The GCF of all the composite numbers under 100 is 1. For e.g 4 & 9 are composite numbers & their GCF is 1, so if we take other composite numbers along with 4 & 9 the GCF will be 1.
1, 4, 9, 16, 25, 36, 49, 64, 81, 100
There are 720 of them. The three digit counting numbers are 100-999. All multiples of 5 have their last digit as 0 or 5. There are 9 possible numbers {1-9} for the first digit, There are 10 possible numbers {0-9} for each of the first digits, There are 8 possible numbers {1-4, 6-9} for each of the first two digits, Making 9 x 10 x 8 = 720 possible 3 digit counting numbers not multiples of 5.
numbers 1-9 is the answer
Prime squares: 4, 9, 25 and 49
20
11 (9, 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99) Those numbers divisible by 9 are the multiple of 9; thus need to know how many multiples of 9 there are between 1 and 100: 100 ÷ 9 = 11 r 1 ÷ last multiple of 9 is 11 × 9 (= 99) → There are 11 - 1 + 1 = 11 numbers between 1 and 100 which are divisible by 9. (They are 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99.)
Just the one which is 9 but there are 20 numbers from 1 to 100 that contain the digt 9
11 times
34 of them.
Twenty of them.
There are nine square numbers including 1 below 100 (1x1...9x9). So, there are 99 - 9 'not square numbers' below 100 which is 90
9
45 of them.
1*9+2*90+3*1=192
Twenty of them from 9 to 99
1 to 9 = 9 10 to 99 = 9o x 2 = 180 100 = 3 9 + 180 + 3 = 192