Find the sum of the digits of $$1000^{20}-20$$ written in decimal notation.
What exactly do they mean by decimal notation? How will the answer be any different if it is not in decimal form?
[Math] Sum of digits of $1000^{20}-20$ in decimal notation
arithmeticdecimal-expansion
Related Solutions
The answer depends on just how you store your number.
If you are storing it as a rational number or as a fraction $\frac ab$ in lowest terms, or if you have a routine to convert your number to this format, here is how you can do it.
Find the prime factorization of $b$. Let's say it is
$$b=2^r5^sp^t\ldots$$
If $b$ has a prime factor other than $2$ or $5$, the number of decimal places is infinite. Otherwise, the number of decimal places in $\frac ab$ is
$$\max(r,s)$$
Do you need an explanation of why this works?
If your number $x$ is a float or a real number, find the smallest $t$ such that
$$10^t\cdot x=\lfloor 10^t\cdot x \rfloor$$
(The brackets are the greatest integer function.) Then $t$ is the number of decimal places. If no such $t$ exists, the number of decimal places is infinite.
I am not answering the question but the post asks for clues so here it is a couple of ideas.
If $a_0,a_1,a_2, \cdots , a_{477} $ are the decimal digits of $3^{1000}$ then the numbers $$b_i=a_{6i}+a_{6i+1}\cdot 10+a_{6i+2}\cdot 10^2+a_{6i+3}\cdot 10^3+a_{6i+4}\cdot 10^4+a_{6i+5}\cdot 10^5$$ for $i=0,1,2,\cdots, 79$ are the digits of $3^{1000}$ in base $10^6$ ($80$ is the nearest integer above $477 \div 6$ so there are $80$ digits numbered $0,1,2,\cdots,79$)
In other words: $$ 3^{1000} = b_0 + b_1 \cdot 10^6+ \cdots + b_{79} \cdot (10^6)^{79}$$
Now, if we resort to modular arithmetic we see that $$ 3^0=1, 3^1=3 , 3^2=2, 3^3=6, 3^4=4, 3^5=5, 3^6=1 $$ (all the equalities taken modulo $7$).
Also $$3^{1000}=(3^6)^{166}\cdot 3^4= 1 \cdot 4 = 4$$(all the equalities taken modulo $7$).
Now if we note that $10^6=1$ (modulo 7) the expression of $3^{1000}$ in base $10^6$ reads (modulo 7) $$ 4=b_0+b_1+ \cdots +b_{79}$$
So we can assert that the sum of digits of $3^{1000}$ in base one milion gives a remainder of $4$ when divided by $7$.
Another partial result comes from the decimal expansion read modulo 7:
$$3^{1000}= a_0+ a_1 \cdot 10 + \cdots +a_{477} \cdot 10^{477} =1 = a_0+3 a_1+ 2a_2 + 6 a_3 + 4 a_4 + 5 a_ 5 + \cdots $$
So, given that $a_0=1$ we can say that this particular linear combination of the remaining digits is divisible by $7$.
Best Answer
$(10^n)^m =10^{nm}$ and this is, in decimal, a one followed by $nm$ zeroes, so the sum of the digits is one.