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(This article was first published on <strong><a href="https://xianblog.wordpress.com/2017/04/14/optimultiplication-a-riddle/"> R – Xi'an's Og</a></strong>, and kindly contributed to <a rel="nofollow" href="https://www.r-bloggers.com/">R-bloggers)</a> </div>
<p><strong>T</strong>he <a target="_blank" rel="nofollow" href="https://fivethirtyeight.com/features/can-you-outsmart-our-elementary-school-math-problems/">riddle of this week</a> is about an optimisation of positioning the four digits of a multiplication of two numbers with two digits each and is open to a coding resolution:</p>
<blockquote>
<p><em>Four digits are drawn without replacement from {0,1,…,9}, one at a time. What is the optimal strategy to position those four digits, two digits per row, as they are drawn, toward minimising the average product?</em></p>
</blockquote>
<p>Although the problem can be solved algebraically by computing <strong>E</strong>[X⁴|x¹,..] and <strong>E</strong>[X⁴X³|x¹,..]  I wrote three R codes to “optimise” the location of the first three digits: the first digit ends up as a unit if it is 5 or more and a multiple of ten otherwise, on the first row. For the second draw, it is slightly more variable: with this R code,</p>
<pre>second&lt;-function(i,j,N=1e5){draw
drew=matrix(0,N,2)
for (t in 1:N)
  drew[t,]=sample((0:9)[-c(i+1,j+1)],2)
conmean=(45-i-j)/8
conprod=mean(drew[,1]*drew[,2])
if (i&lt;5){ #10*i
 pos=c((110*i+11*j)*conmean,
       100*i*j+10*(i+j)*conmean+conprod,
       (100*i+j)*conmean+10*i*j+10*conprod)}else{
 pos=c((110*j+11*i)*conmean,
       10*i*j+(100*j+i)*conmean+10*conprod,
       10*(i+j)*conmean+i*j+100*conprod)
 return(order(pos)[1])}
</pre>
<p>the resulting digit again ends up as a unit if it is 5 (except when x¹=7,8,9, where it is 4) or more and a multiple of ten otherwise, but on the second row. Except when x¹=0, x²=1,2,3,4, when they end up on the first row together, 0 obviously in front.</p>
<p>For the third and last open draw, there is only one remaining random draw, which mean that the decision only depends on x¹,x²,x³ and <strong>E</strong>[X⁴|x¹,x²,x³]=(45-x¹-x²-x³)/7. Attaching x³ to x² or x¹ will then vary monotonically in x³, depending on whether x¹>x² or x¹&lt;x²:</p>
<pre>fourth=function(i,j,k){
comean=(45-i-j-k)/7
if ((i&lt;1)&amp;(j&lt;5)){ pos=c(10*comean+k,comean+10*k)}
if ((i&lt;5)&amp;(j>4)){ pos=c(100*i*comean+k*j,j*comean+100*i*k)}
if ((i>0)&amp;(i&lt;5)&amp;(j&lt;5)){ pos=c(i*comean+k*j,j*comean+i*k)} 
if ((i&lt;7)&amp;(i>4)&amp;(j&lt;5)){ pos=c(i*comean+100*k*j,j*comean+100*i*k)} 
if ((i&lt;7)&amp;(i>4)&amp;(j>4)){ pos=c(i*comean+k*j,j*comean+i*k)}
if ((i>6)&amp;(j&lt;4)){ pos=c(i*comean+100*k*j,j*comean+100*i*k)} 
if ((i>6)&amp;(j>3)){ pos=c(i*comean+k*j,j*comean+i*k)}
return(order(pos)[1])}</pre>
<p>Running this R code for all combinations of x¹,x² shows that, except for the cases x¹≥5 and x²=0, for which x³ invariably remains in front of x¹, there are always values of x³ associated with each position.</p>
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