Difference between revisions of "2016 New Spec"
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'''Example:''' calculating the additive persistence of 87 | '''Example:''' calculating the additive persistence of 87 | ||
| − | 8 + 7 = 15 | + | 8 + 7 = 15 |
1 + 5 = 6 | 1 + 5 = 6 | ||
After 2 steps the method results in a one digit answer so the additive persistence of 87 is 2. | After 2 steps the method results in a one digit answer so the additive persistence of 87 is 2. | ||
| Line 11: | Line 11: | ||
2 * 7 = 14 | 2 * 7 = 14 | ||
1 * 4 = 4 | 1 * 4 = 4 | ||
| − | After 3 steps the method results in a one digit answer so the multiplicative persistence of 39 is 3. | + | After 3 steps the method results in a one digit answer so the multiplicative persistence of 39 is 3. |
=Figure 4= | =Figure 4= | ||
| − | OUTPUT "Enter integer (0-99): " | + | OUTPUT "Enter integer (0-99): " <br/> |
| − | INPUT Value | + | INPUT Value <br/> |
| − | OUTPUT "Calculate additive or multiplicative persistence (a or m)? " | + | OUTPUT "Calculate additive or multiplicative persistence (a or m)? " <br/> |
| − | INPUT Operation | + | INPUT Operation <br/> |
| − | Count = 0 | + | Count = 0 <br/> |
| − | WHILE Value > 9 | + | WHILE Value > 9 <br/> |
| − | :IF Operation = "a" THEN | + | :IF Operation = "a" THEN <br/> |
| − | ::Value = (Value DIV 10) + (Value MOD 10) | + | ::Value = (Value DIV 10) + (Value MOD 10)<br/> |
| − | :ELSE | + | :ELSE<br/> |
| − | ::Value = (Value DIV 10) * (Value MOD 10) | + | ::Value = (Value DIV 10) * (Value MOD 10)<br/> |
| − | :ENDIF | + | :ENDIF<br/> |
| − | :Count = Count + 1 | + | :Count = Count + 1<br/> |
| − | ENDWHILE | + | ENDWHILE<br/> |
| − | OUTPUT "The persistence is: " | + | OUTPUT "The persistence is: "<br/> |
| − | OUTPUT Count | + | OUTPUT Count<br/> |
| − | The MOD operator calculates the remainder resulting from an integer division, for example, 10 MOD 3 = 1. The DIV operator calculates integer division, for example 10 DIV 3 = 3. | + | The MOD operator calculates the remainder resulting from an integer division, for example, 10 MOD 3 = 1. The DIV operator calculates integer division, for example 10 DIV 3 = 3. |
=What you need to do= | =What you need to do= | ||
Latest revision as of 13:08, 29 November 2016
AS Question 5
The algorithm, represented using pseudo-code in Figure 4, describes a method to calculate the additive or multiplicative persistence of a two-digit integer. The examples below illustrate how additive and multiplicative persistence are calculated.
Example: calculating the additive persistence of 87
8 + 7 = 15 1 + 5 = 6
After 2 steps the method results in a one digit answer so the additive persistence of 87 is 2.
Example: calculating the multiplicative persistence of 39
3 * 9 = 27 2 * 7 = 14 1 * 4 = 4
After 3 steps the method results in a one digit answer so the multiplicative persistence of 39 is 3.
Figure 4
OUTPUT "Enter integer (0-99): "
INPUT Value
OUTPUT "Calculate additive or multiplicative persistence (a or m)? "
INPUT Operation
Count = 0
WHILE Value > 9
- IF Operation = "a" THEN
- Value = (Value DIV 10) + (Value MOD 10)
- Value = (Value DIV 10) + (Value MOD 10)
- ELSE
- Value = (Value DIV 10) * (Value MOD 10)
- Value = (Value DIV 10) * (Value MOD 10)
- ENDIF
- Count = Count + 1
ENDWHILE
OUTPUT "The persistence is: "
OUTPUT Count
The MOD operator calculates the remainder resulting from an integer division, for example, 10 MOD 3 = 1. The DIV operator calculates integer division, for example 10 DIV 3 = 3.
What you need to do
- Write a program for the algorithm in Figure 4.
- Test the program by showing the result of entering 47, followed by m when prompted by the program.
- Test the program by showing the result of entering 77, followed by a when prompted by the program.
