3
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Find the divisor and all the digits of the sum.

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source : New scientist Magazine

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2
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it is obvious:

$m=0$ since $w-w=0$ from the beginning and $g+w=10$ since $m=0$

so

$f=w+1$

moreover we know that all letters represents different values, then

From $i-g = f$ we can conclude that $10+i-g=f$ since $m-g=i$ (previously $m-g=w$) and $w=i+1$

After that

we have found $i$ has to be $1$ since the difference between $3$ digit number - $2$ digit number is equal to $2$ digit number. so $w=2$ and $f=3$.

since we know that

$i=1$, so if $10+j-q=1$ that makes $j=0$ and $q=9$. then $130-k9=91$ so $k$ becomes $3$ which is not possible since we know that $f=3$.

so

we know that $j=q+1$ and $13j-kq=q1$ then $13-k=q$ and $k+q=13$, so for example, $q=4$, $j=5$ and $k=9$ is a possibility. all possibilities become $(4,5,9)$, $(5,6,8)$, $(7,8,6)$, $(8,9,5)$.

Moreover,

From $qig-fhp=qw$ we know that $q=f+1$ and we know that $f=3$ so $q=4$. and from the possibility table above, $(q,j,k)$ becomes $(4,5,9)$ and $41g-3hp=42$ and the possible values for $(h,g,p)$ are only $6,7,8$ since the rest is used already so it is easily seen that $(h,g,p)$ will be $(7,8,6)$.

Our letter values are:

$p=6,q=4,w=2,g=8,f=3,m=0,i=1,j=5,k=9,h=7$

our number becomes:

$302158$

To find the divisor, we need to find all values substracted from the values from the number,$wgw$,$igg$, $kq$ and $fhp$, which are:

$wgw=282$, $igg=188$, $kq=94$ and $fhp=376$.

I believe the first number ($pqwg$) represents the result so the answer becomes:

$47$.

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  • $\begingroup$ 302158/47=6428 remainder 42, it's a long division (without the dividend shown) $\endgroup$ – boboquack Sep 4 '17 at 8:51
  • $\begingroup$ @boboquack 302158/94=3214 remainder 42, it is like double of the result. $\endgroup$ – Oray Sep 4 '17 at 8:53
  • $\begingroup$ sorry, mixed it up - note that 47|94 $\endgroup$ – boboquack Sep 4 '17 at 8:53

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