Base Ten Grid Paper: Fill & Download for Free

It is ironic that, on the one hand, insomuch as elementary number theory is not studied in US middle and high schools, the elementary differential and integral calculus is.On the other hand, in the rest of the society the army of folks who, as a hobby, pursue recreational elementary number-theoretic entertainment vastly outnumbers the army of folks who, as a hobby, pursue recreational elementary differential and integral calculus entertainment.I have no actual numbers to back up the last claim. This is just my hunch. Let us name it the Andronov’s Conjecture.If it is true then what does it tell us?Definition: a number composed of all the digits [math]1[/math] is called a repunit, presumably as an abbreviation of the concatenation of the words repetitive and unit or repeated and unit or replicated and unit.In general, the value of a repunit depends on the base [math]B[/math] in which the said number has been rendered, where [math]B[/math] is a natural number such that [math]B>1[/math].Examples: the numbers [math]1, 11, 111, 1111, 11111[/math] and so on are all repunits.However, what does, say, [math]111[/math] actually stand for? Without supplying a concrete base we can not reasonably answer that question.As such, in base ten the number [math]111[/math] stands for [math]111[/math], in base two the number [math]111[/math] stands for [math]7[/math] (in base ten), in base three the number [math]111[/math] stands for [math]13[/math] (in base ten), in base four the number [math]111[/math] stands for [math]21[/math] (in base ten) and so on.In other words, given a base [math]B[/math], a base-ten rendition of a repunit [math]R(B, n)[/math] of [math]n[/math] digits long is given by:[math]R(B, n) = \dfrac{B^n - 1}{B - 1} \tag{1}[/math]Repunits have many interesting properties. We will look at one such property here - a product of two repunits for small [math]n[/math]s.As a kid, I grew up with knowing how to do long multiplication because this is what was taught in my elementary school. I hope that long multiplication is second nature to most readers but if you are not familiar with how the long multiplication works then you could read this wikihow instruction in which they refer to what we will use below as standard long multiplication.While the mechanics of standard long multiplication are not horribly complicated, the proof that it works, the explanation of why it works and the invention of such a procedure in vacuum from scratch to begin with is another matter.Note that the wikihow instructions use the square-ruled paper as a willy-nilly background. In our discussion, however, the square-ruled grid will play an important role because we will organize our digits and numbers into rows and columns. As such, because the kids will be dealing with numbers in rows and columns, this discussion can be used as a super gentle introduction to or as a hint at the subject of linear algebra where the said numbers in rows and columns can be treated as a stand-alone mathematical object.Suppose that we want to multiply two repunits of presumably different length - a longer one [math]R_m[/math] and a shorter one [math]R_n[/math]. By definition, then, the repuint [math]R_m[/math] consists of [math]m[/math] digits [math]1[/math], while the repunit [math]R_n[/math] consists of [math]n[/math] digits [math]1[/math] with [math]m>n[/math].Multiplying [math]R_m[/math] by [math]R_n[/math] using the rules of long multiplication, we arrange the longer repunit above the shorter repunit on a square-ruled grid one unit per square and right-align the two numbers (Fig. 1):Here we took [math]m = 5[/math] and [math]n = 4[/math].As the rules of long multiplication prescribe, we begin the process with the rightmost digit [math]1[/math] of [math]R_4[/math], shown in blue in Fig. 1, and multiply it by all the digits of [math]R_5[/math]. The result of that multiplication is [math]m = 5[/math], also blue, digits long - all these digits are recorded as a new blue row aligned flush with the multiplicands.Ponder - why there are [math]m=5[/math] digits in the blue row? In other words, why is the blue row [math]m = 5[/math] columns wide?Moving right-to-left through the digits of [math]R_4[/math], we make the next, second from right, digit [math]1[/math] our current digit and we shift the result of its multiplication with all the digits of [math]R_5[/math] leftward by exactly one square with respect to the rightmost edge of the repunits [math]R_5[/math] and [math]R_4[/math] (Fig. 2):Above, we have shown the current digit of [math]R_4[/math] and the result of its multiplication with [math]R_5[/math] in the same, blue, color. Also note that we dispense with writing zeros - add them if that makes you feel comfortable.Repeat.Moving right-to-left through the digits of [math]R_4[/math], we make the next, third from right, digit [math]1[/math] our current digit and we shift the result of its multiplication with all the digits of [math]R_5[/math] leftward by exactly two squares with respect to the rightmost edge of the repunits [math]R_5[/math] and [math]R_4[/math] (Fig. 3):One more, and last, time.Moving right-to-left through the digits of [math]R_4[/math], we make the next, fourth from right, digit [math]1[/math] our current digit and we shift the result of its multiplication with all the digits of [math]R_5[/math] leftward by exactly three squares with respect to the rightmost edge of the repunits [math]R_5[/math] and [math]R_4[/math] (Fig. 4):Now that we have exhausted all the digits in [math]R_4[/math], we are done with the multiply-and-shift process.To record the final result of multiplying [math]R_5[/math] by [math]R_4[/math], we draw another horizontal line under the towers of ones and vertically sum all the digits that are located in the same column starting from the rightmost one (Fig. 5):Above we have recorded the sum of all the digits [math]1[/math] in the rightmost column.Next, we sum all the digits [math]1[/math] in the second rightmost column (Fig. 6):Next, we sum all the digits [math]1[/math] in the third rightmost column (Fig. 7):and so on until we sum all the digits in all the remaining columns (Fig. 8):We see that the result of multiplying [math]R_5[/math] by [math]R_4[/math] is a palindrome* - a string of, in this case, digits which read left-to-right in exactly the same way as they read right-to-left.We also see that the if we were to tentatively graph the digits of our palindrome over their positions in it then at first they form a finite (ascending) arithmetic progression with a common difference of [math]1[/math], then they plateau for some finite distance and then they diminish by forming a finite (descending) arithmetic progression with a common difference of [math]-1[/math] (Fig. 9):*Now that we have a, reasonably visual, way of multiplying two repunits and we understand how it works, we can answer some relevant questions assuming that [math]m\geqslant n[/math] and that [math]9\geqslant n[/math] - with these assumptions in place we are guaranteed that no digits will be carried over when we sum them vertically.Question: how many rows will be trapped between the two horizontal lines in Fig. [math]8[/math]?Answer: [math]n = 4[/math] - one row per each digit in the shorter repunit [math]R_n[/math] (Fig. 10):Question: how many shifted rows will be trapped between the two horizontal lines in Fig. [math]8[/math]?Answer: [math]n - 1 = 3[/math] - because the first row below the first horizontal line is not shifted and remains flush with the repunits [math]R_m[/math] and [math]R_n[/math] (Fig. 11):Question: how many columns, read - digits, will be in the product of two repunits [math]R_m[/math] and [math]R_n[/math] (with the [math]m\geqslant n[/math] and [math]9\geqslant n[/math] constraints)?Answer: [math]m [/math][math][/math][math]+ n - 1 = 8[/math] - because to the number of columns in the repunit [math]R_m[/math], which is [math]m[/math], we add the number of horizontal rightward shifts of the intermediate results of multiplication which, in this particular case is equal to the number of shifted rows, which we already calculated in the previous answer to be [math]n - 1[/math] (Fig. 12):Note that the number of horizontal shifts above, counted horizontally, is equal to the number of shifted rows, counted vertically. Thus - the formation of the blue square.Question: what will be the largest digit in the product of our repunits?Answer: [math]n = 4[/math] - because the number of rows whose digits are summed vertically is, as we already computed, equal to [math]n[/math] and, since we shift the intermediate rows by exactly one unit, which ensures that no digit is lost or skipped, there will always be at least one tower or at least one stack that is [math]n[/math] units tall.Because we require that [math]n\leqslant 9[/math], no digits will be carried over into the columns located to the left of the current one.Question: how many times will the largest digit in the product of our repunits be repeated?Answer: [math]m - n [/math][math][/math][math]+ 1 = 2[/math] - because from the total width of the product of our repunits, which we lready computed to be [math]m+n-1[/math], we need to subtract the number of columns which are not fully filled with units.How many such columns will there be?Well, there will be [math]n - 1[/math] such columns, shown in blue below, to the left of the group of digits [math]n[/math], shown in green, and another [math]n - 1[/math] such columns, shown in blue, to the right of the group of digits [math]n[/math] (Fig. 13):Thus:[math]x = m+n-1 -(n-1)-(n-1) = m - n [/math][math][/math][math]+ 1 \tag{2}[/math]It follows, then, that if we want to fix the value of the largest (or central) digit [math]n[/math] and grow the number of such digits in the output (the product of two repunits) then all we have to do is stretch the longer repuint [math]R_n[/math] even longer - like the bellows of an accordion (Fig. 14):For example, when we multiply [math]R_6[/math] by [math]R_4[/math] then the number of digits [math]n = 4[/math] will be equal to [math]3[/math]:[math]111111\cdot 1111 = 123444321 \tag*{}[/math]and when we multiply, say, [math]R_9[/math] by [math]R_4[/math] then the number of digits [math]n = 4[/math] will be equal to [math]6[/math]:[math]111111111\cdot 1111 = 123444444321 \tag*{}[/math]and so on.Conversely, if we want to fix the value of the largest (or central) digit [math]n[/math] and diminish the number of such digits then all we have to do is squeeze the longer repuint [math]R_n[/math] and make it shorter (Fig. 15):How short can we make the repunit [math]R_n[/math] and still ensure that the digit [math]n=4[/math] is present in it?According to (2) - we can keep shortening the length of [math]R_n[/math] until it is as short as [math]R_m[/math]. But then the operation of multiplication of two numbers of equal value becomes squaring:[math]1111\cdot 1111 = 1111^2 = 1234321 \tag*{}[/math]and we now understand how our little factory can produce palindromes of our liking.To produce an entertaining answer to your problem we just have to do a finite computation: because you demand that the central digit must be [math]7[/math], it follows that in our notation it must be the case that [math]m = n = 7[/math]:[math]1111111\cdot 1111111 = 1111111^2 = 1234567654321 \tag*{}[/math]Thus, an answer: a product of two numbers each of which is [math]1111111[/math] is equal to [math]1234567654321[/math].That answer, however, is not unique because the number [math]1111111[/math] itself is a product of two primes:[math]1111111 = 239\cdot 4649 \tag*{}[/math]Therefore, other pairs of natural numbers that multiply out to a number requested exist. For example:[math]239\cdot 5165555039 = 1234567654321 \tag*{}[/math]or[math]4649\cdot 265555529 = 1234567654321 \tag*{}[/math]We will make no comments about the entertaining value of the above factorizations.We also note that:[math]11\cdot 11 = 11^2 = 121 \tag*{}[/math][math]111\cdot 111 = 111^2 = 12321 \tag*{}[/math][math]1111\cdot 1111 = 1111^2 = 1234321 \tag*{}[/math][math]11111\cdot 11111 = 11111^2 = 123454321 \tag*{}[/math]and so on. How long can we keep doing that? Until [math]n[/math] reaches [math]9[/math]:[math]111111111\cdot 111111111 = 12345678987654321 \tag*{}[/math]A square of any repunit with [math]n>9[/math] will no longer be a palindrome.For the computationally-oriented readers: in this post of the Quora space By::Analogy we have shown an algorithm to reverse the order of characters in a (finite) string.It follows, then, that in order to detect if a string is a palindrome, numerical or otherwise, all we have to do is replace the action in the string reversal algorithm with a test, see lines 11–13 in the code listing below where we bail as soon as we detect that the two elements of a string equidistant from the string’s vertical axis of symmetry are distinct:#include <stdio.h> #include <string.h>   extern int isPalindrome( const char* str, size_t n ) {  size_t i;   for ( i = 0; i < n/2; i++ )  {  if ( str[ n - 1 - i ] != str[ i ] )  {  return 0;  }  }   return 1; }  extern int main( int argc, char* argv[] ) {  size_t n = strlen( argv[ 1 ] );  int ip = isPalindrome( argv[ 1 ], n );   printf( "%d\n", ip );   return 0; } Assuming that we have saved the above code in the ispalindrome.c disk file, build it like so:gcc -g -o ispalindrome ispalindrome.c and test it like so:./ispalindrome 1234321 1 or like so:./ispalindrome 12345321 0 Our program emits [math]1[/math] to standard out if a given string is a palindrome and [math]0[/math] otherwise.

The image from the question is a Skeuomorphy or a simulation of a Cattle Grid.Facing the fear of the classic Cattle Grid.Seen in farming areas around the world, the classic Cattle Grid (aka, Cattle Guards) is designed to motivate most animals with hoof feet to not cross the Grid for fear of the unstable walking area. The animals seem to have an aversion to the Grids with a fear that a hoof or entire leg may pass through the open space if they were to navigate the pass.This genius invention received a patent in January, 1915 by William J. Hickey. One day William noticed that his cattle would not pass over some fencing pipe he had laying on the path to the watering area on his Nevada farm. This sparked William to experiment with this as a barrier system for his cattle, and it worked.Many variations to the classic Cattle Grid system exist around the world. Most use a simple grid of pipe/bars or formed sheet metal interspersed by open areas with an average distance of about 4 inches compromising about 12 pipes. They are primarily used where traditional fencing will not be practical. Road surfaces are a typical and most common example. They are a very common occurrence in the open cattle regions of Texas in the US and Australia.Lines On The RoadThere is no clear history on who first started to use painted lines on the road to replace the classic Cattle Grid. Many have staked a claim as to being the first. It seems plausible that someone in the Cattle country of Texas noticed that some cattle would not cross over some painted stripes on a road. As the Folklore goes, this unnamed genius tested a hypothesis by painting white lines of equal distance horizontally across the road, and it worked.Why Do Painted Lines Fool Some Animals?Current animal psychology theory suggest that the lines present visually the same issues of navigation that the classic Cattle Grid presents. A study by Dr. Ted Friend (http://animalscience.tamu.edu/facultystaff/faculty/friend.htm), PhD. Faculty Fellow and Professor, Animal Behavior and Welbeing at Texas A & M University demonstrated that not only do animals that have been exposed to the classic Cattle Grid respond to the simulated Cattle Grid, but also animals that have never been exposed to the classic Cattle Grid also respond to the simulated Cattle Grid. The conclusion is that the fear of getting caught or falling off an edge moderates the animals.A very surprising aspect to the simulated Cattle Grid is the herd will begin to ignore a simulated Cattle Grid if at least one member of a small herd knowingly or unknowingly passes through the area. This does not require a herd leader, it just takes one, to break the spell of the painted lines. The herd size needs to be small for it to be effective. Although no deep studies have been conducted in this area, this opens up the question of how this information is transmitted.The best, rather unconventional, and controversial theory is from Rupert Sheldrake, PhD. (http://www.sheldrake.org/), a Fellow of Clare College, Cambridge, Director of Studies in biochemistry and cell biology and Rosenheim Research Fellow of the Royal Society. Rupert has postulated the concept of Formative Causation (http://www.sheldrake.org/Articles&Papers/papers/morphic/formative.html):According to my hypothesis of formative causation, organisms inherit habits from previous members of their species. This collective memory, I suggest, is inherent in fields, called morphic fields, and is transmitted through both time and space by morphic resonance, a process which takes place on the basis of similarity. From this point of view, cattle confronted for the first time by grids, or by things that look like grids, would tend to avoid them because of morphic resonance from other cattle that had learnt by experience not to try to cross them.Rupert is basing this on a number of studies, he has written 80 scientific papers and ten books on the subject, including regular articles to the Skeptical Inquirer Magazine. Rupert suggests:Morphic resonance would provide an explanation for such a rapid evolutionary process (aversion on non aversion to Cattle Grids). The main alternative would be a neoDarwinian speculation of doubtful plausibility. Assuming that mutant genes for avoiding cattle grids arose by chance, natural selection might have favoured them because animals lacking these genes would tend to injure themselves by trying to cross grids; and even if they did not die of broken legs, breeders would tend not to use them as parents, choosing instead uninjured, grid-avoiding cows and bulls. But, at least at present, there is no empirical evidence for the existence of mutant genes for grid-avoiding; nor that cattle grids have imposed a significant selection pressure.It is very important to note that although there are a number of Scientist and non-Scientist critics of the work that Dr. Sheldrake has performed for the Royal Society, thus far there has been no empirical studies that counter his work. Nor has any other member of the Royal Society produced any compelling alternative with robust data to support it. As more research produces better datasets we may find the deeper construct that is at play here.Our GridsEvery single animal on the planet exhibits a similar response to obstacles in some form. We may look at the simulated Cattle Grid and find some humor about how the “simple” brain of that animal works. We can also gain insight about how we all see the world and perhaps how we are contained by our very own Cattle Grids.After initial hesitation, this dog overcomes a fear of the Cattle Grid.

Lets look at some figures for the UK.The UK population is about 64 million. Assuming 4 people per home that 16 million homes.Currently about 2,500 home units are installed a month (Domestic solar installs hit new monthly low in October) and the industry employs 35,000 people (More than half of jobs in UK solar industry lost in wake of subsidy cuts). So thats 30,000 units a year. Already there are 670,000 domestic solar homes in the UK, http://www.solar-trade.org.uk/wp-content/uploads/2015/08/STA-Solar-Facts-and-Statistics-August-2015.pdfA typical domestic installation is rated at about 3–4 kW, and the current domestic capacity is 2.3 GW. This contrasts with ground mounted units in fields with a capacity of 6GW. (http://www.solar-trade.org.uk/wp-content/uploads/2016/11/Making-Solar-Pay-The-Future-of-the-UK-Solar-PPA-Market-1.pdf)If there was an increase the domestic installed base ten fold to 6.7 million homes might increase capacity tenfold as well to 23 GW. The solar trade association thinks there could be another million more solar homes by 2020.The real growth sector is in the ground mounted units. A 4 fold increase would give the same 23 GW capacity increase which we could easily see in 5 years.For a quora question this is surprisingly within the realm of reality. The impact of everyone installing solar PV will be less than we what would expect to see from the whole PV industry in the next five years!How would this affect the UK energy mix? Solar is known to be strongly hourly and seasonally dependant. So the headline rate of 11.5 GW installed capacity only corresponds to a peeks at about 7GW and in the summer averages to 1.6 GW generation in a day. A unit in shefield is estimating the Solar generation PV_Live Journal PaperWe can see live statistics for all UK grid electricity at GridWatch. The average UK demand in ranges from 30–40GW. Nuclear is pretty steady at about 8 GW, wind can go from zero upto 8GW and solar goes up to 4.7GW. Most of the difference is made up of Closed Circuit Gas Turbines and coal. CCGT is quite flexible and can be switched on and off quite rapidly, but on average nearly 50% of UK electricity comes from these sources.The basic impact of increasing installed solar capacity is that the gas turbines don’t need to be switched on quite as much. The modest increase in domestic solar PV would not really change the dynamics of this.Cost wise the Solar Trade association estimates it would add £1.70 on consumer bills in 2020 to deliver more than 1 million more solar homes in 2020.A few links from the US Elon Musk’s plan to refit 5m homes with solar roofs.In the US the solar industry is estimated at 200,000 people A good job if you can get it: America's solar workforce is heating up