# Solving $5^n > 4,000,000$ without a calculator

If $n$ is an integer and $5^n > 4,000,000.$ What is the least possible value of $n$? (answer: $10$)

How could I find the value of $n$ without using a calculator ?

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Some people are able to multiply by 5 with pencil and paper, without a calculator. –  Gerry Myerson Jul 31 '12 at 0:43
Solve $5^k>2000$ then your $n$ must be $2k$ or $2k-1$ –  Thomas Andrews Jul 31 '12 at 0:44
@GerryMyerson I could do that but I wanted to know if there was some other way of doing that –  MistyD Jul 31 '12 at 0:44
$5^3=125$, $5^4=625$, $5^5=3125$, a bit over $3000$. So $5^{10}$ should be around $10$ million, $5^9$ too small. –  André Nicolas Jul 31 '12 at 0:48
I've read about this, @Gerry :) $$warning: \ \ link \ \ to \ \ text$$ –  The Chaz 2.0 Jul 31 '12 at 1:24

\begin{eqnarray} & 5^n &>& 4.000.000\\ \Leftrightarrow & 5^n &>& 5^6 \cdot 2^8 \\ \Leftrightarrow & 5^{n-6} &>& 256.\\ \end{eqnarray} Then, $n=10$.

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Fine, but how to solve $5^{n-6} > 256$ ? –  wap26 Jul 31 '12 at 7:06
@wap26 $5^3=125$, $5^4=625$, therefore $n = 4+6 = 10$ –  Alex L Jul 31 '12 at 7:11
good old euler theorem –  pyCthon Aug 29 '12 at 5:44
Shouldn't it be $n>10$ ? –  Silent Oct 26 '13 at 6:23
@Sush $n$ can be $\gt 10$ in addition to $10$, so shouldn't it be $n \ge 10$ –  Cole Johnson Jan 20 '14 at 23:07

$5^n > 4,000,000$ find integer $n$.

• take $\log10$ of both sides
• $n\log5>2\log2+6\log10=2\log2+6$
• $n > (2\log2+6)/\log5$ (recall from log paper log2~0.3 and log5~0.7 within a few %.
• $n > 6.6/0.7$ ~ $9.4$ thus rounded up
• $n = 10$
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I remember log10(2) as 0.3010 and hence log10(5) is 0.6990 so 0.3 and 0.7 are much closer than "within a few %"! –  StarNamer Aug 1 '12 at 0:22
your memory has more significant figures than mine.. bravo.. I knew it was< 0.1% but close enough for government work. –  Tony Stewart Aug 2 '12 at 13:39

Taking square roots of both sides we solve $5^r>2000=25\cdot80$. The right side is approximated from below by $5^2\cdot5^2\cdot3$ so we want $5^{r-4}>3$ or $r=5$, so $n=2r=10$. Check $n=9$ is too small: $5^9<2^9\cdot 3^9<2^9\cdot 3^5\cdot 3^4=512\cdot 243\cdot 81<125,000\cdot 100<4,000,000$.

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The easiest way to multiply by $5$ without a calculator is to multiply by $10$ and then divide by $2$, i.e.: $$1: 5\times 5 = 50/2 = 25$$ $$2: 250/2 = 125$$ $$3: 1250/2 = 625$$ $$4: 6250/2 = 3125\ldots$$ Won't take you very long to get to $10$.

True. And for large power, use approximations :

$$5^{n} = \frac{10^n}{2^n}$$

$$5^9 = \frac{10^9}{2^9} = \frac{1,000,000,000}{512} \cong \frac{1,000,000,000}{500} \cong 2*10^6 < 4*10^6$$

$$5^{10} = \frac{10^{10}}{2^{10}} = \frac{10,000,000,000}{1024} \cong \frac{10,000,000,000}{1000} \cong 10^7 > 4*10^6$$

It's not a mathematical way to prove, but it's a way to find the result using approximation.

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cough typo cough –  Joshua Shane Liberman Jul 31 '12 at 20:20
may you explain the comment please ? Something bad in the answer ? –  Al_th Jul 31 '12 at 21:10
@Al_th: It looks as though you are saying $10^7>4 \cdot 10^7$. Otherwise, the above comment seems unhelpful (trust me - I would know!) –  The Chaz 2.0 Jul 31 '12 at 22:00
Oww ... I wanted to state that $10^7 > 4.10^6$ ... Thanks for pointing this out ! :> –  Al_th Jul 31 '12 at 23:09

It helps if you remember that $ln(2) \approx 0.7$ and $ln(10) \approx 2.3$. (These are common bases to work in, so they're generally useful numbers.)

\begin{align} 5^n &> 4\ 000\ 000\\ \ln(5^n) &> \ln(4\ 000\ 000)\\ n (\ln 10 - \ln 2) &> 2 \ln 2 + 6 \ln 10\\ n (2.3 - 0.7) &> 2 \times 0.7 + 6 \times 2.3\\ 1.6 n &> 1.4 + 13.8\\ 1.6 n &> 15.2\\ 1.6 n &> 16 - 0.8\\ n &> 10 - 0.5\\ n &> 9.5\\ n &= 10 \end{align}

A bit much for mental arithmetic, but quite doable just typing into this here box.

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$$\log_{10}(5^n)=n\cdot \log_{10}(5)\approx n\cdot 0.7$$

$$\log_{10}(4000000)=\log_{10}(4)+6\approx 6.6$$

$$7\cdot 9=63\ \text{so that }\ \boxed{n=10}\$$

$\log_{10}(2)\approx 0.3$ was only used giving $\log_{10}(5)=\log_{10}(10)-\log_{10}(2)$ and $\log_{10}(4)=2\cdot\log_{10}(2)$

(if non integer values are allowed $n\approx \frac {6.60206}{0.69897}$)

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The easiest way to multiply by 5 without a calculator is to multiply by 10 and then divide by 2. ie: 1: 5x5 = 50/2 = 25. 2: 250/2 = 125. 3: 1250/2 = 625. 4: 6250/2 = 3125... Won't take you very long to get to 10.

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I dunno, this is a tough one, especially without a calculator.

Here is the Python program I used to figure this one out:

for n in range(1,11):
print "5^%s-4,000,000 = %s" % (n, pow(5,n)-4000000)


Here is the output:

5^1-4,000,000 = -3999995.0
5^2-4,000,000 = -3999975.0
5^3-4,000,000 = -3999875.0
5^4-4,000,000 = -3999375.0
5^5-4,000,000 = -3996875.0
5^6-4,000,000 = -3984375.0
5^7-4,000,000 = -3921875.0
5^8-4,000,000 = -3609375.0
5^9-4,000,000 = -2046875.0
5^10-4,000,000 = 5765625.0


It looks like $n=10$ is the answer.

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Not a tough one without a calculator, and definitely too easy with a Python Program :P –  Patrick Da Silva Jul 31 '12 at 1:57
Pretty sure a computer counts as a calculator (unless you have a Python interpreter built in your brain, if so, fair enough..) –  Thomas Jul 31 '12 at 2:40
You could also have done import math; math.ceil(math.log(4000000,5)) –  Alex L Jul 31 '12 at 8:42
True, but it wouldn't have had that excitement of discovery :-). –  copper.hat Jul 31 '12 at 8:49
People have no sense of humoUr... –  copper.hat Aug 6 '12 at 3:13

By logarithm rules: $$5^{n}>4\cdot10^{6}\iff n>\log_{5}2^{2}2^{6}5^{6}=\log_{5}2^{8}+\log_{5}5^{6}=\log_{5}2^{8}+6=\log_{5}256+6$$

Since these are relatively small numbers I assume it is ok to write : $5^{3}=125$ thus clearly $3<\log_{5}256<4$ hence the minimal $n$ that satisfies this inequality is $4+6=10$

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$4,000,000 = 2^2 \times 10^6 = 2^8 \times 5^6$, so you want $5^{n-6} > 2^8 = 256$. Well, $5^3 = 125\ldots$.

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Divide 4000000 by 5, without a calculator, getting 800000. Divide again; 160000. Again; 32000. Then 6400, then 1280, then 256, then 51 (rounding), then 10, then 2. So $2\times5^9$ is about 4000000, so $5^{10}$ exceeds 4000000.

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Now do it for $5^n > 6.24*10^{310}$ ;) –  Polynomial Jul 31 '12 at 14:04
@Polynomial - Can you easily solve that without a calculator? I get to 5^(n-308) > 2^312*3*13 or so and then need to estimate log(2)/log(5) part to enough precision that I can multiply by a three digit number and know I'm not off by 1 or introduce approximations 624 ~ 625. I can sort of estimate values (e.g., Newton's method) to get within about ~1 (I find N=444 instead of 445), but would be very hard to prove. –  dr jimbob Jul 31 '12 at 20:13
@Polynomial, why you pickin on me? Your comment would apply just as well to any of the other answers (except the Python program). Anyway, everyone knows $624=5^4$ (roughly), and $\log_{10}5=.7$ (roughly), so $n=4+(308/.7)=444$ should be about right, close enough for government work. –  Gerry Myerson Aug 1 '12 at 3:14
@GerryMyerson The intent wasn't to pick on you, I was just joking. I actually upvoted you. The fact that you're dividing by two makes this a very good answer, because you have to take it to ridiculous extremes (like I did) before this method becomes implausible to use. –  Polynomial Aug 1 '12 at 8:10