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

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

### 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.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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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

I love Maths and Music ... dunno which more

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

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

### 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.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #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

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

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

### 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.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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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

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

I love Maths and Music ... dunno which more

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

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

### 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.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #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

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

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

### 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.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #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

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

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

### Re: Play with numbers

Hi smiyc86;

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.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #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 ..

D = 10

I love Maths and Music ... dunno which more

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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

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

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

### 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.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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## #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

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