 | 5 |
321 | 6 |
322 | 7 |
330 | 6 |
331 | 7 |
332 | 8 |
333 | 9 |
400 | 4 |
410 | 5 |
411 | 6 |
420 | 6 |
421 | 7 |
422 | 8 |
430 | 7 |
431 | 8 |
432 | 9 |
433 | 10 |
440 | 8 |
441 | 9 |
442 | 10 |
443 | 11 |
444 | 12 |
500 | 5 |
510 | 6 |
511 | 7 |
520 | 7 |
521 | 8 |
522 | 9 |
530 | 8 |
531 | 9 |
532 | 10 |
533 | 11 |
540 | 9 |
541 | 10 |
542 | 11 |
543 | 12 |
544 | 13 |
550 | 10 |
551 | 11 |
552 | 12 |
553 | 13 |
554 | 14 |
555 | 15 |
600 | 6 |
610 | 7 |
611 | 8 |
620 | 8 |
621 | 9 |
622 | 10 |
630 | 9 |
631 | 10 |
632 | 11 |
633 | 12 |
640 | 10 |
641 | 11 |
642 | 12 |
643 | 13 |
644 | 14 |
650 | 11 |
651 | 12 |
652 | 13 |
653 | 14 |
654 | 15 |
655 | 16 |
660 | 12 |
661 | 13 |
662 | 14 |
663 | 15 |
664 | 16 |
665 | 17 |
666 | 18 |
700 | 7 |
710 | 8 |
711 | 9 |
720 | 9 |
721 | 10 |
722 | 11 |
730 | 10 |
731 | 11 |
732 | 12 |
733 | 13 |
740 | 11 |
741 | 12 |
742 | 13 |
743 | 14 |
744 | 15 |
750 | 12 |
751 | 13 |
752 | 14 |
753 | 15 |
754 | 16 |
755 | 17 |
760 | 13 |
761 | 14 |
762 | 15 |
763 | 16 |
764 | 17 |
765 | 18 |
766 | 19 |
770 | 14 |
771 | 15 |
772 | 16 |
773 | 17 |
774 | 18 |
775 | 19 |
776 | 20 |
777 | 21 |
800 | 8 |
810 | 9 |
811 | 10 |
820 | 10 |
821 | 11 |
822 | 12 |
830 | 11 |
831 | 12 |
832 | 13 |
833 | 14 |
840 | 12 |
841 | 13 |
842 | 14 |
843 | 15 |
844 | 16 |
850 | 13 |
851 | 14 |
852 | 15 |
853 | 16 |
854 | 17 |
855 | 18 |
860 | 14 |
861 | 15 |
862 | 16 |
863 | 17 |
864 | 18 |
865 | 19 |
866 | 20 |
870 | 15 |
871 | 16 |
872 | 17 |
873 | 18 |
874 | 19 |
875 | 20 |
876 | 21 |
877 | 22 |
880 | 16 |
881 | 17 |
882 | 18 |
883 | 19 |
884 | 20 |
885 | 21 |
886 | 22 |
887 | 23 |
888 | 24 |
900 | 9 |
910 | 10 |
911 | 11 |
920 | 11 |
921 | 12 |
922 | 13 |
930 | 12 |
931 | 13 |
932 | 14 |
933 | 15 |
940 | 13 |
941 | 14 |
942 | 15 |
943 | 16 |
944 | 17 |
950 | 14 |
951 | 15 |
952 | 16 |
953 | 17 |
954 | 18 |
955 | 19 |
960 | 15 |
961 | 16 |
962 | 17 |
963 | 18 |
964 | 19 |
965 | 20 |
966 | 21 |
970 | 16 |
971 | 17 |
972 | 18 |
973 | 19 |
974 | 20 |
975 | 21 |
976 | 22 |
977 | 23 |
980 | 17 |
981 | 18 |
982 | 19 |
983 | 20 |
984 | 21 |
985 | 22 |
986 | 23 |
987 | 24 |
988 | 25 |
990 | 18 |
991 | 19 |
992 | 20 |
993 | 21 |
994 | 22 |
995 | 23 |
996 | 24 |
997 | 25 |
998 | 26 |
999 | 27 |
When I mentioned the big multiplication system, above, to someone in the mental maths arena, he said that the maths can be done mentally without needing mnemonics. I know that; but some people can visualise numbers better than others.
When a 2 digit number is multiplied by a 2 digit number, some people do a lot of adding digits of a column of the answer as part of their technique; I think that rote learning the result of adding any three digits helps that step.
Another rote learning idea is to learn flash cards where two digits are in the middle and a digit of the answer is written above and below the two digits. For instance, 7x8=56; so the flash card would be the number 78 but the 5 is written above it and the 6 is written below it. I think that this would be a nice way of recalling parts of a times table subcalculation while doing a multiplication of a 2 digit number by a 2 digit number. School times tables might slow down that process because perhaps there is a delay in visualising the 7x8 before mentally seeing the 56: the school lesson would present 7 x 8 = 56 but recall might be 7x8 and a short while later the 56.
The Super Hero Letter Pairs article contains some of the answers to reciprocals maths; and I intend for the 100 History People (new version coming in late 2022) to contain reciprocal maths as well.
Sometimes, an approximate answer to a maths problem is useful. Eg. Logarithms.
Regarding division, if I know (approximately) what 5/7 is (0.714...) and what 6/7 is (0.857..) ten I can guess what 5.5/7 is because it has to be somewhere between those two answers. Or 5000/75 can be guessed at by using the 5/7 and 5/8 results and moving the decimal point a little: 5/8 is 0.625; so somewhere between 0.714 and 0.625 is the answer to 5/7.5; so let's say 0.65. 5000/75 is like 5.000 / 7.5; so the answer is close to 0.65 but with the decimal point moved twice: 65. The answer is really 66.666 recurring. So it is not too bad.
An item 57 could have a clue in its image to the answer to 'what is 5/7'.
An item 58 could have a clue in its image to the answer to 'what is 5/8'.
In the article about people from 00 to 99, I have put in surnames for the people which, using the 'Person 0 to 9' article rules, represent 3 digit numbers. This is a way to learn the three digits that come after the decimal point of a division question. Eg. 15 is 1/5 is 0.2 or or 200. So 200 is represented as surname Hoo where H = 2, O = 0, O = 0. You have to mentally work out the whole number that is to the left of the decimal point. Eg. 52 is like 5/2 and so you mentally know that 5/2 is 2 remainder something.
Another area of numbers which would be nicely represented by mnemonics is conversion tables. Eg. Converting from pounds to stones or vice versa.