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So I am working out the odds for a lottery, picking 4 numbers between 1-35.

The equation is:

$$\mbox{odds}=\frac{35\cdot 34\cdot 33\cdot 32}{1\cdot 2\cdot 3\cdot 4}=52360$$

Yes, I can work this out on a calculator with ease.

However, how can I work this out on pen and paper, or in my head with ease?

Are there any type of methods or cheats I could use to calculate this quickly?

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  • $\begingroup$ Do you want to find the exact answer or is approximate ok? $\endgroup$
    – Nick H
    Dec 22, 2014 at 16:43
  • $\begingroup$ Either or @NickH $\endgroup$
    – Jonathan Davies
    Dec 22, 2014 at 16:46

6 Answers 6

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Update: I got a very good Idea from comments so I updated the answer: $$T=\large\require{cancel}\frac{35\cdot34\cdot\color{purple}{\cancel{33}^{11}}\cdot\color{blue}{\cancel{32}^{\color{red}{\cancel{8}^{4}}}}}{1\cdot\color{red}{\cancel{2}}\cdot\color{purple}{\cancel{3}}\cdot\color{blue}{\cancel{4}}}=35\cdot34\cdot44$$ Now a good fact is square of a two digit number can be done by multiplying the tens digit by its successor and just appending 25 at the end like $35^2=\widehat{3\cdot4}\;25=1225$ $$\large35\cdot34\cdot44=(35^2-35)\cdot44=(1225-35)\cdot44=1190\cdot44$$

So: $$\large T=1190\cdot44\\\large=1190\cdot4\cdot11\\\large=(4000+400+360)\cdot11\\\large=4760\cdot11\\\large=(47600+4760)=52360$$

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    $\begingroup$ Pretty typesetting :) $\endgroup$
    – ClassicStyle
    Dec 22, 2014 at 16:51
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    $\begingroup$ Since it's $34$ and $35$ I think it would be faster to multiply $35\times35$ and subtract 35 from the result. for all numbers of the form $n5^2$ where $n$ is the tens digit, the answer is $n\times(n+1)$ for the first digits and 25 for the last: $35^2 \Rightarrow 3\times 4 = 12$, put $25$ behind and you get $1225$. $1225-35 = 1190$. $\endgroup$
    – Frank Vel
    Dec 22, 2014 at 16:57
  • $\begingroup$ Very nice use of colour. +1 $\endgroup$
    – beep-boop
    Dec 22, 2014 at 17:22
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    $\begingroup$ @fvel: Alternatively: 35 × 34 = 70 × 17 = 700 + 490 = 1190. $\endgroup$
    – ruakh
    Dec 23, 2014 at 5:11
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    $\begingroup$ @fvel just added the good Idea $\endgroup$
    – RE60K
    Dec 24, 2014 at 5:42
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If you're happy with something very approximate, that you can do in your head, $35\times 34\times 33\times 32$ is about $33^4$. $33^2$ is about $1,100$, so the numerator is about $1.2$ million. The denominator is $1\times 2\times 3\times 4$ is about $25$, which is $100/4$ so the answer is about $1.2$ million divided by a hundred, times four, which is $48,000$.

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$\frac{35\cdot34\cdot33\cdot32}{2\cdot3\cdot4}=35\cdot17\cdot11\cdot8=595\cdot11\cdot8=6545\cdot8=52360$

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On pen and paper, this is just a few multiplies. It is good to cancel factors first, getting $35*34*11*4$ Then look for easy ones. I would now go to $140*34*11$. Since multiplying by $11$ is just an addition, I would next to $140*34$

In your head, you probably won't get an exact answer, so look for approximations. I would say $35/4! \approx 1.5$ and multiplying two of the other factors gives about $1000$, so we have $1.5*34*1000 \approx 51000$ Knowing lots of arithmetic facts helps a lot.

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I tend to use "completing the squares" a lot for exact mental calculations.

$35.34.44 = 35.(39-5).(39+5)=35.(39.39-5.5)$ $= 35.(40.40-40-39-25)=35.(1521-25)=35.(1500-4)$ $=7.5(1500-4)=7.(7500-20)$

$=49000+3500-140=52500-140=52360$

Yes, superficially this looks quite long. However:

  • Each step is very tiny
  • Very little working memory is used
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$${35\cdot34\cdot33\cdot32\over1\cdot2\cdot3\cdot4}={5\cdot7\cdot2\cdot17\cdot3\cdot11\cdot8\cdot4\over3\cdot8}=10\cdot7\cdot17\cdot11\cdot4$$

The factor of $10$ will just give an extra $0$ at the end. The "difficult" multiplications are

$$7\cdot17=70+49=119$$

$$119\cdot11=1190+119=1309$$

and

$$1309\cdot4=5236$$

so the final answer is $52360$.

The trickiest part (for me) was getting the carry correct when adding $1190+119$. Alternatively, you can do the first two multiplications as

$$17\cdot11=170+17=187$$ and $$7\cdot187=7(200-13)=1400-91=1309$$

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