Some operations feed yesterday’s output into today’s step. Others define a new value from an earlier one, or reveal an intended pattern through examples. Build a short state table, name the rule precisely, and then jump ahead without losing the order.
In all three, several small stages reveal the long behavior. The safe habit is to record each state, including the starting state, before trying to jump ahead.
1*9*9*⋯*9
The current output becomes the next left input. Parentheses or a stated left-to-right convention control the route.
a⊙1=a,
a⊙n=2[a⊙(n−1)]+a
A base case starts the process. Every later value refers to an earlier value.
6*2=6+66,
2*3=2+22+222
The examples show the worked example’s intended pattern, but a finite list is not itself a complete mathematical definition.
Look for a base case. A list of examples alone is not a unique definition.
| Step | Previous state | Apply *9 | New state |
|---|
The starting 1 is step zero. Each *9 adds one digit 9.
One step is x next = (b+1)x+b. Add 1 to both sides: x next + 1 = (b+1)(x+1). Therefore every step multiplies the shifted state by b+1. After n steps the shifted state is (starting state + 1) × (b+1)n. Subtract 1 only at the end.
The old output is the complete new left input.
Adding 1 before and after turns the recurrence into repeated multiplication.
| Step | Calculation | State | Shifted state |
|---|
Track x+1; it multiplies by b+1 at every step.
The worked example allows integer a. This slider shows positive examples only. The coefficient calculation works for every integer a, and 1013a=3039 forces the positive answer a=3.
| Stage | New right input | Coefficient recurrence | Current expression | Value for chosen a |
|---|
Track the coefficient c of a. Each next input ka changes c to 2c+k.
The coefficients start 1, 3, 7, 15, … . Add 1 to each and they become 2, 4, 8, 16, … . Doubling the previous coefficient and adding 1 means the shifted coefficient doubles each time. At step n it is 2n, so the original coefficient is 2n−1.
Add 1 to the coefficient sequence 1,3,7,15 to reveal powers of 2.
Two summands: one 6, then two 6s.
Three summands, increasing the run by one digit each time.
Four summands under the same intended pattern.
Write every summand before adding. The final summand has the specified number of repeated digits.
Here a is one digit from 1 through 9, b is a positive integer, and aa means two joined copies of the digit, not a×a. For the worked example equation 1*x=123456789, the definition is explicit. We may generate the values in order and stop when the target appears.
| x | Summands | 1*x | Target status |
|---|
Check the next state after the target: carrying can change the visible digit pattern.
Start at 2 and add 3 consecutive numbers.
Start at 5 and add 4 consecutive numbers.
This is the intended rule used for the questions.
Use exactly b consecutive terms, starting at a and ending at a+b−1.
Correct all eight questions to complete the workshop. Every question repeats its current definition so no symbol meaning carries over accidentally.
Answers:19999;374;26;93;3;492;123456;40. For the coefficient chain use1→4→11→26. For recursion use2⁵−1=31, then multiply by3. Repeated digits give4+44+444=492; the consecutive sum is6+7+8+9+10=40.
Answers:recursive definition;19999999999;3;510;86415. Ten *9 operations append ten9s to the starting1. The coefficient1013 givesa=3039/1013=3. From15m=30 getm=2, then255m=510. Finally7+77+777+7777+77777=86415.
This certifies that the learner can distinguish repeated operations, recursive definitions, and inferred patterns; build state tables; follow left-associated chains; and use repeated-digit and consecutive-sum structures accurately.