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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:

• It has to be a four digit number
• The first digit has to be between 3 and 9 (inclusive) so there are 7 options
• The units digit can be any number so there are 10 options

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?

Feb 5, 2021

#1
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There are 7*48*10 = 3360 possible four-digit numbers.

Feb 5, 2021
#2
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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%

Feb 6, 2021
#3
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Sorry for the typo! The probability should read: 4,970 / 9,000 =55.22%

Guest Feb 6, 2021