Problem 443

Since the second column includes the number $1$, then step (ii) was never used on the
second column, otherwise each entry would be at least $2$.

To generate the $1$, $3$ and $2$ in the second column, we thus need to
have used step (i) $1$ time on row
$1$, $3$ times on row $2$, and $2$ times on row $3$.

This gives:

$1$
$1$
$1$

$3$
$3$
$3$

$2$
$2$
$2$

We cannot use step (i) any more times, otherwise the entries in
column $2$ will increase. Thus, $a = 1 + 3 + 2 = 6$.

To obtain the final grid from this current grid using only step (ii), we
must increase each entry in column $1$ by $6$ (which means using step (ii) $3$ times) and increase each entry in
column $3$ by $4$ (which means using step (ii) $2$ times). Thus, $b = 3 + 2 = 5$.

Therefore, $a + b = 11$.

Since the second column includes the number $1$, then step (ii) was never used on the
second column, otherwise each entry would be at least $2$.

To generate the $1$, $3$ and $2$ in the second column, we thus need to
have used step (i) $1$ time on row
$1$, $3$ times on row $2$, and $2$ times on row $3$.

This gives:

$1$
$1$
$1$

$3$
$3$
$3$

$2$
$2$
$2$

We cannot use step (i) any more times, otherwise the entries in
column $2$ will increase. Thus, $a = 1 + 3 + 2 = 6$.

To obtain the final grid from this current grid using only step (ii), we
must increase each entry in column $1$ by $6$ (which means using step (ii) $3$ times) and increase each entry in
column $3$ by $4$ (which means using step (ii) $2$ times). Thus, $b = 3 + 2 = 5$.

Therefore, $a + b = 11$.

The three different integers selected from $1$ to $6$ and whose sum is $7$ must be the integers $1$, $2$, $4$. Thus, the vertical column contains the
integers $1$, $2$, $4$
in some order.

(Can you see why no other combination of three of the given integers has
a sum of $7$?)

The three different integers selected from $1$ to $6$ and whose sum is $11$ must be $1$, $4$, $6$
or $2$, $4$, $5$
or $2$, $3$, $6$.

If the integers in the horizontal row are $1$, $4$, $6$, then there are two integers in common
with those in the vertical column, namely $1$ and $4$.

Since there have to be five different integers used in the squares, then
there cannot be two integers in common between the two lists, and so
$1$, $4$, $6$
cannot appear in the horizontal row.

Similarly, $2$, $4$, $5$
cannot appear in the horizontal row.

Thus, the horizontal row must contain the integers $2$, $3$, $6$
with $2$ appearing in the centre
square since it is the integer in common between the two lists.

The integer not appearing in any square is $5$.

The figure shows a possible arrangement of the integers.