An inverted repeating number triangle prints digit i exactly i times, counting i down from rows so the first row is widest.
Remember
Rule: for i = n..1, print i exactly i times
55555
4444
333
22
1 ← 5 rows
In C# you solve it with two nested for loops: the outer loop picks the digit, the inner loop repeats it with Console.Write, then WriteLine() ends the row. Program 10 is the growing twin (5, 44, 333…).
Approach
How to Solve It
Two ways to emit the same shape — start with nested loops, then optionally shorten with new string.
Method
Idea
Best for
Nested loops
Outer picks digit n..1; inner repeats it i times
Learning, interviews, exams
new string
Build a whole repeated-digit row in one call
Shorter demos once loops click
Pseudocode
Pseudocode
for i from rows down to 1:
for j from 1 to i:
print i (no newline)
print newline
Change the row count and the inverted triangle updates instantly — including the total digit count.
Whole numbers from 1 to 9 (single digits). Tap a chip or type a value — the preview redraws as you go.
Live result5 rows · 15 digits
55555
4444
333
22
1
Trace
Worked Walkthrough — rows = 4
Trace each outer-loop value of i, how many times the inner loop runs, and the printed row.
i
Inner j
Repeats
Printed row
4
1..4
4
4444
3
1..3
3
333
2
1..2
2
22
1
1..1
1
1
Total digits are 4+3+2+1 = 10 — the triangular number n(n+1)/2.
Code
C# Programs
Three complete programs: fixed height, row-count input, and a new string shortcut. Use View Output for sample results.
Example 1 — Fixed rows = 5
Outer loop picks the digit; inner loop repeats it exactly i times.
C#
using System;
namespace MyApp
{
class Program
{
static void Main(string[] args)
{
int rows = 5;
int i, j;
for (i = rows; i >= 1; i--)
{
for (j = 1; j <= i; j++)
{
Console.Write(i);
}
Console.WriteLine();
}
}
}
}
Output
55555
4444
333
22
1
How It Works
1. Count down.for (i = rows; i >= 1; i--) yields 5, 4, 3, 2, 1 — the digit for each row.
2. Match repeats to digit. When i = 5, the inner loop runs five times (55555); when i = 1, just once (1).
3. Newline.WriteLine() after the inner loop starts the next shorter row.
Example 2 — User Input
Read the row count at runtime. Prefer int.TryParse in real apps so bad input does not throw.
C#
using System;
namespace MyApp
{
class Program
{
static void Main(string[] args)
{
int rows;
int i, j;
Console.Write("Enter the number of rows: ");
if (!int.TryParse(Console.ReadLine(), out rows) || rows < 1)
{
Console.WriteLine("Please enter a positive whole number.");
return;
}
for (i = rows; i >= 1; i--)
{
for (j = 1; j <= i; j++)
{
Console.Write(i);
}
Console.WriteLine();
}
}
}
}
2. Same nested loops. Four rows start at 4444 and end at 1.
Example 3 — new string Shortcut
Build each repeated-digit row in one call. Best for single-digit rows (1–9).
C#
using System;
namespace MyApp
{
class Program
{
static void Main(string[] args)
{
int rows = 5;
for (int i = rows; i >= 1; i--)
{
Console.WriteLine(new string(i.ToString()[0], i));
}
}
}
}
Output
55555
4444
333
22
1
How It Works
1. Repeat a char.new string(i.ToString()[0], i) builds a string of the digit character repeated i times.
2. Same shape. Identical output to Example 1 — keep the two-loop version for exams that ask for both bounds.
Edge Cases & Pitfalls
Check these before calling the solution done.
wrong twin
for (j = rows; j >= i; j--)
That is Program 10 (5, 44, 333…). Keep j from 1 to i so repeats equal the digit.
WriteLine early
WriteLine(i) inside the inner loop
That puts each digit on its own line. Use Write(i); call WriteLine() only after the row finishes.
off-by-one
j < i instead of j <= i
You drop one repeat on every row (e.g. 55555 becomes 5555).
rows = 1
Single digit
Output is just 1 — a good sanity check.
rows ≤ 0
Empty output
The outer loop never runs. Validate and prompt again for clearer UX.
multi-digit
rows > 9
Write(10) prints two characters. Labs usually stick to 1–9; the new string shortcut also assumes a single digit char.
Analysis
Time and Space Complexity
Program
Time
Extra space
Nested loops (Examples 1–2)
O(n²)
O(1)
new string shortcut (Example 3)
O(n²)
O(n) per row for the built string
There are n rows printing n+(n-1)+…+1 digits, so total work is n(n+1)/2.
Remember
Key Takeaways
Rule: for i = n..1, print digit i exactly i times.
Inner bound:j from 1 to i matches repeats to the digit and shrinks each row.
Write vs WriteLine: repeats stay on the line; WriteLine advances after each row.
Next step: Program 12 shrinks a different repeating sequence.
One line: for each digit from n down to 1, print it i times with Write, then WriteLine.
Frequently Asked Questions
The outer loop decreases i from rows down to 1, and the inner loop runs j from 1 to i. Row i repeats digit i exactly i times, so the widest row is first and each next row is shorter.
for (j = 1; j <= i; j++) Console.Write(i) runs the inner body i times. When i is 5 you get 55555; when i is 1 you get a single 1.
Console.Write stays on the same line. Console.WriteLine ends the current line. Repeated digits use Write(i); the row break uses WriteLine after the inner loop.
Program 10 grows repeat count as the digit shrinks (5, 44, 333, 2222, 11111) using for (j = rows; j >= i; j--). Program 11 repeats digit i exactly i times (55555, 4444, 333, 22, 1) using for (j = 1; j <= i; j++).
O(n²) where n is the number of rows. Total Console.Write calls equal n+(n-1)+…+1 = n(n+1)/2.
Yes. Console.WriteLine(new string(i.ToString()[0], i)) prints a full single-digit row in one call. Nested loops are better for learning; the string constructor is a handy shortcut later.
Prefer int.TryParse(Console.ReadLine(), out rows) so bad input does not throw FormatException.
The outer loop never runs, so nothing is printed. Validate and prompt again if you want a clear user message.
🤔
Did you know?
Row digit i repeats exactly i times. The outer loop counts down from rows, so the first row is widest (rows copies of rows) and each line shortens — still O(n²) total prints.
