142857

A truly wonderful number that I came across while going through a book

If you multiply 142857 with anything between 1 - 6, you will get the same digits circularly shifted in their positions for equal number of places, irrespective of their direction of shift
Illustrating,
142857 x 2 = 285714
142857 x 3 = 428571
142857 x 4 = 571428
142857 x 5 = 714285
142857 x 6 = 857142
So here the question is why this happens??
The book provided hint in the form that this arrangement might be understood from identifying that this is the recurring decimal digits of 1/7, which might be used in summing the digits appearing in their respective infinite series, followed by manipulating the sides of the identities thus created
But a slight research with a wiki hint helped me to understand it in an even more independent style
Critically looking at 142857 reveals that this lovely number is consisting of double digits forms that either is a multiple of 14 or > a multiple of 14 by 1
Illustrating,
considering f = 14
42 = 3f
28 = 2f
85 = 6f + 1
57 = 4f + 1
71 = 5f + 1
Accordingly, this particular arrangement may be represented in the following form
142857 = 10⁴f + 2.10²f + (4f + 1)
Interestingly,
7f + 2 = 100
These give,
2 x 142857 = 2.10⁴f + 4.10²f + (8f + 2)
= 2.10⁴f + 4.10²f + 10² + f [remembering (7f + 2) = 100]
= 2.10⁴f + (4f + 1)10² + f
= 285714
Similarly,
3 x 142857 = 3.10⁴f + 6.10²f + (12f + 3)
= 3.10⁴f + 6.10²f + 10² + (5f + 1)
= 3.10⁴f + (6f + 1)10² + (5f + 1)
= 428571
Remaining products are left solely for the interested

Lights

When the housewife helps and the two primary school goers assist, electrical engineering can get indigenous.....

The recipe for LED chain is like this!!

You will need

1. LEDs, obviously!!
2. Power Supply, to supply the current to allow the LED to glow
3. Resistors, to limit the excess current that the Power Supply may indulge in supplying if not restricted!
4. Switches, again obvious to its literal meaning, to toggle between the ON/OFF states
5. Wires, a whole lot of them
6. Accessories will include soldering iron and solder wires, tapes, screwdrivers, etc.

Design steps

You need first to decide how many LED's you require in a chain and how many chains will be there.

For me, there were 5 chains as I got that numbers of differently coloured LEDs and decided each chain to have 10 LEDs each. Also, our decision was to connect the LED's in parallel, as though series connection would have been a bit simpler, but the philosophy would be that if a single LED gets disconnected, the entire chain must not stop working, much the same line of thought as applied to domestic housewiring.

Now, if each LED is allowed to consume ~ 4 mA of current, then each chain will drain 40 mA current from the source and when all the 5 are connected, it will amount to 200 mA current net. The DC power supply current rating must be well beyond that. 24 V, 2 A power supplies are available in the market and it was chosen.

Now, the scheme will be like this

V1 is the power source and Lx are the chains. The scheme of a chain is like the following

Each chain consists of 10 LEDs (DLx1-DLx10) connected in parallel, the entire assembly being series connected to a resistor Rx1 and a switch SWx1 and the terminals marked for +ve (+x) and -ve (-x) polarities. Now,  one thing must be kept in mind. LED, unlike normal bulbs are polar devices, i.e., it has a +ve and a -ve terminal. These markings are engineered while fabrication and please be careful to connect all +ve leads with one wire and the -ve leads with another. But don't worry, nothing will blow spontaneously if connected incorrectly, but some will remain OFF and some will glow with excessive brightness leading to its failure, if connected in the wrong way.

Now, coming to the choice of Rx1, it is a bit tricky than simply choosing a single resistor from the market. Actually, these comes in various standard ratings and you have to choose among them. For this, we need to know the supply voltage, the drop across the LED and the current that will flow through it. The drop across the LED is ~ 0.7 V (as this is a diode and their nature is to have tentatively this amount of voltage across them while conducting) and the power supply chosen has output rating of 24 V. So, the remaining (24 - 0.7) V  = 23.3 V is to be dropped across the resistor. Each chain is allowed 40 mA current. This will correspond to a resistance of (23.3/40) kΩ = 582.5 Ω. Its power dissipation will be 23.3 x 40 mW = 932 mW. Now, standard resistances in this range is 560 Ω, 680 Ω with dissipation rating of 250 mW, 500 mW. So, to enhance the power rating and matching the resistance value, this single resistor is comprised of multiple resistors in parallel combinations. For instance, if we parallel 4 pieces of 2.2kΩ/250mW resistors, the equivalent resistance will come as (2.2/4) kΩ = 550 Ω, allowing (23.3x23.3/2200) W = 246.8 mW dissipation in each resistor thus keeping it within the range.

The switch is used to isolate each chain from the other so that each can be disconnected in case a shorted fault occurs in any of them,

The best advantage of this circuit is that it is supplied with 24 V DC which can be further downgraded and the mains supply is limited to the wires to the power supply terminals, keeping remaining circuit in the safer range. But, it is electrical, so be safe and be supervised by some technical person while making it operational

Ready for lights camera action

Puzzled

It all started with an apparently innocent looking puzzle which ran as

11 x 11 = 4

22 x 22 = 16

33 x 33 = ?

I thought that the progression will be (3 + 3) x (3 + 3) = 36 which is satisfying the other two and posted it as a reply to the person who sent me. I also posted in on other groups. Several replied with 36 but two persons replied with 18. Their logic was

11 x 11 = sum of the digits of the product = 1 + 2 + 1 = 4

Likewise

22 x 22 = 4 + 8 + 4 = 16

which gives

33 x 33 = 1 + 0 + 8 + 9 = 18

Well I was adamant in agreeing to it and though it too complicated but could not find an error to the logic. So my logic of simpleness as a solution was not holding ground. I would have silently ignored the replies, putting a deaf ear to the logical replies of the solvers who came with 18, allowing my false belief in my logic of simple vs complicated to cloud the logical part of my grey cells, had it not been for one of the solvers who personally messaged me and requested me to think over the solution from his perspective. Additionally, being humble in the face of my obstinate replies, he additionally asked me that he multiplied the numbers and added it while we added and then multiplied and the latter result was twice of the former. So he wanted to find whether this is a general case, where the latter operations will always return a significantly higher value than the former. I consider myself fortunate that his insistence led me to the following where not only I could find a rationale behind the results coming like he hypothesized, considering special cases of the repeated digits only, but could also find the incompleteness in the actual puzzle statement.

What follows just onwards is my findings. Once again my acknowledgement to the person, my senior in my locality, who actually triggered me to carry out this interesting work

Case36 (think this is the apt way to label this cases)

aa x aa = (a+a) x (a+a) = 4a², for a = 1, 2, 3, …., 9

Case18 (likewise!)

aa x aa = sum of digits of (10a + a)² =  sum of digits of 121a², for a = 1, 2, 3, …., 9. So let us see how many digits it can go up to. The maximum number of digits will come from max(121a²) =  9801 i.e. it will consist of 4 digits in the extreme. The sum of the highest value of 4 digits = 9 + 9 + 9 + 9 (though this will never occur) = 36.

So it may safely be said that if

4a² > 36

or,  a > 3

then Case36 will always return a higher value than Case18.

Two special situations arise for a < 3

Case36 (1) = 4

Case18 (1) = sum of digits of 121 = 4

Case36(2) = 16

Case18(2) = sum of 484 = 16

But from a = 3 onwards the values will differ. So it can also be said that the puzzle is incomplete / will have logically multiple correct ans if only conditions of a = 1, a = 2 are specified.

The Home Made Theory to Estimating AC tonnage

Why I tried this?

After buying the 1^st AC, it intrigued me that how the ac tonnage is calculated. I’d done that with the various online calculators available in. Browsing thru several sites I cudn’t gather the exact calculation and as is always the case I tried the hard method viz., starting from the 1^st principles. I don’t know whether it’s correct but I think this will give a rough estimate, at least the min requirement, while buying ur AC. In case u think otherwise, plz help me out refining the model.

A bit of a history

The ton that is referred in AC manuals is actually a measure of power which is that power required by a short ton (i.e. 2000lb) of ice to be converted to water at 0⁰C in 24hr. Thus,

1 ton = mass of ice x latent heat of ice / time = 3.5kW.

Calculations

The room where the AC is to be installed must be measured for its length (L), breadth (B) and height (H). Thus the total volume (V) of air, considering it to be an ideal rectangular parallelopiped devoid of any furnitures (crazy right!) will be given by

V = LBH.

If u know the density (r_air) of air, assume a value that is denser nearer to room temp., which will give u a conservative value (a good thing for a safer engg design, though the term safe is relative!), u may calculate the mass of air needed to be cooled by

m_air = r_air x V.

For my design I assumed it to be 1.2kg/m³.

In general for any temp diff (DT) in Kelvin (the temp diff will be same in ⁰C and K), the amount of heat (DH_air) to be absorbed will be

DH_air = m_airS_airDT,

where,

S_air is the heat capacity of air.

Now generally u require AC to cool down to not more than a temp diff of 10⁰C.

Again a good AC assures that the cooling to any temp will be within 2min. Thus the power required for the cooling in Dt will be.

P_air = DH/Dt.

Now the model may be refined by adding other power devices (P_other) that u expect in the room. So if T TR be the tonnage of the AC,

T = (P_air + P_other)/3.5kW.

Remember the only hitches

U have to know certain physical constants and the SI system of units. If that’s too much then follow the shortcut given below.

Feedback

So plz tell me this helped!

Shortcut

For a 10' x 10' x 10' room u will require 0.818 TR according to my logic. Taking this apply the unitary method for any other volume.

Thanx

www.wikipedia.org

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