How many digits are in 7^54?

[kavehmo]

How many digits are in 7^54?

How do we solve this type of questions?

[Topoltergeist]

That is an interesting question. I'm not yet sure of a good way to solve the problem. Here are my thoughts.

Off the bat, since 7<10, the number of digits in 7^54 must be somewhat less than 54. Also since 7 is greater than the square root of 10 (which is somewhere between 3 and 4) then the number of digits in 7^54 is greater than 27. Along the same vein, 7 is greater than 10^(2/3) (which is between 4 and 5) so 7^54 must have more than 36 digits. For a closer estimate, I'd do the following calculation:

The prime factorization of 54 is $$(2)*(3)^3$$, so we can write
$$7^{54} = 7^{2*3^3} = ((49^3)^3)^3$$
Here we say that 49 is pretty close to 50 and compute a bit. The cube of 50 is easy, and we estimate that
$$49^3 \approx 1.25 * 10^5$$
We need to cube this value twice more. The 10s are easy to cube, and we have
$$50^{27} = (1.25^3)^3 * 10^{45}$$
You can use the binomial approximation to estimate that 1.25^3 is about 1.75, which is less than 2. The cube of 2 is less than ten, so 1.25^9 is less than 10 and does not increase the number of digits. So 7^54 has about 45 digits. Since we used 50 to approximate 49, this is an upper bound.

It turns out that $$7^{54} \approx 4.318 * 10^{45}$$ so this upper bound is very sharp.

There will be some problems on the test which can actually be solved by clever estimation. It is a multiple choice test, and you can use this to your advantage.


[EDIT]: In retrospect, this method might not be so bad. The overall plan is to multiply everything out, and when things start to get complicated, make estimates to make your computation easier. By using the binomial approximation
$$(1 + \epsilon)^n \approx 1 + n \epsilon$$
you can get a good first order approximation of your error terms. Since you are going for order of magnitude, this should be good enough.

[kavehmo]

Tanx, but your way is confusing (at least for me!)

this question is similar to this one from GRE (GR0568)

14- what is the units digit in the standard decimal expansion of the number 7^25

a) 1
b) 3
c) 5
d) 7
e) 9

[Ell]

this question is similar to this one from GRE (GR0568)

14- what is the units digit in the standard decimal expansion of the number 7^25

a) 1
b) 3
c) 5
d) 7
e) 9

Actually, this problem is far easier to solve. It is merely asking about the units digit and not the number of digits.
You may notice that 7^1=7, the units digit is 7.
7^2 = 49, the units digit is 9.
And without evaluating the powers, one can see that the unit digit of 7^3 is 3, and that of 7^4 is 1.
Now if you continue, you will end up with the same values in the same order, 7, 9, 3, 1.
In fact, 7^n = 7 iff n = 4k+1. Similarly, it equals 9 iff n = 4k+2, 3 iff n = 4k+3, and 1 iff n = 4k.

25 being congruent to 1 mod 4, 7^25 has unit digit equal to 7.

[Topoltergeist]

I agree that my way is confusing. I was just fiddling around with numbers and eventually got a result. I suggest looking at this link: http://mathforum.org/library/drmath/view/62942.html
If anyone has a sleek solution to the 7^54 problem, I'd really like to see it.

As for the second problem I like solving it this way:

Since
$$7 \equiv 2 \mod 5$$
then we can substitute and group some terms
$$7^{25} \equiv 2^{25} \mod 5$$
$$7^{25} \equiv (2^5)^5 \mod 5$$
Then using Fermat's little theorem we find that
$$7^{25} \equiv 2 \mod 5$$

This last result means that the unit digit of 7^25 is either 2 or 7. Since the power of an odd number is odd, then the unit digit has to be 7.

[robertcardona]

Look at it $$\pmod{10}$$. Note that $$7^1 \equiv 7$$, $$7^2 \equiv 9$$, $$7^3 \equiv 3$$, and $$7^4 \equiv 1$$ and they repeat after. So look at the remainder of 54 when divided by 4, it is 2. Thus it is equivalent to $$7^2 \pmod{10} \equiv 9$$

[Topoltergeist]

robertcardona, you are misunderstanding the problem, see what Ell wrote:

this question is similar to this one from GRE (GR0568)

14- what is the units digit in the standard decimal expansion of the number 7^25

a) 1
b) 3
c) 5
d) 7
e) 9

Actually, this problem is far easier to solve. It is merely asking about the units digit and not the number of digits.

[robertcardona]

My bad: Memorize Log(2), Log(3) and Log(7) all in base 10. The rest follow from basic rules.

$$\log_{10}(7^{54}) = 54 * \log_{10}(7)$$. Knowing that $$\log_{10} 7 \approx 0.8450$$ we can say $$54 * \log_{10}(7) = 54 * .8450$$. Do the multiplication by hand and get $$45.630$$. Now the trick to these questions is to take the Floor of the log, base 10, of the number you want and then add one. So given the number $$n$$ and your asked to find the number of digits, do $$\text{fl } (n)+ 1$$. So in our case, we do $$\text{fl } (45.630) + 1 = 45 + 1 = \boxed{46}$$.

Again, memorize
log 2 = .3010
log 3 = .4771
log 7 = .8450

Log(9) = log(3 * 3) = log(3) + log(3)
log(8) = log(2^3) = 3log(2) = 3 * .3010 = .903

Again, sorry for my misunderstanding!

[nikkoo]

That is an interesting question
but is very complicated for me

[DDswife]

We cannot use a calculator, can we? If this is the case, to memorize a few logs, squares, etc is a good piece of advice

Thanks.