Equation is … X=10+Log(20*X)

Loop 1 is =10+LOG(20*D4) the D4 is the guessed number above it. (Cell D4 where I guess 100)

Eash loop uses X for the number above it and before 20 loops it stops changing.

Guess a X 100 Changed =

Loop 1 13.30102999566400000000 -86.698970004336000

Loop 2 12.42491526851930000000 -0.876114727144710

Loop 3 12.39532343141620000000 -0.029591837103039

Loop 4 12.39428785897010000000 -0.001035572446172

Loop 5 12.39425157414140000000 -0.000036284828671

Loop 6 12.39425030272310000000 -0.000001271418261

Loop 7 12.39425025817260000000 -0.000000044550490

Loop 8 12.39425025661160000000 -0.000000001561048

Loop 9 12.39425025655690000000 -0.000000000054699

Loop 10 12.39425025655500000000 -0.000000000001917

Loop 11 12.39425025655490000000 -0.000000000000068

Loop 12 12.39425025655490000000 0.000000000000000

Loop 13 12.39425025655490000000 0.000000000000000

Loop 14 12.39425025655490000000 0.000000000000000

Loop 15 12.39425025655490000000 0.000000000000000

Loop 16 12.39425025655490000000 0.000000000000000

Loop 17 12.39425025655490000000 0.000000000000000

Loop 18 12.39425025655490000000 0.000000000000000

Loop 19 12.39425025655490000000 0.000000000000000

Loop 20 12.39425025655490000000 0.000000000000000

I did this easy and true computation about 3 years ago.

To compute X, when X is on both sides of the equation,

but one side, the X, and maybe other numbers are in a Log,

you can easily compute the X easy.

Example, find the X to several digits when X=10+Log(20*X).

Just guess an X and compute the =10+Log(20*X).

Then use the solution for the next X as a new loop.

Then if you guess the first X as 1 to 1000 the X can be

computed to many digits in about 8 loops.

So compute this X to about 10 digits.

Example: First use any number for the first X.

If you use X=1 for the first X the first loop will give 11.301029995664

If you use X=100 for the first X the first loop will give 13.301029995664

Then use the first loop X, for the next X loop. After some loops the

X will stop changing. But if you need about 100 digits, it might take

about 30 loops

Tags:

Easy and True
A Mr. Piyush has posted the following:

Find the missing number

5 : 24 :: 8 : x

Options are:

a. 65

b. 63

c. 62

d. 64

Three recipents R(0), R(1) and R(2) each contain an integer volume v_{0}(0) ≥ v_{0}(1) ≥ v_{0}(2) ≥ 1. Each recipient is large enough to contain the combined volumes. You are allowed to transfer some liquid from one recipient to another, only if the receiving one doubles its volume. Show that there is always a way to empty out one recipient in finitely many steps.

Example: If the initial volumes are 17, 8, 5 the sequence of volumes could be

R(0) R(1) R(2)

17 8 5

17 3 10

17 6 7

17 12 1

16 12 2

14 12 4

14 8 8

14 16 0 => R(2) is finally empty.

A function f takes a positive integer and returns another one by moving the leftmost digit to the right.

For example f(12345) = 23451

What is the smallest strictly positive integer n such that f(n) = 1.5 n

In a test involving yes/no answers, the probability that the official answer is correct is t, the probability of getting the real correct answer is b for a boy and g for a girl. If the probability that a randomly chosen boy or girl of getting the official answer to a question is 1/2, then what is the ratio of boys to girls who took the test?

Find all solutions in positive integers a, b, c to the equation

a! b! = a! + b! + c!

916238457 is a nine-digit number that has the property that has the digits 1 to 9 exactly once and the digits 1 to 5 are in counting order but the digits 1 to 6 are not. How many such numbers are there?

I have a very large plot of land on which I have stacked bales of hay, each measuring 1m x 1m x 1m. They are stacked 10 bales high, and span a couple thousand wide and deep. To keep them out of the weather I have also placed a tarp on top that cannot be seen through and should not be walked on or cut open in any way.

While stacking the hay I also stored several boxes of gold for safe keeping, but did not mark down where. However, to make it a little easier for my future self I grouped 8 boxes together so that they occupy a 2m x 2m x 2m space.

I now need some of my gold, but cannot retrieve it myself. I am hiring YOU to help me, and in return you will get to keep 1 gold bar from each of the boxes you return to me.

I need the gold available in a week, so time is somewhat important. You may work whenever you like, and at whatever pace you like. I do not have a minimum or maximum requirement.

Some other information:

- You must enter through the ‘side’ as to not disturb my tarp or the ground.

- You should remove at least 2 bales high so you can walk through, more if you wish.

- Bales above those removed will remain in place (magic?) – same with the tarp. [I suppose in theory you could remove all adjacent bales and 1 might ‘float’ right in front of you.]

- Once a bale is removed, a face of the 5 adjacent (but not diagonal) bales/boxes are revealed. You may remove any of which you can see the face.

Thoroughness is partially important (finding ALL of the gold within the search area), but Efficiency will yield the most gold found per hay bale removed.

The question: What is the most Efficient method of finding my gold? (ex: remove all bales, remove in a checker-board pattern, make a long 5-bale-high tunnel, etc.)

This info might also be helpful: When I stacked the hay I placed them in 16 x 16 x 10 piles each day with one group of gold boxes ‘randomly’ placed within the pile. The next day I placed another 16 x 16 x 10 pile directly adjacent along with another group of gold boxes somewhere within, and so on.

A class has six pairs of twins. The teacher wishes to set up teams for a quiz, but doesn’t want to put any pair of twins in the same team.

1) In how many ways can they be split into two teams of six?

2) In how many ways can they be split into three teams of four?