19 കൊണ്ടൊരു മാജിക്ക്

>> Monday, October 19, 2009


ഇന്ന് ഒക്ടോബര്‍ 19. നമുക്കൊരു കളിയാകാം. അല്ലേ? കളിയും പഠനവും ഒത്തൊരുമിപ്പിച്ച ഒരു മാജിക് എന്നും ഇതിനെ പേര് വിളിക്കാം. ഈ മനോഹരമായ ഗണിതവിസ്മയം നമുക്കയച്ചു തന്നത് കോഴിക്കോട് വടകര, അരീക്കുളം കെ.പി.എം.എസ്.എം.എച്ച്.എസിലെ അധ്യാപകനായ N.M വിജയന്‍ സാറാണ്. വിജയന്‍ സാറിന് നന്ദി പറഞ്ഞു കൊണ്ട് നമുക്കീ കളിയിലേക്ക് കടക്കാം.

19 കൊണ്ട് ഹരിക്കാം കളിക്കാം...
ഇതോടൊപ്പം നല്‍കിയിരിക്കുന്ന ചിത്രം കണ്ടോ? അതില്‍ നിങ്ങള്‍ക്ക് വിവിധ സംഖ്യകള്‍ ഉള്‍​പ്പെടുന്ന മൂന്ന് ഓര്‍ബിറ്റുകള്‍ കാണാന്‍ കഴിയും. അകത്തെ ഓര്‍ബിറ്റില്‍ 19 മാത്രമേ ഉള്ളു. അതിന് പുറത്തുള്ള രണ്ടാം ഓര്‍ബിറ്റില്‍ ഘടികാരദിശയില്‍ (Clockwise direction) 1, 10, 5, 12, 6, 3.............. എന്നും ഏറ്റവും പുറമെയുളള മൂന്നാം ഓര്‍ബിറ്റില്‍ 0, 5, 2, 6, 3, 1, 5........... എന്നും കണ്ടല്ലോ.

ഇനി നിങ്ങള്‍ക്ക് ഈ ചിത്രം നോക്കി 1/19, 2/19, 3/19, 4/19, .... , 18/19 എന്നിവ എളുപ്പത്തില്‍ കണ്ടെത്താം. നിങ്ങള്‍ക്ക് 1/19 ആണോ വേണ്ടത്? എഴുതിക്കോളൂ.. ..052631578947368421

6/19= .315789473684210526
14/19= .736842105263157894

ഉത്തരമെഴുതാന്‍ രണ്ടാം ഓര്‍ബിറ്റില്‍ നിന്നും അംശം (Numerator) കണ്ടെത്തുക. അതിനു നേരെ മൂന്നാം ഓര്‍ബിറ്റിലുള്ള സംഖ്യ മുതല്‍ ഘടികാരദിശയില്‍ എഴുതുക. മുകളിലെ മൂന്നുദാഹരണങ്ങളും ഉപയോഗിച്ച് ഉത്തരം ശരിയല്ലേയെന്ന് നോക്ക്യേ... വിശ്വാസമായില്ലേ? കാല്‍ക്കുലേറ്ററുണ്ടോ കയ്യില്‍. ഇല്ലെങ്കില്‍ മൊബൈലിലോ കമ്പ്യൂട്ടറിലോ നോക്കൂ. 9/19 എത്രയാണെന്ന് കാല്‍ക്കുലേറ്ററില്‍ നോക്കിയ ശേഷം 9 നു നേരെ മൂന്നാം ഓര്‍ബിറ്റിലുള്ള സംഖ്യമുതല്‍ ഘടികാരദിശയില്‍ എഴുതി നോക്കിക്കോളൂ. എങ്ങനെയുണ്ട് വിദ്യ?

എങ്ങനെയാണ് ഈ ടേബിള്‍ നിര്‍മ്മിച്ചത്? വളരെയെളുപ്പം. 1 എന്ന് രണ്ടാം ഓര്‍ബിറ്റില്‍ എവിടെയെങ്കിലും എഴുതുക. ഘടികാരവിപരീതദിശയില്‍ (Anticlockwise direction) അതിന്റെ ഇരട്ടി എഴുതുക. അതിന്റെ ഇരട്ടി തൊട്ടപ്പുറം ഇതേ പോലെ തന്നെ എഴുതുക. എപ്പോളിത് 19 ലും മുകളില്‍ പോകുന്നുവോ അപ്പോള്‍ എഴുതേണ്ടത് ആ സംഖ്യയില്‍ നിന്ന് 19 കുറച്ച ശേഷം മാത്രം. എഴുതിയതിന്റെ ഇരട്ടി തൊട്ടപ്പുറം എഴുതുക. അങ്ങനെ രണ്ടാം ഓര്‍ബിറ്റില്‍ 18 സംഖ്യകളാകുമ്പോള്‍ നിര്‍ത്തുക.

ഉദാഹരണത്തിന് 16 ന്റെ ഇരട്ടി എഴുതുമ്പോള്‍ 32. 32 ല്‍ നിന്നും 19 കുറക്കുക. ഉത്തരം 13 അത് ഓര്‍ബിറ്റില്‍ ഘടികാരവിപരീതദിശയില്‍ (ആന്റി ക്ലോക്ക്​വൈസ് ഡയറക്ഷനില്‍) എഴുതുന്നു. വീണ്ടും 13 ഇരട്ടിയാക്കുമ്പോഴോ 26. 19 കുറക്കുക, ഉത്തരം 7.

മൂന്നാം ഓര്‍ബിറ്റും വളരെയെളുപ്പം തയ്യാറാക്കാം. രണ്ടാം ഓര്‍ബിറ്റിലുള്ളത് ഇരട്ടസംഖ്യയാണെങ്കില്‍ നേര്‍പകുതിയാണ് അതിന്റെ നേരെ മൂന്നാം ഓര്‍ബിറ്റില്‍ എഴുതേണ്ടത്. ഒറ്റസംഖ്യയാണെങ്കിലോ, സംഖ്യയില്‍ നിന്നും 1 കുറച്ച് നേരെ പകുതി എഴുതുക. അതായത് രണ്ടാം ഓര്‍ബിറ്റില്‍ 1 ആണ് ഉള്ളതെങ്കില്‍ അതിന് നേരെ 0 എഴുതുക. 3 ആണെങ്കില്‍ 1 (മൂന്നില്‍ നിന്ന് 1 കുറച്ച് പകുതി), 11 എങ്കില്‍ 5, 12 എങ്കില്‍ 6 (ഇരട്ടസംഖ്യയ്ക്ക് നേര്‍പകുതി)

എങ്ങനെയുണ്ട് 19 നെ മെരുക്കുന്ന വിദ്യ? ഇപ്രകാരമുള്ള ടേബിളുകള്‍ 1/7, 1/29, 1/39, 1/49 എന്നിവ കണ്ടെത്താന്‍ ഉണ്ടാക്കിയെടുക്കാം. ശ്രമിക്കില്ലേ?

58 comments:

Anonymous October 19, 2009 at 6:14 AM  

Interesting

Geetha

JOHN P A October 19, 2009 at 6:15 AM  

thank you vijayan sir . I am trying to make similar aiths other numbers. Also an effort to think how it works.Also I will try to provide other plays with numbers

Anonymous October 19, 2009 at 6:23 AM  

Wow,
We're waiting for it, John Sir!


Geetha

Anonymous October 19, 2009 at 7:49 AM  

wonderful ..there may be some more maths behind it..

Unknown October 19, 2009 at 8:08 AM  

thanks and hearty congrats....
really wonderful experiment.

vijayan October 19, 2009 at 8:21 AM  

draw 18 diametres in the diagram
you can notice one more speciality

you add the two numbers, you will get 19&9, which is on the opposite sides
eg:(1) 8,16,3,1 falls in one line
8+1=9,16+3=19
(2): 4,9 10,5 falls in another line 4+5=9,9+10=19.
lllrly you get 18 different pairs.
try ,try you will get more &more.....vijayanlarva

AZEEZ October 19, 2009 at 9:50 AM  

Fentastic

Now we are waiting for John Sir.

JOHN P A October 19, 2009 at 12:49 PM  

I found some surprising result from this ready reckner.Shall I call this as O M D
the orbit model of division ?
I calculated 1/7 2/7 3/7 4/7 5/7 6/7
The same digits 1 4 2 8 5 repeats
1/17 ,2/17 , 3/17 .........16/17
the asme digits 58823529411764 repeatsin all caases
1/29 2/29 .............28/29 gives the same digits ( may be) did not get time to check
I think these are remarkable results
ARE ALL PRIMES HOLD THIS PROPERTY ? I DONT KNOW .IT MAY BE
THERE IS A INVISIBLE PROPERTY BEHIND THIS

Anonymous October 19, 2009 at 5:03 PM  

Wonderful Vijayan Sir & John Sir. It's really amazing.

Lalitha
Samooham HS
N Paravur

JOHN P A October 19, 2009 at 6:43 PM  

again
1/89 = .0112358(13)(21)(34)(55)(89)(144)(233)........
I made a small omission
Yet , it makes a FIBONACCI sequenceafter decimel

vijayan October 19, 2009 at 7:04 PM  

I request all of my well wishers to go through 6 th comment and draw pictures that is enough to a 5 th std student.1/7,1/17, 1/29/ 1/39, 1/49.....are very very easy.these numbers have other properties also.group the numbers into two.add you will get 9999999........so spend one night and publish all your comments .thank you every body

vijayan October 19, 2009 at 8:34 PM  

dear john sir ,1/89= .01123595505617977528089887640449438202247191 again the same group (44digits)repeats.one way to find
(remember) this number. write 1 ,extremeright.multiply it with 9,then write left to it.then multiply 9 with 9, then write 1 left side.repeat till you get zero .that is 1/89. you can have diagramatic representation also .
one thing more :u can note the last eight number group doubles to its left side again and again the group doubles.

JOHN P A October 19, 2009 at 10:23 PM  

ok now we reach Fibonacci squence.Shall I ask one question
What is the sum of the FEBONACCI SEQUENCE
1,2,3,5 ...............1597?

JOHN P A October 19, 2009 at 10:27 PM  

sorry
1, 1, 2,3 ,5,8 ............1597
Find the sum

muraleedharancr October 19, 2009 at 11:10 PM  

very good sir
Do you know about the following number
105263157894736842
if the rightended digit '2'is placed to left ended the number become twiced
ie
105263157894736842*2=210526315789473684
Also
157894736842105263*2=315789473684210526
210526315789473684*2=421052631578947368
263157894736842105*2=526315789473684210
315789473684210526*2=631578947368421052
368421052631578947*2=736842105263157894
421052631578947368*2=842105263157894736
473684210526315789*2=473684210526315789

muraleedharancr October 19, 2009 at 11:17 PM  

there is a correction inthe last no.
ie 473684210526315789*2=947368421052631578

vijayan October 20, 2009 at 5:57 AM  

dear murali sir ,your number group is also in the picture.that is nothing but 2/19.if you multiply it by 2,4,6,...you will get the other group.

JOHN P A October 20, 2009 at 6:10 AM  

Shall i give a number Puzzle and hint to solve it. Otherwise readers may find it difficult
MY PHON NUMBER HAS SEVEN DIGITS.IT IS VERY DIFFICULT TO REMEMBER THAT NUMBER. MY FRIEND ,HE IS GOOD IN MATHEMATICS TOLD ME " if the last four digits are placed in front of the remaining three we getone more than twice that number
WHAT IS THAT NUMBER?

number = 10000 x + y
Changed no 1000y + x
20000x + 2y + 1 = 1000y + x
998y -19999x= 1
19999 = 998 * 20 +39
998 = 39 * 25 +23
...................
................
....................
ans
8717435 = 1+2*4358717
THIS IS A GOOD MATHS CLUB ACTIVITY

Anonymous October 20, 2009 at 6:27 AM  

Excellent!
I learned a lot from the comments of these Math Legends! (Vijayan, John, Murali, Azeez, etc.)
I'm not a math teacher, but interested!

Geetha

muraleedharancr October 20, 2009 at 6:54 AM  

998y-19999x=1
the above eqn is solved using 'continued fraction'
am I correrct ?

AZEEZ October 20, 2009 at 12:52 PM  

Find a 10 digit number such that:

The first digit tells how many zeros are in the number,

The second digit tells how many ones are in the number,

The third digit tells how many twos are in the number, etc.

vijayan October 20, 2009 at 2:53 PM  

why are you deviating the content?
dear john sir ,I think febinocci series and 1/89 are entirely different.febinnocci is ' 0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765....
11/89= .01123595505617977528089887640449438202247191.(the group repeats).
these two are like 'sambar and payasam' .don't try to mix both tasty foods. if you mix ,both may lost their taste and become another taste.
so let geetha,lalitha ,murali,azees and etc .play with the beauty of 19.
I ,along with bloggers, expect more from you john sir.......thank you.

vijayan October 20, 2009 at 2:53 PM  

why are you deviating the content?
dear john sir ,I think febinocci series and 1/89 are entirely different.febinnocci is ' 0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765....
11/89= .01123595505617977528089887640449438202247191.(the group repeats).
these two are like 'sambar and payasam' .don't try to mix both tasty foods. if you mix ,both may lost their taste and become another taste.
so let geetha,lalitha ,murali,azees and etc .play with the beauty of 19.
I ,along with bloggers, expect more from you john sir.......thank you.

vijayan October 20, 2009 at 3:32 PM  

just received
" total of first n terms in febinnocci series=(n+2)
nd term-1"

find the sum of first 6 terms=0+1+1+2+3+5=12
8 th term= 13
13-1=12
find sum of first 18 terms=4180.
2o th term is 4181.
4181-1=4180

AZEEZ October 20, 2009 at 8:07 PM  

VIJAYAN SIR & JOHN SIR

WHERE IS MY 10 DIGIT NUMBER?

VERY SIMPLE

TRY

vijayan October 20, 2009 at 9:31 PM  

mr azeez,don't deviate from the beauty of 19 .your answer is "9000000000".if you accept the answer find the relation between sambar and payasam.

JOHN P A October 20, 2009 at 10:57 PM  

The sum of the digits of Ramanujan number is 19 . At the same time the Ramanujan number is divisible by 19

muraleedharancr October 21, 2009 at 5:29 AM  

Azees sir.Your reqwuired is
6210001000,?

muraleedharancr October 21, 2009 at 5:31 AM  

Azees sir.Your required no.is
6210001000,?

AZEEZ October 21, 2009 at 1:01 PM  
This comment has been removed by the author.
AZEEZ October 21, 2009 at 1:24 PM  

Dear Muralidharan Sir,

your answer is right.

Dear Vijayan Sir.

Pls check your answer.

AZEEZ October 21, 2009 at 1:26 PM  

Now try this.

Consider the fraction 1/6=.1666........

This is in the form of A/B=.ABBBBBBB.........

CAN YOU FIND ONE MORE FRACTION WITH THE SAME PROPERTY?

Areekkodan | അരീക്കോടന്‍ October 21, 2009 at 2:58 PM  

ഇവിടെ ആരും മലയാളം പറയില്ലേ?കണക്കും അതില്‍ തൂങിയുള്ള മറ്റു കണക്കുകളും ഇഷ്ടപ്പെട്ടു.

Anonymous October 21, 2009 at 3:12 PM  

അരീക്കോടന്‍ സാര്‍,
അധ്യാപകര്‍ക്കിടയില്‍ അത്രയൊന്നും പരിചിതമല്ലാത്ത ഒന്നാണ് ബ്ലോഗെഴുത്ത്. പഠനത്തോടൊപ്പം തന്നെ ആശയവിനിമയത്തിനുള്ള ഒരു സമ്പ്രദായമായ ഈ ഇന്റര്‍നെറ്റ് സംവിധാനത്തിലേക്ക് അധ്യാപകരെ കൈ പിടിച്ചു കൊണ്ടു വരുന്നതിനുള്ള ഒരു ഉപാധി മാത്രമാണ് ഈ ബ്ലോഗ്. അവര്‍ മലയാളം പഠിച്ചു വരുന്നതേയുള്ളു. ഒരു ആറു മാസം കാത്തിരുന്നോളൂ സാര്‍, അധ്യാപക സമൂഹം മലയാളത്തില്‍ ചര്‍ച്ച ചെയ്യുന്നത് അങ്ങേയ്ക്ക് കാണാന്‍ കഴിയും. ഞങ്ങളുടെ ഒപ്പമുണ്ടാകണം. ഉമേഷും കാല്‍വിനും സത്യാന്വേഷിയും സ്വതന്ത്രനും വിനീതനും അടക്കമുള്ള പ്രമുഖ ബ്ലോഗെഴുത്തുകാര്‍ക്കൊപ്പം താങ്കളും തെറ്റുകുറ്റങ്ങള്‍ ചൂണ്ടിക്കാട്ടാന്‍ ഞങ്ങള്‍ക്കൊപ്പമുണ്ടാകണം.

JOHN P A October 21, 2009 at 5:41 PM  

shall I try to answer AZEEZ sirs last question algebrically
A/B = .ABBBBBBBB.....
= (A/10)+ (B/ 100) +(B/ 1000) +......
= (A/10)+ B[( 1/100) +(1/1000) +...
=(A/10) + B [ ( 1/100)/1-10] NOTE***
sum to infinity of G P
'''''''''''''''''''''''''
.......................
A[(10-B)/10B] =B/90
When B = 6 we get A= 1
Try for other digital value of A for some other digital value of B
Good project for 9 th standard children
in the chapter 3

Anonymous October 21, 2009 at 8:51 PM  

There is a correction
The sum to th terms of an infinite G P with common ratio less than 1 is a/1-r
here sum =( 1/100 )/ 1- 1/10
see the step in the previous comment
all other stepe are correct

Note
1/6 is the only possibily
we can arrive this conclusion logically ,not by direct checking

JOHN P A October 21, 2009 at 8:53 PM  

There is a correction
I forgot to give name
see last comment

vijayan October 21, 2009 at 9:46 PM  

dear john sir, what about my 19?
i have posted the sum of febinnocci series . am i RIGHT?

YOUR FEBINNOCCI and 1/89 is different.if you get time ,go through my comments related to your doubts. expect it tonight itself.
to azees , watch your mail and you get all possibilities that i posted related to your qn.

JOHN P A October 21, 2009 at 9:52 PM  

Yes Vijayan sir
Fibonacci sum is correct

AZEEZ October 22, 2009 at 9:46 AM  

A/B=.ABBBBBB.........
HAS ONE MORE POSSIBILITY.

A LITTLE TRICKY.

TRY TO IND IT.

AZEEZ October 22, 2009 at 12:06 PM  

CATS+HATE=DOGS


3415+6410=9825
2573+6570=9143
2794+3790=6584

ALL THESE CATS HATE DOGS.

nOW TRY THIS

ONE+TWO+TWO+THREE+THREE=ELEVEN

(DON'T FORGET MY A/B)

muraleedharancr October 22, 2009 at 12:45 PM  

Dear Azeez sir,
I have a doubt
is A & B are single digit numbers ?
ie 0 less than A,B less than 10?
If so
A/B=A/10 +B/100+B/1000+.....
=A/10+B/100(1+1/10+....)
=A/10+B/100{1/(1-1/10)}
=A/10+B/100{10/9}
=A/10+B/90
ie 90A=9AB+B*B
ie B*B+9AB-90A=0
Put 9A=T we get
B*B+TB-10T=0
Since A & B are intigers the discriminent should be a perfect square
ie T*T+40T should be a perfect square
ie T(T+40) should be a perfect square
Since T is a multiple of 9 we can take values 9, 18, 27 etc
If T=9 ,Discriminent =21*21 & A=1, B=6
If T=81 Discriminent =99*99 & A=9,B=9
ie we can write 9/9=.9999999....
MURALEEDHARAN.C.R
GVHSS VATTENAD

muraleedharancr October 22, 2009 at 12:55 PM  

Dear Azeez sir,
I have a doubt
is A & B are single digit numbers ?
ie 0 less than A,B less than 10?
If so
A/B=A/10 +B/100+B/1000+.....
=A/10+B/100(1+1/10+....)
=A/10+B/100{1/(1-1/10)}
=A/10+B/100{10/9}
=A/10+B/90
ie 90A=9AB+B*B
ie B*B+9AB-90A=0
Put 9A=T we get
B*B+TB-10T=0
Since A & B are intigers the discriminent should be a perfect square
ie T*T+40T should be a perfect square
ie T(T+40) should be a perfect square
Since T is a multiple of 9 we can take values 9, 18, 27 etc
If T=9 ,Discriminent =21*21 & A=1, B=6
If T=81 Discriminent =99*99 & A=9,B=9
ie we can write 9/9=.9999999....


MURALEEDHARAN.C.R
GVHSS VATTENAD

Anonymous October 22, 2009 at 1:13 PM  

Please send me the steps for malayalam comments

VIJAYAN N M October 22, 2009 at 9:41 PM  

1f A=-1,B=6;" -1/6=-.16666666"satisfies your property. if you agree pl find:ABC*DE/FG*HI=AEG*DB/FC*HI=A.

VIJAYAN N M October 22, 2009 at 9:42 PM  

1f A=-1,B=6;" -1/6=-.16666666"satisfies your property. if you agree pl find:ABC*DE/FG*HI=AEG*DB/FC*HI=A.

AZEEZ October 23, 2009 at 1:20 PM  

Dear Muraligharan Sir

u r right

9/9=0.999.............is the other possibility.

Thanks

JOHN P A October 23, 2009 at 9:56 PM  

Dear Azeez sir
kindly read murali sirs comment and my comment. 9/9 seems to be .99999999
but not exactly.All possibilites are not exactly crrrect . We can arrive it logically. see 36 th comment. It will be a good project for higher seconday children after studying convergence of infinite gp

AZEEZ October 24, 2009 at 11:25 AM  

Dear John Sir & Murali Sir,

Take x=0.99999......
Then 10x=9.99999.....

By subtracting we can get

9/9=0.999999.....

1=.9999........

AZEEZ October 24, 2009 at 12:10 PM  

Dear John sir, Vijayan sir,Murali sir and All other Sirs,

Comment about my above post.

Is it right or not?

JOHN P A October 24, 2009 at 12:31 PM  

Dear Azeez sir
It seems to be correct.This is because of the FREEDOM IN MATHEMATICS.I shall explain
x = .99999999999 ...
= 9/10 +9/100+9/1000+ .....
=9[ 1/10 +1/100 +1/1000.....]
=9/10 [ 1+1/10 +1/100+...........]
9/10 * 1/[1-(1/10)]
summation of infinite G P with modulus of r less than 1
9/ 10 * 10/9 = 1
so x = 1
The real problem lies in the formula a/1-r . This formula is established by negleting the higher powers of 1/10 in a stage by considering convergence of infinite series. That approximation makes slight in the result.so .99999... is not exactly 1 but convergence to 1. It is an accepted freedom

JOHN P A October 24, 2009 at 12:36 PM  

correction
(slight change) line 15
converges to 1 ( last line)

JOHN P A HIBHS October 24, 2009 at 1:53 PM  

There is a scope to an article titled "FREEDOM IN MATHEMATICS"

vijayan October 24, 2009 at 8:49 PM  

Dear Azees, now Iremember one old story.
one day a boy went to a tea shop and had tea for 10 Re.he had only 10 Paise and he proved 1 Re= 1Paise.the shop keeper agreed with his mathematics and accepted with the boy .if he knows something in maths the boy would have been get severe punishment.
the proof was like this
" 1 Re= 100 Paise.
=10Ps*10Ps
= 1/10Re*1/10Re
=1/100Re
= 1Paise
Therefore 10Re=10Paise." tahnk you.

vijayan October 24, 2009 at 9:07 PM  

take X=0.99
10X=9.9
10X-x=9.9-0.99
9x= 8.91
x= 8.91/9= 0.99

so X is not equal to 1.it is always 0.99( IN INDIA ,I dont know about QATAR)

muraleedharancr October 24, 2009 at 9:24 PM  

dear john sir
is .99999...... exactly = 1 or tends to 1 ?
we know a theorm(?} 'for any 2 no.s there exist a no. in between them
if .99999.....& 1 are different,I think,
there does not exist any no. between them
let k be such no.
ie .99999.. less than k less than 1
it is not believable because the no.
.99999..... overtake the no. k

AZEEZ October 25, 2009 at 9:51 AM  

Dear Vijayan Sir,

When I was studying in 9th std. my maths teacher proved the above problem in the above mentioned method.It is not Qatar Maths.It is Pure Indian Maths.
Can you say where is the mistake in the proof?

You take only two digits after decimal.

Take it to infinity and solve the problem.

Thanks

vijayan November 10, 2009 at 7:19 PM  

we heard a lot about 19.how to preapare a table of 1/7,29,39,49,59,89....? is it possible to make a table like1/ 19? comment?expect it from those who started commenting in this post.
1/19 is .052631578947368421
add 7894 in between 5&7we get
526315789478947368421.multiply it by 19,the answer is 10000000000099999999999.[11zero&11 nines]
perhaps you may think what is special with this number.
109,1000999,10000099999,1(7)0(7)9,1(9)0(9)9 are prime.but 1(11)0(11)9 is not prime.
simillarly 19,1999,199999,19999999 are prime.1999999999 is not prime.

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