Eikonal Blog

2012.03.14

Pretty little tables

Filed under: mathematics, number theory, puzzles — Tags: , — sandokan65 @ 14:59

Recently I have seen in an math forum this:

    Wonder #1

and just a few days later this one, too:

    Wonders 1-4

Pretty little tables, aren’t they? How could they be so regular? Can they be generalized somehow?

Answers are: yes, you will see, and yes.

Wonder #1

Let’s first take a look at the first table:

Table 1
  • 1 \times 8 + 1 = 9,
  • 12 \times 8 + 2 = 98,
  • 123 \times 8 + 3 = 987,
  • 1,234 \times 8 + 4 = 9,876,
  • 12,345 \times 8 + 5 = 98,765,
  • 123,456 \times 8 + 6 = 987,654,
  • 1,234,567 \times 8 + 7 = 9,876,543,
  • 12,345,678 \times 8 + 8 = 98,765,432,
  • 123,456,789 \times 8 + 9 = 987,654,321,

In order to understand it one has to work not with the specific numbers (digits), but with their abstract representations. For this, we will work with a number system of base B (\in {\Bbb N}), which in the orginal tables is B=10. Then we can rewrite several first members of the Table 1 as follows:

  • 1 \cdot B^0 \times (B-2) + 1 = (B-1) B^0,
  • (1 \cdot B^1 + 2 B^0) \times (B-2) + 2 = (B-1) B^1 + (B-2)B^0,
  • (1 \cdot B^2 + 2 B^1 + 3 B^0) \times (B-2) + 3 = (B-1)B^2 + (B-2)B^1 + (B-3)B^0,
  • etc

Ok, we see some regularity here. To proceed further, rewrite the n^{th} row of that table in the form a mathematical equation y_n :\equiv x_n \cdot \hbox{some number} + \hbox{some other number}, transforming the first beautifully looking number (x_n) into second beautifully looking number y_n.

Here the series \{x_n\} is:

  • x_1 = 1_B = 1\times B^0,
  • x_2 = 12_B = 1\times B^1 + 2 \times B^0,
  • x_3 = 123_B = 1\times B^2 + 2 \times B^1 + 3 \times B^0,
  • x_n = 123...n_B = 1\times B^{n-1} + 2 \times B^{n-2} + \cdots + n \times B^0 = \sum_{k=1}^n k B^{n-k}.
    A side note:
    Note that x_n =  \sum_{s=0}^{n-1} (n-s) B^s. It can be explicitly summarized as follows:

      x_ n = (n- B \partial_B) \sum_{s=0}^{n-1} B^s = (n-B\partial_B) \frac{B^n-1}{B-1} = \frac{B(B^n-1)-n(B-1)}{(B-1)^2}.

    For example, for B=10 that formula yields x_n = \frac{10(10^n-1)-9n}{81}: x_1 = 1, x_2 = 12, …, x_5 = 12345, etc.

Let’s go back to the main line of discussion.

Now we are interested in the following derivative series y_n :\equiv x_n \cdot (B-2) + n. The straightforward manipulation leads to the anticipated result:

    y_ n = B^n + \sum_{s=1}^{n-1} (s+1-n)B^s - n
    = (B-1)B^{n-1} + B^{n-1} + \sum_{s=1}^{n-2} (s+1-n)B^s - n
    = (B-1)B^{n-1} + (B-2) B^{n-2} + \sum_{s=1}^{n-3} (s+1-n)B^s - n

    \cdots

    = \sum_{r=1}^{m} (B-r)B^{n-r} + B^{n-m} + \sum_{s=1}^{n-m-1} (s+1-n)B^s - n

    \cdots

    = \sum_{r=1}^{n-2} (B-r)B^{n-r} + B^{2} + \sum_{s=1}^{1} (s+1-n)B^s - n
    = \sum_{r=1}^{n-2} (B-r)B^{n-r} + B^{2} + (2-n)B^1 - n
    = \sum_{r=1}^{n-1} (B-r)B^{n-r} + B^{1} - n
    = \sum_{r=1}^{n} (B-r)B^{n-r}.

i.e. y_n = (B-1)(B-2)...(B-n+2)(B-n+1)(B-n)_B. The initial pyramid of simple results holds for every base B.

Example: for B=5 we have x_n = \frac{5(5^n-1)-4n}{16}, so x_1 = \frac{5\cdot 4 - 4 \cdot 1}{16} = 1, x_2 = \frac{5\cdot 24 - 4 \cdot 2}{16} = 7_{10} = 12_5, etc. Then, for example y_2 = x_2 \cdot 3 + 2 = 7\cdot 3 + 2 = 21 + 2 = 23_{10} = 43_5.

Wonder #2

Let’s look at the Table 2:

Table 2
  • 1 \times 9 + 2 = 11,
  • 12 \times 9 + 3 = 111,
  • 123 \times 9 + 4 = 1,111,
  • 1,234 \times 9 + 5 = 11,111,
  • 12,345 \times 9 + 6 = 111,111,
  • 123,456 \times 9 + 7 = 1,111,111,
  • 1,234,567 \times 9 + 8 = 11,111,111,
  • 12,345,678 \times 9 + 9 = 111,111,111,
  • 123,456,789 \times 9 + 10 = 1,111,111,111,

Here the first (i.e. the independent) variable x_n is the exactly same as the one used for Table 1. The second (i.e. the dependent) variable z_n is new one, determined by defining equation:

    x_n = \sum_{s=0}^{n-1}(n-s)B^s,

Then, using steps similar to these used in analysis of the Table 1, we get:

    z_n :\equiv x_b \cdot (B-1) + (n+1) =
    = \sum_{s=0}^{n-1} (n-s) B^{s+1} - \sum_{s=0}^{n-1} (n-s) B^s + (n+1) =
    = \sum_{s=1}^{n} (n+1-s) B^{s} - \sum_{s=0}^{n-1} (n-s) B^s + (n+1) =
    = B^n + \sum_{s=1}^{n-1} (n+1-s - n +s) B^{s} - n + (n+1) =
    = B^n + \sum_{s=1}^{n-1} B^{s} +1 =
    = \sum_{s=0}^{n} B^{s} =
    = 1\cdots 1_B,

where there are (n+1) copies of digit 1.

Nice. Easy.

Wonder #3

Table 3
  • 9 \times 9 + 7 = 88,
  • 98 \times 9 + 6 = 888,
  • 987 \times 9 + 5 = 8,888,
  • 9,876 \times 9 + 4 = 88,888,
  • 98,765 \times 9 + 3 = 888,888,
  • 987,654 \times 9 + 2 = 8,888,888,
  • 9,876,543 \times 9 + 1 = 88,888,888,
  • 98,765,432 \times 9 + 0 = 888,888,888,

That series of relations has form v_n = u_n \cdot 9 + (8-n) where the dependent variable is
v_n = \underbrace{8\cdots8}_{n+1} in the normal decimal system (B=10).

For general basis B this generalizes to the:
v_n = u_n \cdot (B-1) + (B - n -2). Here the independent variable u_n is

    u_n = \sum_{k=1}^n (B-k) B^{n-k} =
    = \sum_{s=0}^{n-1} (B- n -s) B^s =
    = (B - n + B \partial_B) \sum_{s=0}^{n-1}B^s =
    = (B - n + B \partial_B)\frac{B^n-1}{B-1} =
    = \frac{B(B-2)(B^n-1)+n (B-1)}{(B-1)^2}.

Now

    v_ n :\equiv u_n (B-1) + (B-n-2) =
    = (B-1) B^n - \sum_{s=1}^{n-1} B^s - 2 =
    = (B-2) B^n + (B-1) B^{n-1} - \sum_{s=1}^{n-2} B^s - 2 =
    = (B-2) B^n + (B-2) B^{n-1} + (B-1)B^{n-2} - \sum_{s=1}^{n-3} B^s - 2 =

    \cdots

    = (B-2) B^n + \cdots + (B-2) B^{n-m+1} + (B-1)B^{n-m} - \sum_{s=1}^{n-m-1} B^s - 2 =

    \cdots

    = (B-2) B^n + \cdots + (B-2) B^{3} + (B-1)B^{2} - B^1 - 2 =
    = (B-2) B^n + \cdots + (B-2) B^{3} + (B-2)B^{2} + B^2 - B^1 - 2 =
    = (B-2) B^n + \cdots + (B-2) B^{2} + (B-1)B^{1} - 2 =
    = (B-2) B^n + \cdots + (B-2) B^{1} + B - 2 =
    = \sum_{r=0}^{n} (B-2) B^r.

For B=10 that covers all examples in the Table 3.

Note: Even more, we can add two more members to it, corresponding to n=9 and n=10:

  • u_9 = 987,654,321 corresponds to v_9 = x_9 \cdot 9 + (-1) = 8,888,888,888,
  • u_{10} = 9,876,543,210 corresponds to v_{10} = x_{10} \cdot 9 + (-2) = 88,888,888,888.

Also, the u_0=0 provides one more line, which is prepended to this more complete table 3:

Table 3*
  • 0 \times 9 + 8 = 8,
  • 9 \times 9 + 7 = 88,
  • 98 \times 9 + 6 = 888,
  • 987 \times 9 + 5 = 8,888,
  • 9,876 \times 9 + 4 = 88,888,
  • 98,765 \times 9 + 3 = 888,888,
  • 987,654 \times 9 + 2 = 8,888,888,
  • 9,876,543 \times 9 + 1 = 88,888,888,
  • 98,765,432 \times 9 + 0 = 888,888,888,
  • 987,654,321 \times 9 - 1  = 8,888,888,888,
  • 9,876,543,210 \times 9 - 2  = 88,888,888,888.

Wonder #4

Table 4
  • 1^2 = 1,
  • 11^2 = 121,
  • 111^2 = 12,321,
  • 1,111^2 = 1,234,321,
  • 11,111^2 = 123,454,321,
  • 111,111^2 = 1,2345,654,321,
  • 1,111,111^2 = 1,234,567,654,321,
  • 11,111,111^2 = 123,456,787,654,321,
  • 111,111,111^2 = 12,345,678,987,654,321.

Here the independent variable is

    p_n :\equiv 1\cdot B^{n} + 1\cdot B^{n-1} + \cdots + 1\cdot B^0 = \sum_{i=0}^n B^i = \frac{B^{n+1}-1}{B-1}.

The resulting variable is

    r_n :\equiv p_n^2 = \sum_{i=0}^n \sum_{j=0}^n B^{i+j} =
    = \sum_{k=0}^{2n} (\sum_{i=0}^n \sum_{j=0}^n \delta_{i+j,k}) B^k.

Now the sum in brackets can be transformed as follows:

    \sum_{i=0}^n \sum_{j=0}^n \delta_{i+j,k} = \sum_{i=0}^n  \Theta(0 \le  k-i \le n) =
    = \sum_{i=0}^n \Theta(k-n \le i \le k) =  \sum_{i=\hbox{max}(0,k-n)}^{\hbox{min}(n,k)} 1 =
    = \hbox{min}(n,k) - \hbox{max}(0,k-n) + 1 = *

which has three posible simplifications:

  • * = k - 0 + 1 = k + 1 for k < n,
  • * = n + 1 for k = n,
  • * = n - (k-n) + 1 = 2n - k + 1 for k > n.

So, now we can write the final form for r_n as following:

    r_n = \sum_{k=0}^{n-1} (k+1) B^k + (n+1) B^n + \sum_{k=n+1}^{2n} (2n-k+1) B^k =
    = 1\cdot B^0 + 2\cdot B^1 + 3\cdot B^2 + \cdots + (n+1) B^n + \cdots 2\cdot B^{2n-1} + 1\cdot B^{2n} =
    = 123\cdots(n+1)\cdots321_B.

That is it.


Note that one can get some regularities for degrees higher than 2. For example, for degree 3 one has:

    1^3 = 1
    11^3 = 1,331
    111^3 = 135,531

    1,111^3 = 13,577,531
    11,111^3 = 1,357,997,531
    111,111^3 = 135,79b,b97,531
    1,111,111^3 = 13,579,bdd,b97,531

    \cdots

    {\underbrace{1\cdots1}_{[B'/2]+1}}^3 = 1357\cdots B'B'\cdots 7531.

up to the last member (of that series) where the two central digits are the highest single-digit number B' allowed in the number system of base B (i.e. B'=B-2 if B is odd, and B'=B-1 if B is even).

Note: Here a = 10_{10}, b = 11_{10}, c = 12_{10}, d = 13_{10}, e = 14_{10}, f = 15_{10}, g = 16_{10}, h = 17_{10}, i = 18_{10}, etc.

For the degree 4 the similar pyramid/table is:

    1^4 = 1
    11^4 = 14,641
    111^4 = 1,48a,841

    1,111^4 = 1,48a,cec,841

    \cdots

Here one can also work some more (and I did not do that work yet) to establish which is the last member of that table (as a function of the base B), and what is the innermost digit in that last member.
This could be a homework for you. 🙂

For the degree 5:

    1^5 = 1
    11^5 = 15885
    111^5 = 15ciic51

    \cdots


Similar here: More simple math wonders – https://eikonal.wordpress.com/2012/03/14/more-simple-math-wonders/ | Mental calculation of cube root of a six-digit number – https://eikonal.wordpress.com/2010/01/14/mental-calculation-of-cube-root-of-a-two-digit-number/ | Squares with just two different decimal digits – https://eikonal.wordpress.com/2010/01/05/squares-with-just-two-different-decimal-digits/ | Number theory finite concidental sums – https://eikonal.wordpress.com/2010/01/05/number-theory-finite-considental-sums/

2011.06.03

Collatz conjecture

2010.06.20

Rubik’s cube

Filed under: mathematics, puzzles — Tags: — sandokan65 @ 23:07

2010.01.05

Squares with just two different decimal digits

Filed under: mathematics, number theory, puzzles — Tags: — sandokan65 @ 16:00

I used to state here

There exist only finitely many squares with just two decimal digits” providing a really short list

  • 38^2 = 1,444
  • 88^2 = 7,744
  • 109^2 = 11,881
  • 173^2 = 29,929
  • 212^2 = 44,944
  • 235^2 = 55,255
  • 3,114^2 = 9,696,996

plus infinite classes:

  • 10^{2n},
  • 4\cdot 10^{2n},
  • 9\cdot 10^{2n}.

This is not correct. Thanks to Bruno Curfs for pointing this out, as well as for finding the reference [2] cited below.

  • Namely, one can see that squares of all single digit numbers satisfy targeted condition:
    • 1^2 = 01,
    • 2^2 = 04,
    • 3^2 = 09,
    • 4^2 = 16,
    • 5^2 = 25,
    • 6^2 = 36,
    • 7^2 = 49,
    • 8^2 = 64,
    • 9^2 = 81,
  • Then, there are squares of following two-digit numbers:
    • 10^2 = 100,
    • 11^2 = 121,
    • 12^2 = 144,
    • 15^2 = 225,
    • 20^2 = 400,
    • 21^2 = 441,
    • 22^2 = 484,
    • 26^2 = 676,
    • 30^2 = 900,
    • 38^2 = 1,444,
    • 88^2 = 7,744,
  • For squares of 3-digit numbers one finds:
    • 100^2 = 10,000,
    • 109^2 = 11,881
    • 173^2 = 29,929
    • 212^2 = 44,944
    • 235^2 = 55,255
    • 264^2 = 69,696 (thanks to Bruno Curfs)

Of the higher cases, there are known only infinitely long classes:

  • 10^{2n}
  • 4\cdot 10^{2n}
  • 9\cdot 10^{2n}

and following two “sporadic” occurrences:

  • 3,114^2 = 9,696,996
  • 81,619^2 = 6,661,661,161.

Author of page “1. Squares of 2 different digits” (http://www.asahi-net.or.jp/~KC2H-MSM/mathland/math02/three01.htm) provides an implementation of the algorithm to perform exhaustive check. According to his numerical experiments, there are no other sporadic occurrences up to at least y=10^{48} (i.e. x=10^{24}).


Reference:


Similar here: More simple math wonders – https://eikonal.wordpress.com/2012/03/14/more-simple-math-wonders/ | Mental calculation of cube root of a six-digit number – https://eikonal.wordpress.com/2010/01/14/mental-calculation-of-cube-root-of-a-two-digit-number/ | Squares with just two different decimal digits – https://eikonal.wordpress.com/2010/01/05/squares-with-just-two-different-decimal-digits/ | Number theory finite concidental sums – https://eikonal.wordpress.com/2010/01/05/number-theory-finite-considental-sums/

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