#include
#include
void main()
{
clrscr();
for (int i=1;i<=5;i++) // this for loop controls the line
{
for(int
j=1;j<=i;j++) // this for loop controls the values(1,12,123..)
{
printf("%d",j);
}
printf("\n");
}
getch();
}
REMPATTERNS (not necessary) CLS FOR I = 1 TO 5 STEP +1 FOR J = 1 TO I STEP+1 PRINT J NEXT J PRINT NEXT I END Hope you find this helpful :)
====== 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
write a program to print the series 1/12+1/22+.........+1/n2 ?
num%=9 CLS PRINT "The square of: "; num%; " = "; num%*num% END
// example of 1..12x12 table for($i = 1; $i <= 12; $i++) { for($j = 1; $j <= 12; $j++) { print ($i * $j) ." "; } print "\n"; }
The sequence you provided is known as the powers of 11. Each term represents the coefficients of the binomial expansion of ((a + b)^n) where (n) is the term's index. The next terms in the sequence would be 161051 (for (n=5)) and 1771561 (for (n=6)). Thus, the complete sequence is 1, 11, 121, 1331, 14641, 161051, 1771561.
1, 11, 121, 1331.
1, 11, 121, 1331.
1212 = 14,641 The factors of 14,641 are 1, 11, 121, 1331, and 14641. The prime factors of 14,641 are 11, 11, 11, and 11. The prime factorization of 14,641 is 11 x 11 x 11 x 11 = 114
1, 2, 11, 22, 121, 242, 1331, 2662.
1,331 = 1 x 1,331and1,331 = 11 x 121
To calculate 121 x 11, you can use the distributive property of multiplication over addition. First, multiply 121 by 10 to get 1210. Then, multiply 121 by 1 to get 121. Finally, add these two results together to get the final answer of 1331.
buy aprinter and somepaper and then go on wordput numbers in and click print and click color and youhave your numbers
The positive integer factors of 2662 are: 1, 2, 11, 22, 121, 242, 1331, 2662
1, 11, 13, 121, 143, 169, 1331, 1573, 1859, 17303, 20449, 224939
#include<stdio.h> #include<conio.h> void main() { int i,j,sp=0; clrscr(); for(i=5;i>=1;i--) { for(j=1;j<sp;j++) printf(" "); for(j=1;j<=i;j++) printf("*"); printf("\n"); s++; } getch(); }
1 x 6655, 5 x 1331, 11 x 605, 55 x 121.