1 12 123 1234 12345
12345 1234 123 12 1
write a program to print the series 1/12+1/22+.........+1/n2 ?
// example of 1..12x12 table for($i = 1; $i <= 12; $i++) { for($j = 1; $j <= 12; $j++) { print ($i * $j) ." "; } print "\n"; }
====== FOR tablesNo%=1 TO 12 CLS PRINT "PROGRAM: "; tablesNo%; "X Tables:-" PRINT FOR timesNo%=1 TO 12 PRINT timesNo%; " X "; tablesNo%; " = "; timesNo%*tablesNo% NEXT PRINT PRINT "Press [SPACEBAR] key to continue..." SLEEP NEXT END
As an example if your number in Malaysia is 012 123 1234 International: +60 12 - 123 1234 Local: 012 - 123 1234 As you notice in Malaysia the - (minus) sign is after the prefix and not as a separator of the number as it is practised in USA.
by:jason pin Jason4
the next one is 1234
Account Number 4000 1234 5678 9010 123 DEBIT VISA Good Thru 12/23
printf ("12345 1234 123 12 1\n"); ... or, did you mean to do it with loops ? ... int i, j; for (i=5; i>0; i--) { for (j=1; j<=i; j++) printf ("%d", j); printf (" "); } printf ("\n");
1 12 123 1234 12345
1+12+123+1234+12345+123456+1234567+12345678+123456789+12345678910+1234567891011+123456789101112 =124703839845238 All the operations are addition so all the addition can happen in one step.
12345 1234 123 12 1
12 x 123 = 1,476
write a program to print the series 1/12+1/22+.........+1/n2 ?
0.0976
14808