+55 Hello! I am new to this website and I am not the best at math, but I like helping people so I will try to help. Please tell me if I get it wrong, and don't be mad. Thanks!
My thinking is that there are exactly five possible even digits, and we can first take a look at the possible 4-digit numbers where the first digit is even. There are 4 choices for the first one because it can't be 0, and 10 for each one following. This gives 4*10*10*10=4000. We can do the same for when the even digit is in the hundereds, tens, and ones place, then add them all up: 4000+5000+5000+5000=19000. So the answer is 19000. Again, I might not be correct. :)
+781 There can only be one even digit...so the other three are odd. Let O represent an odd digit and E an even digit.
The numbers can be like this:
EOOO
OEOO
OOEO
OOOE
In each number, there are 5 possible odd digits(1, 3, 5, 7, 9) and 5 possible even digits(0, 2, 4, 6, 8). In the first case, EOOO, the first digit can't be 0 so there are 4*5*5*5=500 ways.
The other three cases are the same: each one has 5^4=625 ways.
So the total number of integers with only one even digit is 624+500=1125.
Let me know if this is correct.
a=1;b=0;c=0;d=0;p=0; cycle:n=a*1000+b*100+c*10+d;if(a%2==0 and b%2==1 and c%2==1 and d%2==1 or b%2==0 and a%2==1 and c%2==1 and d%2==1 or c%2==0 and a%2==1 and b%2==1 and d%2==1 or d%2==0 and a%2==1 and b%2==1 and c%2==1, goto loop, goto next); loop:printn," ",;p=p+1; next:d++;if(d<10, goto cycle, 0);d=0;c++;if(c<10, goto cycle, 0);d=0;c=0;b++;if(b<10, goto cycle,0);b=0;c=0;d=0;a++;if(a<10, goto cycle,0);print"Total = ",p
OUPUT = 2,375 such numbers.
375 various evens between 1011 and 1998
125 Twos between 2111 and 2999
375 various evens between 3011 and 3998
125 Fours between 4111 and 4999
375 various evens between 5011 and 5998
125 Sixes between 6111 and 6999
375 various evens between 7011 and 7998
125 Eights between 8111 and 8999
375 various evens beween 9011 and 9998
Grand Total =[5 x 375] + [4 x 125] =2,375 integers with exactly one EVEN digit.
