$\overline{1T} \times \overline{1N}$

If we write out the 10s and 100s columns in the cryptogram, it states that:

$(10+T)\times(10+N)=200+10T+T$

where $T$ and $N$ are between 0 and 9. Now we can expand, simplify and factorise to get all the Ts and Ns together:

$100+10T+10N+TN=200+11T$

$TN-T+10N-10=90$

$T(N-1)+10(N-1)=90$

$(T+10)(N-1)=90$

Now we know that $T+10$ and $N-1$ are two integers which multiply to give 90. These are all the pairs of factors which multiply to give 90:

$1,90$

$2,45$

$3,30$

$5,18$

$6,15$

$9,10$

However, for a pair of factors to satisfy the cryptogram, one factor must equal $T+10$ while the other must equal $N-1$ , and since $T$ lies between 0 and 9 a factor must therefore lie between 10 and 19. For instance, if we looked at the pair of factors $(3,30)$ , $T$ would have to equal either -7 or 20, neither of which would make sense in the cryptogram. An example of a pair that works is $(5,18)$ , as $T$ can equal 8 and $N$ can equal 6.

The only pairs containing a factor between 10 and 19 are:

$5,18$

$6,15$

So the answer is that there are only 2 possible solutions for $N$ .

We can check these; if we were to let $N-1$ equal the smaller number in the pair and $T+10$ equal the larger number, we would get

$N=6$ and $T=8$ for the $(5,18)$ pair, so the cryptogram would be

$18\times16=288$ , or

$N=7$ and $T=5$ for the $(6,15)$ pair, so the cryptogram would be

$15\times17=255$

We can see that both of these solutions check out with the cryptogram in the question, and so there are 2 solutions.