kowalski box method (2)

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  • 8/3/2019 Kowalski Box Method (2)

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    By: Steven Kowalski

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    Kowalski Box Method (KBM) vs. Regular Method A preview

    How this method was envisioned Trick to multiplying AB*11

    Digits Normal Digit Notation

    Comma Notation

    Kowalski Box Method

    2-digit by 1-digit 2-digit by 2-digit

    3-digit by 2-digit

    General formula: n-digit by m-digit

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    142x17

    1294+420714

    42x17

    4,3|0,1|47 1 4

    More examplestoward the end

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    Let AB be a 2-digit number such as 2323

    x112 5 3

    (no need to write2a and 2b)

    2 + 3

    Carry the 1 if necessary.

    1

    2a

    1

    2a

    2b

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    Fancy trick!I know, right?Could we do this with other numbers

    besides 2-digit numbers multiplied by 11?How would you do it given any two Whole

    numbers?

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    The Arabic Numeralsare the figures{0,1,2,3,4,5,6,7,8,9}, each with a numericalvalue These are the most common numerals used to

    construct numbers.

    Each number is composed of digits,

    numbers with a certain place value. ones-place, tens-place, hundreds-place, etc.

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    Let A, B, C, D, E be any Arabic numeralsand let ABCDE be the numeral string.

    Then, ABCDE = A*104 + B*103 + C*102 +D*101 + E*100 Ex: A=1, B=5, C=3, D=4, E=8. Then ABCDE =

    15348

    ABCDE = 1*10000 + 5*1000 + 3*100 + 4*10 + 8*1 ABCDE = 1*104 + 5*103 + 3*102 + 4*101 + 8*100

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    ABCDE = 1 5 3 4 8

    Let X be a number with n place values,each of which contains a numeral

    X = Xn Xn-1 X3 X2 X1

    Ones

    place

    Tens

    place

    Hundreds

    place

    Thousands

    place

    Ten-

    thousandsplace

    100splace

    101splace

    102splace

    10n-2splace

    10n-1splace

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    Suppose we want to write a number with ANY numberin the place values (not just the numerals). Then youuse a comma to separate each place value.

    So, given ABCDE = 15348, it can be expressed asfollows: = 1,5,3,4,8 = 1*104 + 5*103 + 3*102 + 4*101 + 8*100 = 15,3,4,8 = 15*103 + 3*102 + 4*101 + 8*100 = 1,5,0,34,8 = 1*104 + 5*103 + 0*102 + 34*101 + 8*100 = 1,4,13,0,48 = 1*104 + 4*103 + 13*102 + 0*101 + 48*100

    = 153,1,38 = 153*102 + 1*101 + 38*100

    (Diagram on next page) So, in using comma notation to write a number, we see

    that in between each comma can be ANY non-negativeinteger (or Whole number).

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    15348 = 1 , 4 , 13 , 0 , 48

    15348 = 153 , 1 , 38

    [100s]Ones

    place

    [101s]Tens

    place

    [102s]Hundreds

    place

    [103s]Thousands

    place

    [104s] Ten-thousands

    place

    Ones

    placeTens

    placeHundreds

    place

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    In general, where X{1,2,3,,n-1,n} are non-negative integers, if X is written in commanotation as:

    X = Xn, Xn-1,, X3, X2, X1Then

    X = Xn*10n-1+Xn-1*10

    n-2++X2*101+X1*10

    0

    And this can be

    expressed by:

    But there has to be an easier way to calculateit!!!!!!!!

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    The Box Method: Given a number in comma notation, suchas X = 12,315,84 ,

    1. Draw a line before the ones-place digit for each place in thecomma notation: 12, 315, and 84 1|2,31|5,8|4

    2. Add the numbers in each box (from right to left), carryingremainders to the next compartment. 1|2,31|5,8|4

    1|2+31|5+8|4 (commas mean plus)1+3|3+1 | 3 |4 (added right to left, carrying remainders)4 4 3 4 (added right to left, no remainders to carry)

    More Condensed 1+3|2,31+1|5,8|4

    4 4 3 4 (added right to left carrying each remainder,while dropping the ones place)

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    This method is useful especially whenmultiplying 3-digit numbers by 3-digitnumbers or less.

    The multiplication can be done from rightto left or from left to right.

    It also takes up less space on paper.

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    Let AB and C bewhole numbers:

    AB*C =

    (10A*C+1B)*(1C)= 10A*C+1B*C

    = A*C,B*C

    ABxC

    A*C,B*C

    25x6 .

    1|2,3|01 5 0

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    Let AB and CD be wholenumbers:

    AB*CD=

    (10A+1B)*(10C+1D)=100A*C+10A*D+10B*C+1B*D

    =100(A*C)+10(A*D+B*C)+1(B*D)

    ABxCD

    A*C,A*D+B*C,B*D

    25x61 .

    1|2,3|2,|51 5 2 5Or

    25x 61 .

    1|2,3|0,|5| |2 | .1 5 2 5

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    Let ABC and DE bewhole numbers:

    (100A+10B+1C)

    *(10D+1E)=1000A*D+100A*E+1

    00B*D+10B*E+10C*D+1C*E

    =1000A*D+100(A*E+B*D)+10(B*E+C*D)+1C*E

    ABCx DE

    A*D,A*E+B*D,B*E+C*D,C*E

    514x42 .

    2|0,1|4,1|8,|8 .

    2 1 5 8 8

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    Let X = Xn*10n-1+Xn-1*10

    n-2++X2*101+X1*10

    0 and

    Y = Yp*10p-1+Yp-1*10

    p-2++Y2*101+Y1*10

    0 where npThen

    But Lets make this one summation:

    Did it backwards but dont want to rewrite it Sorry

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    Ill get back to it tomorrow (promise)

    because Im too tired to keep writing.