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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 ?

  • 58
    Some people are able to multiply by 5 with pencil and paper, without a calculator.2012-07-31
  • 5
    Solve $5^k>2000$ then your $n$ must be $2k$ or $2k-1$2012-07-31
  • 2
    @GerryMyerson I could do that but I wanted to know if there was some other way of doing that2012-07-31
  • 5
    $5^3=125$, $5^4=625$, $5^5=3125$, a bit over $3000$. So $5^{10}$ should be around $10$ million, $5^9$ too small.2012-07-31
  • 12
    [I've read about this](http://downlode.org/Etext/power.html), @Gerry :) $$warning: \ \ link \ \ to \ \ text$$2012-07-31
  • 0
    @The, thanks - I was thinking of that story, but I couldn't remember the name, so didn't try to search for it.2012-07-31
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    @TheChaz: Brilliant! Truly brilliant.2012-07-31
  • 1
    You can't do log base 5 operations in your head?2012-07-31
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    @TheChaz That is one awesome story by the genius of Asimov.2012-07-31

11 Answers 11

67

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

  • 0
    Fine, but how to solve $5^{n-6} > 256$ ?2012-07-31
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    @wap26 $5^3=125$, $5^4=625$, therefore $n = 4+6 = 10$2012-07-31
  • 0
    good old euler theorem2012-08-29
  • 1
    Shouldn't it be $n>10$ ?2013-10-26
  • 1
    @Sush $n$ _can_ be $\gt 10$ in addition to $10$, so shouldn't it be $n \ge 10$2014-01-20
  • 0
    @ColeJohnson, you are right!2014-01-21
44

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.

  • 8
    Now do it for $5^n > 6.24*10^{310}$ ;)2012-07-31
  • 3
    @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.2012-07-31
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    @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.2012-08-01
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    @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.2012-08-01
14

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

11

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 $4+6=10$

9

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.

  • 6
    Not a tough one without a calculator, and definitely too easy with a Python Program :P2012-07-31
  • 0
    These slippery problems can sometimes turn around and bite you in the asp (that's what Cleopatra would have said).2012-07-31
  • 11
    Pretty sure a computer counts as a calculator (unless you have a Python interpreter built in your brain, if so, fair enough..)2012-07-31
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    Hmm @Thomas, don't get technical on me now :-).2012-07-31
  • 3
    You could also have done `import math; math.ceil(math.log(4000000,5))`2012-07-31
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    True, but it wouldn't have had that excitement of discovery :-).2012-07-31
  • 0
    Hmm, I'm picking up a few negative votes here.2012-07-31
  • 0
    Hmm, that's +3 -3 so far...2012-08-02
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    @copper.hat: I don't believe that the OP was looking for a brute force argument carried out by a computer. This explains at least my downvote, just in case you're seriously concerned about them.2012-08-02
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    @NilsMatthes: Thanks Nils! It was actually meant as a joke, since obviously 'without a calculator' would preclude a computer. It would only concern me if someone was upset by it...2012-08-02
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    @copper.hat: No hard feelings on my side.;) I usually don't consider jokes as helpful answers, that's all. Cheers!2012-08-02
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    People have no sense of humoUr...2012-08-06
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    @copper.hat i do.${}{}{}{}{}$2014-03-20
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    @Sabyasachi: :-)!2014-03-20
7

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.

3

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.

  • 0
    *cough* typo *cough*2012-07-31
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    may you explain the comment please ? Something bad in the answer ?2012-07-31
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    @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!*)2012-07-31
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    Oww ... I wanted to state that $10^7 > 4.10^6$ ... Thanks for pointing this out ! :>2012-07-31
2

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.

2

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

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

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

$\log_{10}(2)\approx 0.3$ 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 $n\approx \frac {6.60206}{0.69897}$)

1

$5^n > 4,000,000$ 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
  • $n = 10$
  • 0
    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 %"!2012-08-01
  • 0
    your memory has more significant figures than mine.. bravo.. I knew it was< 0.1% but close enough for government work.2012-08-02
1

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: $5^9<2^9\cdot 3^9<2^9\cdot 3^5\cdot 3^4=512\cdot 243\cdot 81<125,000\cdot 100<4,000,000$.