A multiplication table for a number n lists products n × 1, n × 2, n × 3, … up to a chosen limit (commonly 10).
For n = 5 (limit 10):
5 × 1 = 5
5 × 2 = 10
5 × 3 = 15
…
5 × 10 = 50
Pattern:
for i from 1 to limit:
print n × i = (n * i)| n | i | n × i |
|---|---|---|
5 | 1 | 5 |
5 | 2 | 10 |
5 | 10 | 50 |
Four clear steps — a simple for loop does the work.
n (hard-coded or from input).10).i from 1 to that limit.n × i = n * i.| Method | Idea | Time | Best for |
|---|---|---|---|
| Fixed 1–10 | for (i = 1; i <= 10; i++) | O(1) rows | Classic beginner demos |
| Custom limit | Loop to a parameter limit | O(limit) | Flexible / interview follow-ups |
function PrintTable(n, limit):
for i from 1 to limit:
print n, "×", i, "=", n * iKeep these facts handy while writing the program.
| Topic | Detail |
|---|---|
| Row formula | n * i for i = 1..limit |
| Default limit | 10 in many textbooks |
| Loop | for is the clearest choice |
| Output | printf("%d × %d = %d\n", n, i, n * i) |
| Input | scanf("%d", &n) when asked |
| Classic check | 5 × 10 = 50 |
Enter a number to print its table from 1 to 10.
n = 5, limit 5Each loop step multiplies n by the current i.
i = 1 → 5 × 1 = 5
i = 2 → 5 × 2 = 10
i = 3 → 5 × 3 = 15
i = 4 → 5 × 4 = 20
i = 5 → 5 × 5 = 25
Stop when i exceeds the limitStart with the classic 1–10 table. Then use a custom limit when the problem asks for more (or fewer) rows.
Print n × 1 through n × 10 with a simple for loop.
#include <stdio.h>
int main(void)
{
int n = 5;
int i;
for (i = 1; i <= 10; i++)
{
printf("%d × %d = %d\n", n, i, n * i);
}
return 0;
}n = 5 (or read it with scanf later).i starts at 1 and runs while i <= 10.n * i.i++ moves to the next multiplier.i = 10, the loop stops — ten rows printed.Accept n and a row count so the table length is flexible.
#include <stdio.h>
void print_table(int n, int limit)
{
int i;
if (limit < 1)
{
printf("Limit must be at least 1.\n");
return;
}
for (i = 1; i <= limit; i++)
{
printf("%d × %d = %d\n", n, i, n * i);
}
}
int main(void)
{
print_table(7, 5);
return 0;
}n and the upper limit.limit < 1 so the loop has a sensible range.for pattern prints 1..limit rows.print_table(7, 5) you get five lines ending at 7 × 5 = 35.Check these before calling the solution done.
Validate the upper bound before looping.
Every product is 0 — still a valid table.
i < 10 stops at 9 and misses the last row.
Confirm scanf returns 1 before using the read value.
| Program | Time | Extra space |
|---|---|---|
| Fixed 1–10 | O(1) (10 rows) | O(1) |
| Custom limit | O(limit) | O(1) |
Work grows with the number of rows printed. No extra arrays are required.
i and print n * i.10.One line: for each i from 1 to limit, print n × i = n * i.
Check whether an integer is a natural number.
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