# Math Help - Sum of digits

1. ## Sum of digits

The sum of $2^{1000}$ 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?

Keep Smiling
Malay

2. Hello, Malay!

The sum of $2^{1000}$ 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 $2^{20} \:= \:1,048,576 \:\equiv \:2^2\pmod 9$

. . then: $N \:= \:2^{1000} \:= \:\left(2^{20}\right)^{50} \:\equiv\:\left(2^2\right)^{50}\pmod{9}$

Hence: . $N \:\equiv \:2^{100} \:= \:\left(2^{20}\right)^5 \:\equiv\;(2^2)^5\pmod{9}$

Therefore: . $N\:\equiv\:2^{10} \:=\:1024\:\equiv\:7\pmod{9}$

The one-digit number is $\boxed{7}$

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An alternate solution . . .

Consider the digit roots of consecutive powers of 2.

. . . $\begin{array}{ccccccc}2^1 \equiv 2 \\ 2^2\equiv 4 \\ 2^3 \equiv 8 \\ 2^4 \equiv 7 \\ 2^5 \equiv 5 \\ 2^6 \equiv 1 \\ \vdots\end{array}$

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

Since $2000 \div 6\:=\:166,\;rem\;4$
. . the digit root of $2^{1000}$ is the fourth one: $\boxed{7}$

3. but why have you found modulo 9?

Keep Smiling
Malay

4. Hello, Malay!

but why have you found modulo 9?

It's a bit tricky to explain . . . Given a positive integer $N$,
it turns out that $N$ is equivalent to the sum of its digits, modulo 9.

If $N = abc\hdots n$, then: . $N \,\equiv \,(a+b+c+\hdots+n) \pmod{9}$

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

Let $N\:=\:10^na + 10^{n-1}b + 10^{n-2}c + \hdots + 10m + n$

. . $= \10^na - a + a) + (10^{n-1}b - b + b) + (10^{n-2}c - c + c)" alt="= \10^na - a + a) + (10^{n-1}b - b + b) + (10^{n-2}c - c + c)" /> $+ \hdots + (10m - m + m) + n$

. . $=\10^n - 1)a + a+ (10^{n-1}-1)b + b + (10^{n-2}-1)c + c" alt="=\10^n - 1)a + a+ (10^{n-1}-1)b + b + (10^{n-2}-1)c + c" /> $+ \hdots + (10 - 1)m + m + n$

. . $= \:\underbrace{(10^n-1)a + (10^{n-1} -1)b + (10^{n-2} - 1)c + \hdots + (10 - 1)m}$ $\,+\,(a + b + c + \hdots + m + n)$
. . . . . . . . . . . . This expression is a multiple of 9 *

Therefore: . $N \:\equiv \a + b + c + \hdots +n)\pmod{9}" alt="N \:\equiv \a + b + c + \hdots +n)\pmod{9}" />

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* $10^k - 1$ is of the form: . $999\hdots9$ (k 9's)

5. Originally Posted by Soroban
Hello, Malay!

It's a bit tricky to explain . . . Given a positive integer $N$,
it turns out that $N$ is equivalent to the sum of its digits, modulo 9.

If $N = abc\hdots n$, then: . $N \,\equiv \,(a+b+c+\hdots+n) \pmod{9}$

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

Let $N\:=\:10^na + 10^{n-1}b + 10^{n-2}c + \hdots + 10m + n$

. . $= \10^na - a + a) + (10^{n-1}b - b + b) + (10^{n-2}c - c + c)" alt="= \10^na - a + a) + (10^{n-1}b - b + b) + (10^{n-2}c - c + c)" /> $+ \hdots + (10m - m + m) + n$

. . $=\10^n - 1)a + a+ (10^{n-1}-1)b + b + (10^{n-2}-1)c + c" alt="=\10^n - 1)a + a+ (10^{n-1}-1)b + b + (10^{n-2}-1)c + c" /> $+ \hdots + (10 - 1)m + m + n$

. . $= \:\underbrace{(10^n-1)a + (10^{n-1} -1)b + (10^{n-2} - 1)c + \hdots + (10 - 1)m}$ $\,+\,(a + b + c + \hdots + m + n)$
. . . . . . . . . . . . This expression is a multiple of 9 *

Therefore: . $N \:\equiv \a + b + c + \hdots +n)\pmod{9}" alt="N \:\equiv \a + b + c + \hdots +n)\pmod{9}" />

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* $10^k - 1$ is of the form: . $999\hdots9$ (k 9's)

Ok.
Let $2^{1000}$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

6. Hello, Malay!

Ok.
Let $2^{1000} = N$ . . . It contains $n$ digits.
You have found the sum of digits of $N:\; a+b+c+d+....+n.$

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

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

If $X \equiv a\!\!\pmod{9}$ and $Y \equiv b\!\!\pmod{9}$, then $XY \equiv ab\!\!\pmod{9}$

Since $2^{10} \,= \,1,024 \,\equiv \,7\!\!\!\pmod{9}$
. . then: . $2^{20}\:=\2^{10})(2^{10}) \:\equiv \:7\cdot7 \:\equiv\:49 \:\equiv\:4\!\!\!\pmod{9} " alt="2^{20}\:=\2^{10})(2^{10}) \:\equiv \:7\cdot7 \:\equiv\:49 \:\equiv\:4\!\!\!\pmod{9} " />

And from there, we can work our way up to $2^{1000}$

7. 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

8. 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 $7^9$ 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: $7^9 \,= \,40,353,607$
Then: $4 + 0 + 3 + 5 + 3 + 6 + 0 + 7 \,= \,28\quad\Rightarrow\quad2 + 8 \,=$ $10\quad\Rightarrow\quad 1 + 0 \,= \,1$

But we can get the final digit without expanding $7^9$.

We have: . $\begin{array}{cccc}7 \equiv 7\text{ (mod 9)} \\ 7^2 = 49 \equiv 4\text{ (mod 9)} \\ 7^3 = 343 \equiv 1\text{ (mod 9)}\\ \vdots\end{array}$

Since $7^3 \equiv 1\text{ (mod 9)}:\;\;(7^3)(7^3)(7^3) \equiv 1\cdot1\cdot1\text{ (mod 9)}$

So we have: . $7^9 \equiv 1\text{ (mod 9)}$ . . . see?

The original problem was $2^{1000}$.
We will never see that number ... nor be able to add its digits ... ever!
. . (It begins $10,715,086...$ and has 302 digits.)

So we begin with, say, $2^{20} \equiv 4\text{ (mod 9)}$, and create an expression with $2^{1000}$.

9. what is modulo 9?

10. This comes from the old "casting out nines" technique.

11. Originally Posted by Soroban
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 $7^9$ 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: $7^9 \,= \,40,353,607$
Then: $4 + 0 + 3 + 5 + 3 + 6 + 0 + 7 \,= \,28\quad\Rightarrow\quad2 + 8 \,=$ $10\quad\Rightarrow\quad 1 + 0 \,= \,1$

But we can get the final digit without expanding $7^9$.

We have: . $\begin{array}{cccc}7 \equiv 7\text{ (mod 9)} \\ 7^2 = 49 \equiv 4\text{ (mod 9)} \\ 7^3 = 343 \equiv 1\text{ (mod 9)}\\ \vdots\end{array}$

Since $7^3 \equiv 1\text{ (mod 9)}:\;\;(7^3)(7^3)(7^3) \equiv 1\cdot1\cdot1\text{ (mod 9)}$

So we have: . $7^9 \equiv 1\text{ (mod 9)}$ . . . see?

The original problem was $2^{1000}$.
We will never see that number ... nor be able to add its digits ... ever!
. . (It begins $10,715,086...$ and has 302 digits.)

So we begin with, say, $2^{20} \equiv 4\text{ (mod 9)}$, and create an expression with $2^{1000}$.

I have no doubt in your method. I am pretty sure it works. The only thing I want to know why it works?(As galactus pointed out 'casting out nines' was taught to us in schools, we used it to get lucky number by procesing the same way on date of birth).
I just got. There was some misunderstanding.

Thanks

Keep Smiling
Malay