[https://www.youtube.com/watch?v=zIEXhLZjqLE:embed:cite]
The program computes the greatest common divisor using the
subtraction-based Euclidean algorithm. When GCD(98, 56) is called, it
assigns the input values to two working variables, m and n, and then
repeatedly reduces the larger one by subtracting the smaller one. The loop
continues while m and n are different, and the point that matters most
for this question is the placement of the output statement: the program
prints the pair (m, n) immediately after it updates either m or n,
before it returns to the top of the loop to check whether m != n is still
true. That is why the printed sequence includes every intermediate state
produced by each subtraction step, including the final state where the two
values become equal. At the start, initialization sets m = 98 and n =
56. The loop condition asks whether m and n are unequal. Because 98
and 56 are not equal, the loop body executes. Inside the loop, the program
compares the two values. Since m ＞ n is true at this moment, it updates
m by subtracting n from it. The new value becomes m = 98 − 56 = 42,
while n remains 56. Immediately after this update, the program outputs
the current values, so the first printed pair is 42 56. Only after
printing does control flow return to the loop header. The loop condition is
evaluated again. Now m = 42 and n = 56, so they are still different and
the loop continues. The comparison m ＞ n is checked again, but this time
it is false because 42 is not greater than 56. Therefore, the program takes
the else-branch and updates n instead, subtracting m from n. The new
value becomes n = 56 − 42 = 14, while m stays 42. Right after updating
n, the program outputs the pair, producing the second printed values 42
14. Once again, the condition is not checked until the next iteration
begins. The third iteration starts with m = 42 and n = 14. Since they
are not equal, the loop runs. Now m ＞ n is true, so the program updates
m by subtracting n: m = 42 − 14 = 28. The variable n remains 14.
The program then outputs the current pair, so the third printed pair is 28
14. This reflects exactly what the algorithm is doing: it is shrinking the
larger number while preserving the GCD of the pair. In the fourth
iteration, m = 28 and n = 14 are still not equal, so the loop
continues. The comparison m ＞ n is again true, and m is updated as m =
28 − 14 = 14. Now both variables are equal, with m = 14 and n = 14.
However, because the output statement is executed immediately after the
update, the program still prints this final pair before the loop condition
is checked again. Therefore the fourth printed pair is 14 14. Only after
printing does the program return to the loop header. At that point the
condition m != n becomes false, because 14 equals 14, so the loop
terminates and the function returns m, which is 14. If the answer choices
present the output as one continuous sequence rather than as four separate
pairs, you simply read the printed pairs in order and flatten them. The
program prints 42 56, then 42 14, then 28 14, then 14 14. Flattened
into a single sequence, this becomes 42 56 42 14 28 14 14 14. This is
exactly the sequence that corresponds to option (2), which is why the
correct answer for the question is b), matching option (2). It is also
useful to understand why the algorithm is correct, because that explains
why the final returned value must be the GCD and why the intermediate
outputs make sense. Subtracting the smaller number from the larger does not
change the set of common divisors. If some integer d divides both m and
n, then it also divides m − n (when m ＞ n) because m − n is a
linear combination of m and n. Conversely, if d divides m − n and
also divides n, then it divides (m − n) + n = m. This means the
greatest common divisor is invariant under the transformation (m, n) → (m
− n, n) when m ＞ n, and similarly invariant under (m, n) → (m, n − m)
when n ＞ m. The loop keeps applying one of these transformations,
decreasing the larger value while keeping both values positive, so it must
eventually reach a state where m = n. When that equality occurs, the
common value must be the greatest common divisor of the original inputs,
because the GCD has been preserved at every step and the only common
divisor that can equal both values is their greatest common divisor. In
this execution, the process reaches 14 14, so the function returns 14,
and the printed sequence is simply the trace of these invariant-preserving
reductions.
