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GO Classes DPP | GATE CS | C Programming | Nested Loops

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Initially, $\texttt{count = 0}$.

For $\texttt{i = 1}$:
$\texttt{j = 1}$ gives $\texttt{1 * 1 = 1}$, so add $\texttt{1}$.
$\texttt{j = 2}$ gives $\texttt{1 * 2 = 2}$, so add $\texttt{2}$.
$\texttt{j = 3}$ gives $\texttt{1 * 3 = 3}$, so add $\texttt{3}$.
$\texttt{j = 4}$ gives $\texttt{1 * 4 = 4}$, so add $\texttt{4}$.
Now, $\texttt{count = 10}$.

For $\texttt{i = 2}$:
$\texttt{j = 1}$ gives $\texttt{2 * 1 = 2}$, so add $\texttt{1}$.
$\texttt{j = 2}$ gives $\texttt{2 * 2 = 4}$, so add $\texttt{2}$.
$\texttt{j = 3}$ gives $\texttt{2 * 3 = 6}$, so add $\texttt{3}$.
$\texttt{j = 4}$ gives $\texttt{2 * 4 = 8}$, so $\texttt{break}$ occurs.
Now, $\texttt{count = 10 + 1 + 2 + 3 = 16}$.

For $\texttt{i = 3}$:
$\texttt{j = 1}$ gives $\texttt{3 * 1 = 3}$, so add $\texttt{1}$.
$\texttt{j = 2}$ gives $\texttt{3 * 2 = 6}$, so add $\texttt{2}$.
$\texttt{j = 3}$ gives $\texttt{3 * 3 = 9}$, so $\texttt{break}$ occurs.
Now, $\texttt{count = 19}$.

For $\texttt{i = 4}$:
$\texttt{j = 1}$ gives $\texttt{4 * 1 = 4}$, so add $\texttt{1}$.
$\texttt{j = 2}$ gives $\texttt{4 * 2 = 8}$, so $\texttt{break}$ occurs.
Now, $\texttt{count = 20}$.

Therefore, the output is $\texttt{20}$.

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