It depends with your speed but it can take you 10^18 light years to count from 1 to 10 sextillion.
Locate 6,1Input "Your Name Please ",Nam$LOCATE 8,1FOR Count=1 to 10Print TAB(5) Count " > ";Nam$Next Count
five next time count
10
If you count 11 as 2 instances, the digit 1 appears 18 times if you don't count 10, 19 times if you do. 10,11,12,13,14,15,16,17,18,19,21,31,41,51,61,71,81,91
you cna you can only count backwards form 10-1
It depends with your speed but it can take you 10^18 light years to count from 1 to 10 sextillion.
int i; for (i=1; i<=10; i++) printf ("%d %d\n", i, i*i);
Locate 6,1Input "Your Name Please ",Nam$LOCATE 8,1FOR Count=1 to 10Print TAB(5) Count " > ";Nam$Next Count
public class DoWhileLoopDemo {public static void main(String[] args) {int count = 1;System.out.println("Printing Numbers from 1 to 10");do{System.out.println(count++);}while( count
it is 10. First, you need to count how many 1's there are. 1,2,3,4,5,6,7,8,9,10! There are 10.
1 2 3 4 5 6 7 8 9 10
there are 10 squares if you count 100
Count Duckula - 1988 Castle Duckula Open to the Public 1-10 is rated/received certificates of: Australia:G
Count the zeroes. 10000000000 = 1 x 10^10 (the ^ means exponent).
Using n bits, you can count to 2n - 1. This is for unsigned integers. So 10 bits = 210 - 1 = 1023 14 bits = 214 - 1 = 16383 To count to 511 you need log2(511+1) = log2(512) = 9 bits. To count to 63 you need log2(63+1) = log2(64) = 6 bits.
five next time count