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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 ?

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    @TheChaz That is one awesome story by the genius of Asimov.2012-07-31

11 Answers 11

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\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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    @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.

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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
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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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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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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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    @Sab$y$asachi: :-)!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.

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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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    Oww ... I wanted to state that 10^7 > 4.10^6 ... Thanks for pointing this out ! :>2012-07-31
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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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$\log_{10}(5^n)=n\cdot \log_{10}(5)\approx n\cdot 0.7$

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

$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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$5^n > 4,000,000$ 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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    your memory has more signi$f$icant figures than mine.. bravo.. I knew it was< 0.1% but close enough for government work.2012-08-02
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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: $5^9<2^9\cdot 3^9<2^9\cdot 3^5\cdot 3^4=512\cdot 243\cdot 81<125,000\cdot 100<4,000,000$.