Saturday, May 04, 2019

Fulcrum Numbers

A Square Deal

Niranjan Ramakrishnan

Did you ever notice that the difference between successive squares is arithmetical progression of odd numbers?

It is. 1, 3, 5, 7, 9, 11,...This is just 1²-0², 2²-1², 3²-2², 4²-3², 5²-4², 6²-5²,...

The reason is straightforward - the difference between (n+1)² and n² is 2n+1,  which can also be expressed as n+(n+1). Thus,  given any square m², all you've got to do to get the next square is add m +(m+1) to m²! Pretty simple.

In this manner I managed to get by heart the squares of all the numbers from 1 thru 50.

But along the way I noticed a peculiar pattern.  There is symmetry about the number 25. Later we shall see that there is also an identical on about 50 and 75 too.

Take a look at this :

20² = 400, 30² = 900.
21² = 441, 29² = 841.
22² = 484, 28² = 784.

You can verify that this pattern holds for...  not only 23 and 24, which you would expect,  but all the way down to 1.

(25-24)² = 1, (25+24)² = 2401.

We can see two things:

1. For any two numbers equidistant from 25,  the last two digits of their squares are the same.
2. The difference between the two squares is a multiple of 100, and this multiple is the distance between 25 and either number.

Not only that, but you can also extend the logic to negative territory.  For example,  consider -2 , which is (25-27). If you take 52-squared, which is (25+27)²  (-2)²+2700, or 2704!

Fulcrum Numbers

Actually,  this property of a symmetry of squares of equidistant numbers on each side is common to all numbers ending in 0 or 5. Truly it holds for any number, but for 0 and 5 the multiple is a factor of 10, making the calculation simple.

As we shall see, for a number f about which we are looking at squares of equidistant numbers on each side, the difference between the two squares is the distance d times a constant multiple m. This multiple m is always f*4.

This is why, for the fulcrum number of 25, the multiple is 100. For 50, it will be 200, for 75, 300, etc.

The reason why is simple. Consider a fulcrum f,  and two numbers on either side of it at a distance d.  Obviously, the two numbers are (f+d)  and (f-d).

The difference between the two squares is then (f+d)²-(f-d)².
That is, (f²+2*f*d. + d²)  - (f² - 2*f*d +  d²),
which is

Now, it's easy to mentally reckon squares of numbers. E.g., what's the square of 182? Using 100 as the fulcrum, 182 squared is 82*400+18-squared, i.e.,  32800+324,or 33124.

Monday, July 09, 2018

Divisibility Formula for Numbers Ending in 5  - Proof

Niranjan Ramakrishnan
July 9, 2018

The previous post gave an empirical formula to determine if a number was divisible by a divisor ending in 5. This post offers a proof for the same.

Any number (10r+u) | (10x+5), if (r-x) | n for u=5, or r | n for u=0, where n=(10x+5)/5.

If n=(10x+5)/5, n= (2x+1).

Since we're only considering dividends ending in 0 or 5, it is sufficient to show that ((10r+u) / 5) | n.

When u=5, (10r+u) /5 = (2r + 1).
(2r+1) can be rewritten as (2r -2x +2x + 1).
Thus if (r-x) |  n, then (2r + 1) | n, because n=2x+1.

When u=0, (10r+u) / 5 = 2r.
If r | n, then 2r | n.

Saturday, June 02, 2018

The Divisibility Rule for Numbers ending in 5

Niranjan Ramakrishnan
June 2, 2018

A previous post, Divisibility - A more general approach, provided rules for checking divisibility of a number by any odd number not ending in 5. This post addresses that gap.

In that post I had stated that after trying briefly to discern a pattern for divisors ending with 5, I'd given up when no obvious pattern suggested itself. Well. This morning I gave it some more thought, and was thrilled to f ind a simple, consistent - and in the end almost tautological - pattern for
divisors ending with 5.

The divisibility rule for 5 itself is almost trivial - it divides any number ending with either 5 or 0. The only problem is, this doesn't tell us anything about checking for divisibility by 15, 25, 35, 45, etc.

To consider the problem I first set down the odd multiples of 15, 25, 35, etc., getting:

15: 15, 45, 75, 105, 135,...
25: 25, 75, 125, 175, 225,...
35: 35, 105, 175, 245, 315,...

If we omit the 5 at the end of each number a pattern begins to emerge.

15: 1, 4, 7, 10, 13,...
25: 2, 7, 12, 17, 22,...
35: 3, 10, 17, 24, 31,...

True, it's an arithmetic progression just as was the previous set of series  (hardly a surprise in a multiplication table). But the second batch of number sequences with smaller items is easier to
 comprehend and mine for a pattern.

For a start we see that for 15 the difference between the terms is 3, for 25, 5 and for 35, 7.

Thus, the difference between successive terms in each case is our divisor (recall that we are only dealing with divisors ending in 5) divided by 5.

We are now ready to formulate the rules, but first some definitions.

We shall use the vertical bar, |, to signify divisibility. That is, a|b means a is divisible by b.

Let's call the divisor (10x +5) . Therefore the divisor can also be considered as 5*n (thus, n is 3 for divisor 15, 5 for divisor 25, 7 for divisor 35, etc.).

The dividend in this case is of the form (10r+u), where u can only be 5 or 0.

The formula is simple.
For u=5,
(10r+u) | (10x + 5 )  if (r-x) | n

If u=0,
(10r+u) |  (10x+5) if r|n.

Let's look at some examples, one for u=5 and one for u=0.

Is  7645  |  55?
Here, r=764, u=5, x=5, n=11.
Since u=5, is (r-x) | n?
Is (764-5) | 11?
759 | 11, so 7645 |  55.

Is 15290 |   695?
Here, r=1529, u=0, x =69, n=139.
Since u=0, is 1529 |  139?
It is. So 15290 |  695.

Tuesday, May 15, 2018

Divisibility by Numbers ending in 1, 3, 7, and 9 - Proofs of the Formulas

Niranjan Ramakrishnan
May 15, 2018

The previous post provided a set of four empirical formulas to ascertain if a number is divisible by a divisor ending in 1, 3, 7 or 9. This post provides a proof for each of the formulas. Many grateful thanks to Dr Ranjan Roy, who was kind enough to indicate how the proof might be evolved, most importantly the difference of 1 from a multiple of 10.

Notation: The vertical bar, '|' is used here to signify divisibility. E.g., x|y means x is divisible by y.

Let's start by setting down a few identities.
10 (x)-1(10x+1) = -1             [1]
10(3x+1)-3(10x+3) = 1         [2]
10(3x+2)-3(10x+7) = -1      [3]
10(x+1)-1(10x+9) = 1           [4]

Without losing generality, we can represent the numbers as (10r+ u), (10x + 1), (10x + 3), (10x +7), (10x + 9), etc.

For divisors ending in 1
If (r-m*u)|(10x+1), then (10r+u)|(10x+1).
where m=x.
If (r-m*u) | (10x + 1) then 10(r-m*u) | (10x + 1).
i.e., (10r - 10m*u)| (10x + 1), or, (10r-10x*u)| (10x + 1).
i.e., (10r-(10x+1-1)*u)| (10x + 1).
i.e., ((10r+u) - (10x + 1) * u) | (10x + 1).
Since (10x + 1) * u is divisible by (10x + 1), it follows that (10r+u) ,| (10x + 1).

For divisors ending in 3
If (r+m*u)  | (10x + 3) then (10r+u) | (10x + 3)
where m=(3x + 1).
If (r+m*u)  | (10x + 3) then (10r+10m*u) | (10x + 3).
Substitution for m gives
(10r +  10(3x + 1) *u)| (10x + 3).
But as shown above,,
10(3x + 1) =  3(10x + 3) +  1.
(10r + (3(10x + 3) + 1)*u) | (10x + 3).
i.e., ((10r+u) +3(10x + 3) * u)) | (10x + 3).
Since 3(10x + 3)*u ) | (10x + 3), it follows that
(10r + u ) | (10x + 3).

For divisors ending in 7
If (r-m*u) | (10x +7), then (10r + u) | (10x +7), where m =(3x + 2 ).
If (r-m*u) | (10x +7), then (10r - 10m*u) | (10x +7).
Substitution for m gives
(10r - 10(3x + 2) *m) | (10x +7).
But as noted above,
10(3x + 2)  = 3(10x +7) - 1.
Making this substitution yields
(10r - (3(10x +7) - 1) *u) | (10x +7).
Rearranging terms,
((10r+u) - (3(10x +7)*u)) | (10x +7).
Since the second term (3(10x +7) * u) | (10x +7),
(10r+u) | (10x +7).

For divisors ending in 9
If (r+m*u)  | (10x + 9) then (10r+u) | (10x + 9) where m=x+1.
If (r+m*u) | (10x + 9), then 10(r+m*u) | (10x + 9).
i.e., (10r +10m*u) | (10x + 9).
Substitution for m gives
(10r + 10(x+1)*u) | (10x + 9).
But, as noted above,
10(x + 1) = (10x + 9) +  1.
Substitution yields
(10r + ((10x + 9) + 1 ) * u) | (10x + 9).
Rearranging terms,
((10r + u ) + (10x + 9)) | (10x + 9).
The second term, is obviously divisible by (10x + 9).
Therefore, so must be the first team.
(10r + u) | (10x + 9).

Thursday, May 10, 2018

Divisibility - A more general approach
Niranjan Ramakrishnan
May 8, 2018

Everyone is taught some basic rules for checking if a number is divisible by 2, 3, 4, 5, etc . Any even number is divisible by 2. For 3, if the sum of its digits is divisible by 3, so is the number itself. If the number formed by its last two digits can be divided clean by 4, the number is divisible by 4. Any number ending in either a zero or a 5 is divisible by 5. Any An even number divisible by 3 is divisible by 6. The rule for 8 is similar to that for 4, except that the number tested is that formed by the last three digits - not the last two. The rule for 9 is identical to that for 3 - if the sum of the digits is divisible by 9, so is the number. The test for 10 is trivial - any number ending in a 0. One other rule is taught: if total of the odd positioned and even-positioned digits is either equal or differs by a multiple of 11, the number is divisible by 11.

If still awake a reader might notice that the rules summary is missing any mention of 7. This is because in teaching rules of divisibility the number 7 stood in some bad odor, unamenable to  ready rulemaking. We used to declare, as our elders nodded mournfully, that there was no rule for 7.

But there does exist a divisibility rule for 7, as I learned from the Internet some years ago. Yesterday I stumbled on the fact that the rule for 7 contains an approach that blows open a path to check divisibility more general than the techniques we were taught.

Let's start with the divisibility formula for 7. First some definitions. Three concepts are involved. The number being tested is broken into two parts - the last digit, the number in the units or 1's place, which we shall call 'u', and the rest of the original number, which we shall call, 'r'. That's two of the three concepts.

E.g., if our number is 266, u=6 and r=26.
The algorithm for7 is as follows: if r-2*u is divisible by 7, then so is the original number.
Applying this to 266, r-2*u = 26-2*6 = 26-12 = 14
Since 14 is divisible by 7, so is 266 (7 times 38).
Note that a subtraction result of 0 also means the number is divisible, as zero is divisible by 7 (or by any Natural number).
E.g., 84. r=8, u=4. r-2*u gives 0.

The third concept is the multiplier 'm', 2 in this algorithm for 7. How exactly 'm' is to be calculated will be discussed shortly.
I rested content in this knowledge until yesterday, when a sudden email arrived from my sister, comprising a single line, "What is the formula for divisibility by 17?"

I didn't know, of course. Wondering if there wasn't something for 17 along the lines of the formula for 7, I  began to explore in my mind various combinations and to my surprise chanced upon one which seemed to work. All that was required was to use 5 as the multiplier in place of 2. Thus for 17 itself, r=1, u=7,m=5, and r-m*u = -34. For 102, 10-5*2= 0!

This success emboldened me look at 27, and I found that a multiplier of 8 worked beautifully. It seemed to be a series - 2(7), 5(17) 8(27), 11(37), 14(47), etc. in short, m=3x+2, where x is 0 for 7, 1 for 17, 2 for 27, 3 for 37,...25 for 257, etc.

Next I tackled 13, the next prime number without a divisibility formula (that I knew of). Turned out to have one, but of the form r + m * u, i.e., a plus instead of a minus. For 13, m=4. For 23 m= 7, etc. the general expression being m=3x+1, where x is 0 for 3, 1 for 13, 2 for 23, ...25 for 253, etc.

Then for numbers ending in 9, the formula is m=x+1, where x is again 0 for 9, 1 for 19,... 25 for 259, etc.  We have to check r+m*u,  just like for numbers ending in 3.

Finally,  numbers that end in 1. Here, m=x,  where x is 0 for 1, 1 for 11, 2 for 21, 3 for 31,... 25 for 251, etc.  The number to test is r-m*u.

Deciding to press my luck I ventured into numbers ending in 5. No obvious pattern suggested itself.

But there is an alternating pattern to the four number-endings,.  1 (r-m*u),  3 (r+m*u),  7 (r-m*u)  and 9 (r+m*u).

I'm sure students of mathematics are familiar with these formulas,  but I certainly hadn't encountered rules for 7, 13, 19, etc.  during my several years of science and mathematics in school and college,  which is why the utter simplicity - and generality -  brought so much delight when I uncovered them.

One example each of divisors ending in 1, 3, 7 and 9.

Is 372 divisible by 31?
Formula: r-m*u and m=x.
Here,  r=37, u=2, m(=x)=3.
r-m*u = 37-2*3 = 37-6  = 31. Divisible!

Is 559 divisible by 43?
Formula: r+m*u,  where m=3x+1.
Here,  r=55,  u=9, m(3x+1)=13.
r+m*u = 55+13*9 = 55+117 = 172. 172 = 4 * 43. Divisible!

Is 1164 divisible by 97?
Formula: r-m*u,  where m=3x+2.
Here,  r=116, u=4,m(3x+2)=29.
r-m*u = 116-29*4 = 116-116 = 0. Divisible!

 Is 1157 divisible by 89?
Formula: r+m*u,  where m=x+1.
Here, r=115, u=7, m(x+1)=9.
r+m*u = 115+9*7 = 115+63 = 178. 178 is 2 times 89. Divisible!

Wednesday, September 28, 2011

The Key to Sussex?

Two royals seeing eye to eye

by Niranjan Ramakrishnan

Shortly after news  of Pataudi's death, a friend of mine sent me the following one-line email:

Who was the other cricketer with one eye?........Ranjitsinhji.

I thought this didn't make any sense. I had read somewhere long ago how Ranji's cousin Duleepsinhji, on first going to England, was told by some doctor that he had a problem with his eyesight.  His house master dismissed any such notion saying that no relative of Ranji could possibly have anything wrong with his eyes.  Besides, I reasoned, it's one of those things you expect would be common knowledge if true.

Ranjit Singh
Then it occurred to me my friend might be joking.  He was talking, no doubt, about Maharaja Ranjit Singh, who did indeed have only one eye!  One of the Yehudi-Menuhin-is-a-violinist. Mahatma-Gandhi-is-a...non-violinist  variety, it seemed.

After dashing off a clever note to my friend saying I wasn't aware that Maharaja Ranjit Singh played cricket, something impelled me to read up on Ranji just to be sure. On Wikipedia at first glance, there was lots about his time in England, his cricket of course, and his disputes over the title to his principality of Nawanagar.  There was no prominent mention of any business of making do with one eye, etc.

As I read through the Wikipedia page, though, I found the following passage deep in its bowels:

"When the First World War began in August 1914, Ranjitsinhji declared that the resources of his state were available to Britain, including a house that he owned at Staines which was converted into a hospital. In November 1914, he left to serve at the Western Front, leaving Berthon as administrator.[note 9][209]  Ranjitsinhji was made an honorary major in the British Army, but as any serving Indian princes were not allowed near the fighting by the British because of the risk involved, he did not see active service. Ranjitsinhji went to France but the cold weather badly affected his health and he returned to England several times.[210] On 31 August 1915, he took part in a grouse shooting party on the Yorkshire Moors near Langdale End. While on foot, he was accidentally shot in the right eye by another member of the party. After travelling to Leeds via the railway at Scarborough, a specialist removed the badly damaged eye on 2 August."

Don't ask me how an eye damaged on 31 August needed removing on 2 August. I'm merely quoting Wikipedia verbatim.

This was long after his prime cricketing years. He had played his last test match in 1908, and seems to have last played serious county cricket in 1912. He played for Sussex, even captaining it briefly (as did Pataudi).

For all that it is my friend who will have the last laugh. Wikipedia again:

"Ranjitsinhji's last first-class cricket came in 1920; having lost an eye in a hunting accident, he played only three matches and found he could not focus on the ball properly. Possibly prompted by embarrassment at his performance, he later claimed his sole motivation for returning was to write a book about batting with one eye; such a book was never published.[166]"

Was there any famous cricketer (other than Pataudi) from India who played with a visual handicap? Well, you can bet your right eye on it.

Niranjan Ramakrishnan is a writer living in the USA. He can be reached at

Monday, September 19, 2011

Who controls your food supply?

The Food Bandits
"The number of hungry people has soared to nearly 1 billion, despite strong global harvests...Just four companies control at least three-quarters of international grain trade; and in the United States, by 2000, just ten corporations—with boards totaling only 138 people—had come to account for half of US food and beverage sales. Conditions for American farmworkers remain so horrific that seven Florida growers have been convicted of slavery involving more than 1,000 workers. Life expectancy of US farmworkers is forty-nine years."
Read full article... 

Nation Magazine's upcoming Oct 3 issue carries an anchor piece by Francis Lappe Moore (author of Diet for a Small Planet), along with replies by well-known experts on the topic of food security: Raj Patel, Vandana Shiva, Eric Schlosser, and Michael Pollan.

That this is a vital issue of both individual liberty and national sovereignty is without question. That it is discussed so little is a reflection on our myopia.

Raj Patel in his piece Why hunger is still with us says,
"[W]e’re growing more crops than ever before not for direct human consumption, or even animal feed, but as biofuels, to keep cars on the road. Already, more than a tenth of the world’s total coarse grain output is used for fuel, and the OECD predicts that within a decade a third of all sugar cane grown on earth will be used not for sweetening but for combustion."
  Eric Schlosser places the problem in larger context,
"The corporate monopolies and monopsonies, the contempt for labor unions, the capture of federal agencies, the corruption of elected officials, the lies routinely told to consumers, the disregard for the environment and for public health—none of these things are unique to the food industry. You will find them in the oil, chemical, media and financial industries, among many others. They have become commonplace in the US economy. They are signs of a much larger problem, of a society where a handful of corporations choose the lawmakers, dictate the laws, control production and distribution, widen the gulf between rich and poor."
And increasingly, one might add, none of these things are unique to the United States either. As Vandana Shiva says of the situation in India,
"But the biggest threat we face is the control of seed and food moving out of the hands of farmers and communities and into a few corporate hands. Monopoly control of cottonseed and the introduction of genetically engineered Bt cotton has already given rise to an epidemic of farmers’ suicides in India. A quarter-million farmers have taken their lives because of debt induced by the high costs of nonrenewable seed, which spins billions of dollars of royalty for firms like Monsanto."
Far more significant than who wins in 2012, don't you think?