How many four-digit numbers greater than 2999 can be formed such that the product of the middle two digits exceeds 5?
This is what I know so far:
I just don't know how to count the number the middle two digits smartly (obviously listing all of them wouldn't exactly be smart considering there are many of them). The answer will be 7 * (# of possible middle digit combos) * 10. Can anyone take over from here?
a=3;b=0;c=0;d=0;p=0; cycle:n=a*1000+b*100+c*10+d;if((b * c)>5, 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
OUTPUT==4,970 such integers. Since there are 9,000 4-digit integers between 1,000 and 9,999, then the probability is: 4,970 / 9,000 =49.7%