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#26 2009-04-17 12:28:49

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 83,251

Re: Play with numbers

Hi smiyc86;

Here is my answer to #1, I hope it is what you require.

As you know the way to sum all the digits of any number is to take that number mod 9.
1999 is congruent to 1 mod 9 so 1999^1999 is also going to be congruent to 1 mod 9.
Therefore D = sum of the digits of 1999^1999 = 1

Also for fun:

B=29656
C=28

bobbym

Last edited by bobbym (2009-04-17 12:34:59)


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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#27 2009-04-20 19:31:29

smiyc86
Member
Registered: 2009-03-19
Posts: 78

Re: Play with numbers

i agree to upto an extent ... but there's a part remaining to be proved ....

HINT : according to ur proof
sum of digits of 199 = sum of digits of 28
or 19 = 10

wink


I love Maths and Music ... dunno which more wink

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#28 2009-04-20 20:04:06

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 83,251

Re: Play with numbers

Hi smiyc86;

   Thanks for responding.

   The sum of the digits of 199 = 19 = 1 because 199 is congruent to 1 mod 9, The sum of the digits of B = 1999^1999 = 29656 (by expanding this number, I assumed you didn't want to sum it again else B=1 C=1 D=1) and C = Sum of the digits of B = 28. D = The sum of the digits of C =10. I assumed you would reduce this 10 down to 1. This is as far as I follow the problem please provide more guidance.

Last edited by bobbym (2009-04-20 20:06:09)


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

#29 2009-04-21 18:43:15

smiyc86
Member
Registered: 2009-03-19
Posts: 78

Re: Play with numbers

bobbym wrote:

The sum of the digits of 199 = 19 = 1 because 199 is congruent to 1 mod 9

if A = 199
den b = sum of digits of A = 10 (not 1)

bobbym wrote:

(by expanding this number, I assumed you didn't want to sum it again else B=1 C=1 D=1)

u r absolutely right here ... i didn't want it to sum it again.

bobbym wrote:

This is as far as I follow the problem please provide more guidance.

u r absolutely correct here ... u followed it absolutely correctly...


HINT: Prove that sum of digits of D will be a single digit number

if u find that sum of digits of D will be a two digits number ... then the ans may be one of... 19,28,37,46,55,64,73,82,91
coz all of the above two digit numbers are congruent to 1 under MOD 9 ....

but if the sum of digits of D is a single digit number then the answer will be 1 .... i hope u get wot i say

Last edited by smiyc86 (2009-04-21 18:44:37)


I love Maths and Music ... dunno which more wink

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#30 2009-04-21 19:50:19

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 83,251

Re: Play with numbers

smiyc86;

   You are absolutely right. I was confusing the digital root with the digital sum.
I got the following from:
http://en.wikipedia.org/wiki/Digit_sum

Digital root of 199 =1                                (199 is congruent to 1 mod 9)
Digital sum of 199 =19 but still not 10       (No easy way to compute. Just sum the digits)

Digital root of 84001 = 4                           (84001 is congruent to 4 mod 9)
Digital sum of 84001 = 13                          (No easy way to compute. Just sum the digits)

Based on what I think you require, I am stuck on following the rules for digital sums so:

A=1999^1999
B=29656
C=28
D=10

Thanks a lot for listening, please respond.

Last edited by bobbym (2009-04-22 22:50:00)


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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#31 2009-04-21 20:16:25

smiyc86
Member
Registered: 2009-03-19
Posts: 78

Re: Play with numbers

Hi bobbym,

you found out D .... but in the ques you need to find out the sum of digits of D ... which turns out to be 1+0 = 1

i ll give u the last and most important hint:

let E = sum of digits of D

find a way to prove that E would be a single digit number

then u can easily say that E would be 1

reason:
if the digital root of a number is 1 and the number is single digit number ... then the number is got to be 1 smile

Last edited by smiyc86 (2009-04-21 20:17:45)


I love Maths and Music ... dunno which more wink

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#32 2009-04-21 20:52:58

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 83,251

Re: Play with numbers

Hi smiyc86;

This process:

A=1999^1999
B=29656
C=28
D=10
E=?

is the definition of the digital root. That is computed by 1999 is congruent to 1 mod 9 so 1^1999 =1. This saves all the computing. E=1. How was that?


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

#33 2009-04-21 20:54:02

smiyc86
Member
Registered: 2009-03-19
Posts: 78

Re: Play with numbers

what procedure do you use to find B


I love Maths and Music ... dunno which more wink

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#34 2009-04-21 20:57:10

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 83,251

Re: Play with numbers

I was afraid that you would ask that. I did the calculation using a computer and added every digit up. This is as far as I know, and that wikipedia site agrees, the only way to get the digital sum. The digital root can be done by congruences.

Last edited by bobbym (2009-04-21 20:57:51)


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

#35 2009-04-21 21:09:33

smiyc86
Member
Registered: 2009-03-19
Posts: 78

Re: Play with numbers

well you dont get my point do you

TRY TO PROVE THAT THE SUM OF DIGITS OF 'D' WILL BE A SINGLE DIGIT NUMBER

in this question you don't even have to find out the values of b,c,d


I love Maths and Music ... dunno which more wink

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#36 2009-04-22 01:53:12

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 83,251

Re: Play with numbers

Hi smiyc86;

Found this page:
http://nz.answers.yahoo.com/question/index?qid=20080620225029AAeW6Uh

with this proof:

We know that Iterative Sum of Digits = n mod 9( also called the digital root)

D = (1999^1999) mod 9 = (1999 mod 9)^1999

We know:

1) 1999 mod 9 = 1
2) 1^f = 1, for any value of f.

D = 1^1999 = 1

This is what I have been saying all along.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

#37 2009-04-23 18:17:21

smiyc86
Member
Registered: 2009-03-19
Posts: 78

Re: Play with numbers

but u have taken from the page the only solution which is ncorrect ..

btw, For your information ...

D = 10


I love Maths and Music ... dunno which more wink

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#38 2009-04-23 18:26:30

smiyc86
Member
Registered: 2009-03-19
Posts: 78

Re: Play with numbers

check the solution of manjyome... her only mistake is she found out D as the answer whereas the reqd answer is the sum of digits of D.

one more thing

the number of digits in 'n' is (characteristic of log n) + 1


I love Maths and Music ... dunno which more wink

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#39 2009-04-24 10:34:00

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 83,251

Re: Play with numbers

Hi smiyc86;

   Hard to believe but I am familiar with manjyome's stuff. Anyway why should we consider his/her answer. After all it is only a probable answer( 3 out of 5). Berry's idea, underneath manjyome's post is right on the money. I don't have any way at present to firm up manjyome's work. If you do I would like to see it.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

Offline

#40 2009-04-26 17:20:49

smiyc86
Member
Registered: 2009-03-19
Posts: 78

Re: Play with numbers

i have already given a proof to it


I love Maths and Music ... dunno which more wink

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