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Dropdown 1: 24
Dropdown 2: 5
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
52% (03:08) correct
48%
(04:09)
wrong
based on 58
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The flowchart represents a mathematical algorithm that takes two positive integers as the input and returns a positive integer as the output. Processes are indicated in the rectangular symbols in the flowchart. Each process is symbolized by an equation, such as S = S + x. In this particular process, the current values of the variables S and x are added together and the sum then becomes the value of S. For example, if the value of S is 4 and the value of x is 9 before the process S = S + x is completed, then the value of S is 13 and the value of x is 9 after the process is completed.
Use the drop-down menus to fill in the blanks in the following statements based on the algorithm represented by the flowchart.
If 12 and 20 are entered as the values for x and y, respectively, then immediately after the process x = 3x is completed for the second time, the value of S is .
If 15 and 17 are entered as the values for x and y, respectively, then the process S = S + x is completed times before the algorithm stops.
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ShowHide Answer
Official Answer
Dropdown 1: 24
Dropdown 2: 5
The flowchart represents a mathematical algorithm that takes two positive integers as the input and returns a positive integer as the output. Processes are indicated in the rectangular symbols in the flowchart. Each process is symbolized by an equation, such as S = S + x. In this particular process, the current values of the variables S and x are added together and the sum then becomes the value of S. For example, if the value of S is 4 and the value of x is 9 before the process S = S + x is completed, then the value of S is 13 and the value of x is 9 after the process is completed.
Use the drop-down menus to fill in the blanks in the following statements based on the algorithm represented by the flowchart.
If 12 and 20 are entered as the values for x and y, respectively, then immediately after the process x = 3x is completed for the second time, the value of S is .
If 15 and 17 are entered as the values for x and y, respectively, then the process S = S + x is completed times before the algorithm stops.
Official Solution:
Drop-down 1:
Start with:
S = 0, x = 12, y = 20
Since 20 is not divisible by 3:
S = 0 + 12 = 12
y = 20 - 1 = 19
Since 19 is not divisible by 3:
S = 12 + 12 = 24
y = 19 - 1 = 18
Since 18 is divisible by 3:
x = 3 * 12 = 36
y = 18/3 = 6
This is the first time the process x = 3x is completed.
Since 6 is also divisible by 3:
x = 3 * 36 = 108
This is the second time the process x = 3x is completed. The value of S has not changed since it became 24.
Therefore, the value of S is 24.
Drop-down 2:
Start with x = 15 and y = 17. We only need to count how many times the process S = S + x is completed.
17 is not divisible by 3, so S = S + x is completed for the first time, and y becomes 16.
16 is not divisible by 3, so S = S + x is completed for the second time, and y becomes 15.
15 is divisible by 3, so x becomes 45 and y becomes 5.
5 is not divisible by 3, so S = S + x is completed for the third time, and y becomes 4.
4 is not divisible by 3, so S = S + x is completed for the fourth time, and y becomes 3.
3 is divisible by 3, so x becomes 135 and y becomes 1.
1 is not divisible by 3, so S = S + x is completed for the fifth time, and y becomes 0.
The algorithm then stops.
Therefore, the process S = S + x is completed 5 times.
Correct answer:
Drop-down 1: "24"
Drop-down 2: "5"
General Discussion
abhikc3004
27 Aug 2026, 23:50
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Making a table and simulating the logic is helpful:
Case 1:
Hence, \(S = 24\)
Case 2:
The algorithm stops at \(y = 0\), hence no. of iterations of \(S = S+x\) is \(5\).
Bunuel - this took me around 4 minutes to compute, which is feel is a lot. Is there a shortcut/smarter way to tackling such questions?
Case 1:
| \(S \\ (=S+x)\) | \(x \\ (=3x)\) | \(y \) (= \( y/3 \) or \(y-1) \) |
| 0 | 12 | 20 |
| 24 | 12 | 18 |
| 24 | 36 | 6 |
| 24 | 108 | 2 |
Hence, \(S = 24\)
Case 2:
| Condition Counter | \(S \\ (=S+x) \) | \(x\\ (=3x)\) | \(y \) (= \( y/3 \) or \(y-1) \) |
| 0 | 15 | 17 | |
| 1 | 15 | 15 | 16 |
| 2 | 30 | 15 | 15 |
| - | 75 | 45 | 5 |
| 3 | 75 | 45 | 4 |
| 4 | 120 | 45 | 3 |
| - | 120 | 135 | 1 |
| 5 | 255 | 135 | 0 (end) |
The algorithm stops at \(y = 0\), hence no. of iterations of \(S = S+x\) is \(5\).
Bunuel - this took me around 4 minutes to compute, which is feel is a lot. Is there a shortcut/smarter way to tackling such questions?
