Number __GAME__!!!!!!

We have to consider the total numbers $N$ by stages as follows:

$\begin{matrix} Up \space to & & Numbers \\ 4 & 4= & 4 \\ 44 & 4\times 4= & 16 \\ 444 & 4 \times 4\times 4= & 64 \\ 3444 & 3 \times 4 \times 4\times 4= & 192 \\ 4244 & 1 \times 2 \times 4\times 4= & 32 \\ 4314 & 1 \times 1 \times 1\times 4= & 4 \\ 4321 & 1 \times 1 \times 1\times 1= & 1 \end{matrix}$

Therefore, the total number of natural numbers formed with digits 1, 2, 3 and 4 with repetitions $N = 4+16+64+192+32+4+1 = \boxed{313}$

$The\ question\ says\ that\ we\ have\ to\ find\ natural\ numbers\ using\ digits\ from\ 1\ to\ 4.$ $But\ it\ hasn't\ specified\ on\ the\ number\ of\ digits\ of\ each\ number.$ $Thus\ we\ have\ to\ find\ all\ numbers\ with\ digits\ from\ 1\ to\ 4 \ not\ exceeding\ 4321/$ $We\ will\ do\ this\ casewise.$ $Case\ 1:$ $Single\ digit\ numbers$ $Obviously\ 4\ can\ be\ formed.$ $case\ 2:$ $Two\ digit\ numbers$ $We\ have\ four\ possibilities\ for\ the\ unit\ digit\ and\ four\ possibilities\ for\ the\ tens\ digit$ $Thus,\ 16\ numbers\ can\ be\ formed.$ $Case\ 3:$ $3\ digit\ numbers$ $We\ have\ four\ possibilities\ for\ each\ place\ thus,\ 64\ numbers\ are\ possible.$ $Till\ now\ it\ has\ been\ pretty\ straight\ forward\ but\ the\ next\ case\ has\ some\ complications\ as\ we\ have\ a\ restriction.$ $Case\ 4:$ $Four\ digit\ numbers$ $Subcase\ 1:$ $1\ is\ the\ digit\ in\ the\ thousandths\ place$ $Then\ we\ have\ 4\ possibilities\ for\ each\ place,\ as\ a\ number\ which\ has\ 1\ in\ its\ thousandths\ place\ will\ be\ less\ than\ 4321$ $Thus,\ we\ have\ 64\ numbers.$ $Subcase\ 2:$ $2\ is\ the\ digit\ in\ the\ thousandths\ place$ $Then\ we\ have\ 4\ possibilities\ for\ each\ place,\ as\ a\ number\ which\ has\ 2\ in\ its\ thousandths\ place\ will\ be\ less\ than\ 4321$ $Thus,\ we\ have\ 64\ numbers$ $Subcase\ 3:$ $3\ is\ the\ digit\ in\ the\ thousandths\ place.$ $Then\ we\ have\ 4\ possibilities\ for\ each\ place,\ as\ a\ number\ which\ has\ 3\ in\ its\ thousandths\ place\ will\ be\ less\ than\ 4321$ $Subcase\ 4:$ $4\ is\ the\ digit\ in\ the\ thousandths\ place.$ $If\ we\ have\ 1\ or\ 2\ in\ the\ hundreths\ place\ then\ 16\ +\ 16\ numbers$ $If\ the\ number\ has\ 3\ in\ the\ hundreths\ place\ and\ 1\ in\ the\ tens\ place\ then\ there\ are\ 4\ possibilities.$ $If\ the\ number\ has\ 3\ in\ the\ hundrteths\ place\ and\ 2\ in\ the tenths\ place\ then\ there\ is\ only\ one\ possibility.$ $Adding\ gives\ 313.$

As the maximum number which can be formed is 4444. Hence, we can easily calculate the numbers from 4322 upto 4444 formed with the Digits 1,2,3 & 4 only, which comes out to be 27. Now we can calculate the total numbers which can be made using the digits (1,2,3 &4) which is 4^4(4 digit numbers) + 4^3 (3 digit numbers) +4^2 (2 digit numbers) +4 (single digit number) which comes out to be 340.

Hence the answer is 340-27= $\boxed{313}$

PARI/GP code:

is_ok(n)=m=Vec(Str(n));i=1;while(i<=#m,if(eval(m[i])==(1||2||3||4),i++,return(0);break));return(1)
answer=0;for(i=1,4321,if(is_ok(n),answer++));return(answer)