The sum of is written and the sum of the digits of this number is written and the process is repeated a number of times, till a one digit number is obtained. What is this one digit number?

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- July 16th 2006, 08:22 PMmalaygoelSum of digits
The sum of is written and the sum of the digits of this number is written and the process is repeated a number of times, till a one digit number is obtained. What is this one digit number?

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Malay - July 19th 2006, 08:13 PMSoroban
Hello, Malay!

Quote:

The sum of is written and the sum of the digits of this number is written

and the process is repeated until a one-digit number is obtained.

What is this one-digit number?

The sum-of-digits means we are dealing with "digital roots"

. . . . . that the numbers are reduced modulo-9.

We can reduce the products as we work, rather than waiting until the final product.

Since

. . then:

Hence: .

Therefore: .

The one-digit number is

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

An alternate solution . . .

Consider the digit roots of consecutive powers of 2.

. . .

The digits roots have a six-digit cycle: 2-4-8-7-5-1.

Since

. . the digit root of is the fourth one:

- July 19th 2006, 09:09 PMmalaygoel
but why have you found modulo 9?

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Malay - July 20th 2006, 07:13 AMSoroban
Hello, Malay!

Quote:

but why have you found modulo 9?

It's a bit tricky to explain . . . Given a positive integer ,

it turns out that is equivalent to the sum of its digits, modulo 9.

If , then: .

Let's see if I can devise a proof for this . . .

Let

. .

. .

. .

. . . . . . . . . . . . This expression is a multiple of 9*****

Therefore: .

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

*****is of the form: . (*k*9's)

- July 20th 2006, 08:26 AMmalaygoelQuote:

Originally Posted by**Soroban**

Let is N. It contains n digits. You have found the sum of digits of n(equal to a+b+c+d+....+n)

But how will you find a+b+c+d+..+n so than we can proceed to the next step.

Malay - July 20th 2006, 08:53 AMSoroban
Hello, Malay!

Quote:

Ok.

Let . . . It contains digits.

You have found the sum of digits of

But how will you find so that we can proceed to the next step?

We don't need that sum . . . We can work with smaller quantities.

If and , then

Since

. . then: .

And from there, we can work our way__up__to

- July 20th 2006, 09:00 AMmalaygoel
I think I was not clear in my last post.

Suppose we have a number 256784 which is congruent to 5 modulo 9.

You showed that it is congruent to 2+5+6+7+8+4 modulo 9

now, what will be the next step

Malay - July 20th 2006, 02:47 PMSoroban
Hello again, Malay!

I hope I can finally make this clear . . .

We*don't need the product*to find its digital root.

Your thoughts seem to be along this tline:

Consider and add its digits.

If the sum is greater than 9, add the digits again.

Continue until you get a one-digit number.

Grabbing our calculator:

Then:

But we can get the final digit*without*expanding .

We have: .

Since

So we have: .*. . . see?*

The original problem was .

We will**never**see that number ... nor be able to add its digits ... ever!

. . (It begins and has 302 digits.)

So we begin with, say, , and__create__an expression with .

- July 20th 2006, 02:59 PMQuick
what is modulo 9?

- July 20th 2006, 04:52 PMgalactus
This comes from the old "casting out nines" technique.

- July 20th 2006, 07:36 PMmalaygoelQuote:

Originally Posted by**Soroban**

I just got. There was some misunderstanding.

Thanks

Keep Smiling

Malay