Lab Report Week 4 Lydia Nadeau · Lab Report Week 4 Lydia Nadeau Partner: Molly Kydland September 21, 2015 We will define a function that will tell us the fraction of integers that

$Page 1: Lab Report Week 4 Lydia Nadeau · Lab Report Week 4 Lydia Nadeau Partner: Molly Kydland September 21, 2015 We will define a function that will tell us the fraction of integers that$
Lab Report Week 4

Lydia Nadeau

Partner: Molly Kydland

September 21, 2015We will define a function that will tell us the fraction of integers that are multiples of their units digit.

p[n0_] := Module{n = n0}, T = TableIntegerQn IntegerDigits[k][[-1]], {k, 1, n};

Count[T, True] / n

p[100]

Power::infy : Infinite expression 1

0 encountered.


0 encountered.


0 encountered.

General::stop : Further output of Power::infy will be suppressed during this calculation.

2

5

There is still the problem of dividing by zero, so we will need to fix this. We will try using the Mod

function.

Mod[8, 3]

2

Mod[9, 4]

1

We will try replacing IntegerQ[n/IntegerDigits[n][[-1]]] with Mod[n,IntegerDigits[n][[-1]]].

There is still a problem when the units digit of n is 0, so we will need to find a way to exclude the outputs

of “indeterminate”.

Printed by Wolfram Mathematica Student Edition

n = 100;

T = Table[Mod[k, IntegerDigits[k][[-1]]], {k, 1, n}]

Mod::indet : Indeterminate expression Mod[10, 0] encountered.



General::stop : Further output of Mod::indet will be suppressed during this calculation.

{0, 0, 0, 0, 0, 0, 0, 0, 0, Indeterminate, 0, 0, 1, 2, 0, 4, 3, 2, 1, Indeterminate, 0,

0, 2, 0, 0, 2, 6, 4, 2, Indeterminate, 0, 0, 0, 2, 0, 0, 2, 6, 3, Indeterminate, 0,



0, 2, 0, 0, 2, 3, 0, 8, Indeterminate, 0, 0, 0, 2, 0, 0, 6, 2, 0, Indeterminate}

T = Cases[T, i_ /; i != "Indeterminate"]

{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 4, 3, 2, 1, 0, 0, 2, 0, 0, 2, 6, 4, 2, 0, 0, 0, 2, 0,

0, 2, 6, 3, 0, 0, 1, 0, 0, 4, 5, 0, 4, 0, 0, 2, 2, 0, 2, 1, 2, 5, 0, 0, 0, 0, 0, 0, 4,

4, 6, 0, 0, 1, 2, 0, 4, 0, 6, 7, 0, 0, 2, 0, 0, 2, 3, 0, 8, 0, 0, 0, 2, 0, 0, 6, 2, 0}

Now we have a table that does not include the “indeterminates”.

Next, we will redefine p.

p[n0_] := Module{n = n0, T}, T = Table[Mod[k, IntegerDigits[k][[-1]]], {k, 1, n}];

T = Cases[T, i_ /; i != "Indeterminate"];

Count[T, 0] n

p[10000]





4673

10000

This shows us how many numbers out of 10,000 are multiples of their units digit. This method seems to

work even though we still see “indeterminate”.

Here is the table p(n):

2 Week 4 Lab Report.nb


t1 = SessionTime[];

S = Table[{n, N[p[n]]}, {n, 1, 8000}];

p1 = ListPlot[S]

Print["This took ", SessionTime[] - t1, "seconds"]





2000 4000 6000 8000

0.468

0.469

0.470

0.471

0.472

This took 82.1336638seconds

The higher the range, the longer it takes to compute. On the school’s computer, this took about 57

seconds, but here it took about 82 seconds.

Now, we want to average the values of p(n) and turn them into a graph.

fractions = S[[All, 2]];

fractions[[1 ;; 100]]

{1., 1., 1., 1., 1., 1., 1., 1., 1., 0.9, 0.909091, 0.916667, 0.846154, 0.785714,

0.8, 0.75, 0.705882, 0.666667, 0.631579, 0.6, 0.619048, 0.636364, 0.608696,

0.625, 0.64, 0.615385, 0.592593, 0.571429, 0.551724, 0.533333, 0.548387,

0.5625, 0.575758, 0.558824, 0.571429, 0.583333, 0.567568, 0.552632, 0.538462,

0.525, 0.536585, 0.547619, 0.534884, 0.545455, 0.555556, 0.543478, 0.531915,

0.541667, 0.530612, 0.52, 0.529412, 0.538462, 0.528302, 0.518519, 0.527273,

0.517857, 0.508772, 0.5, 0.491525, 0.483333, 0.491803, 0.5, 0.507937, 0.515625,

0.523077, 0.530303, 0.522388, 0.514706, 0.507246, 0.5, 0.507042, 0.513889,

0.506849, 0.5, 0.506667, 0.5, 0.506494, 0.5, 0.493671, 0.4875, 0.493827, 0.5,

0.493976, 0.5, 0.505882, 0.5, 0.494253, 0.5, 0.494382, 0.488889, 0.494505,

0.5, 0.505376, 0.5, 0.505263, 0.510417, 0.505155, 0.5, 0.505051, 0.5}

Week 4 Lab Report.nb 3


diningtable = Table[Mean[fractions[[1 ;; k]]], {k, 1, Length[fractions]}];

p2 = ListPlot[diningtable, PlotStyle → Red]

2000 4000 6000 8000

0.475

0.480

0.485

0.490

0.495

0.500

Show[{p1, p2}]

2000 4000 6000 8000

0.468

0.469

0.470

0.471

0.472

This doesn’t look right.

September 23, 2015We need to figure out why the last graph looks the way it does. Let’s try switching some things around.



Show[{p2, p1}]

2000 4000 6000 8000

0.475

0.480

0.485

0.490

0.495

0.500

Show[p1, p2]

2000 4000 6000 8000

0.468

0.469

0.470

0.471

0.472

We attempted to fix the graph by switching the order of p1 and p2, and then we tried taking away the

braces. Neither helped. Our guess is that p2 may be wrong because it has much higher values than p1.

We are assuming that p2 should be running through p1.

The graphs of p1 and p2 look like inverse log functions.

September 25, 2015We would like to find the equations for p1 and p2 and also find the limits of both of them. This all

depends on if the internet starts working in the next ten minutes.

For now, we are going to try and find a limiting function for p1 and p2. All numbers that end in 1, 2, or 5

will be divisible by the respective digit, half of numbers that end in 4 are divisible by 4, 1/3 of numbers

that end in 3 or 6 are divisible by their last digit, 1/4 of numbers ending in 8 are divisible by 8, 1/7 of

numbers ending in 7 are divisible by 7, and 1/9 of digits divisible by 9 are divisible by 9. We are going to

take the average of these values.

Week 4 Lab Report.nb 5


3 + .5 + 2 3 + .25 + 1 7 + 1 9 9

0.518959

This does not seem to match up with p1 or p2 which seem to have ranges lower down on the graph,

closer to 4.68.

We are still looking into how to find the equations of the graphs. We will try the fit function later on, we

just need to learn how to use it.



Documents

Lab Report Week 4 Lydia Nadeau · Lab Report Week 4 Lydia Nadeau Partner: Molly Kydland September 21, 2015 We will define a function that will tell us the fraction of integers that