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Chapter 25 · Defined Operations

Work Backward to Recover Inputs and Hidden Parameters

A custom symbol may hide an unknown input, a fixed parameter, or even two linked rules. Replace the symbol first, solve the ordinary equation, then verify the result in the original definition.

inverse equationshidden constantsdomain checksexact fractions
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Use the backward-solving loop

1 · Pin the rule2 · Substitute known inputs3 · Form an ordinary equation4 · Solve5 · Check the domain6 · Verify in the original rule

Do not “undo the symbol” by guessing. The symbol disappears only after its definition is written. Then ordinary arithmetic and equation reasoning take over.

original focus. This lesson rebuilds Chapter 25 Guided Practice 2, Example 5, Guided Practice 5, and related exercises on hidden inputs, parameters, integer parts, and fractional parts. It also includes carefully labeled transfer problems from Test 25.

Verification is part of the answer

F(candidate) = required output
candidate belongs to the stated domain

A value that solves a transformed equation can still fail because it is outside the allowed number set or makes a denominator zero.

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The backward route

Replace the custom symbol before solving

Not complete

Suppose the current problem defines a◇b=2a÷b+3 for positive integers and asks you to solve 256◇x=19. Which route is mathematically valid?

1a◇b=2a÷b+3
pin the current rule
2512÷x+3=19
substitute
3512÷x=16
solve normally
4256◇x ?= 19
verify and check domain
The transformed equation is a tool. The original definition is the final judge.
Backward-route checkpoint
Need a hint?

First replace the custom symbol with the stated rule; then solve the ordinary equation.

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original Guided Practice 2

Recover an input from GCD and LCM

Not complete

Why does the search cover every answer?

For positive x, lcm(6,x) is at least x and gcd(6,x) is at least 1. If their sum is 33, then x + 1 ≤ 33, so x ≤ 32. Searching through 40 therefore covers every possible worked solution.

A short hand solution

Call the GCD g. It divides 6 and the LCM, so it also divides their sum 33. Thus g is 1 or 3. If g=1, the LCM would be 32, which is not a multiple of 6. So g=3 and the LCM is 30. Using GCD × LCM = 6 × x gives 3 × 30 = 6x, hence x=15. Check: gcd(6,15)=3 and lcm(6,15)=30.

a∧b=[a,b]+(a,b), where [a,b] is the least common multiple and (a,b) is the greatest common divisor.
matches1
first match within the displayed search15
GCD there3
LCM there30
xgcd(a,x)lcm(a,x)a∧xstatus
Why search is legitimate. Here x is a positive integer, and every candidate is tested by the complete definition—not by only the GCD or only the LCM.
Complete the worked example equation 6∧x=33.
Need a hint?

The GCD and LCM must both be checked. The LCM cannot be smaller than x.

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Worked example 5

Recover a fixed hidden parameter, then reuse the rule

Not complete
X@Y= 6XYmX+2Y where m is one fixed positive natural number
Known equation1@2=1

12÷(m+4)=1

→
Use the same m again2@8

96÷(2m+16)

m = 8
1@21
distance from 10
2@83
new denominator32
The letter m is not chosen again for the second calculation. It is a fixed part of the operation’s definition.
Complete both stages of the worked example.
Need a hint?

Use the known equation to recover m once; reuse that same m for the new inputs.

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original Guided Practice 5

Factor the rule before solving a nested equation

Not complete
a*b=ab-a-b+1=(a-1)(b-1)

The factorized form reveals the structure of (2*x)*2=0 more clearly than repeated expansion.

x = 2
inner operation2*2=1

2*x=x−1

→
outside operation1*2=0

(2*x)*2=x−2

Factoring is not a new rule. It is an equivalent form of the same definition, chosen because it makes the unknown easier to see.
Solve (2*x)*2=0.
Need a hint?

Expand (a−1)(b−1) to see why it equals the original rule.

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Exercise 2

Reverse a quotient rule and protect the domain

Not complete
a★b=2a÷b+3   and   256★x=19
x = 32
512÷x16
256★x19
distance from 190
domainvalid
1512÷x+3=19
2512÷x=16
316x=512
4x=32
The worked example requires positive integers. The candidate is valid only if it is positive and the denominator is not zero.
Complete and verify the practice problem.
Need a hint?

After subtracting 3, multiply by the nonzero denominator to solve the equation.

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Exercise 13

Recover numbers from integer and fractional parts

Not complete
a+[b]=15.3,   {a}+b=7.8
Local definitions. For nonnegative numbers, [x] is the integer part and {x} is the fractional part. Thus x=[x]+{x} with 0≤{x}<1.

Try a

[a]8
+
{a}0.3

Try b

[b]7
+
{b}0.5
a+[b]15.3
{a}+b7.8
first error0
second error0
Read the decimal parts first: the first equation fixes the fractional part of a, and the second then fixes b.
Recover both numbers. Fractions or exact decimals are accepted.
Need a hint?

The fractional part of the first equation reveals {a}; then use the second equation.

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General inverse-operation laboratory

Search a finite domain and classify the solution set

Not complete

Two domains: the general search uses positive integers up to the displayed limit. The consecutive-sum original mode also requires b > a. Its table rejects a candidate when that extra condition fails. “Unique” always means unique in the stated search domain; a wider conclusion needs a bound or proof.

solutions1
classificationunique
first solution2
checked100
candidatesubstitutionoutputstatus
A finite search proves completeness only inside the displayed domain. The laboratory states that domain explicitly and tests every positive integer in it. The table shows exact matches and a sample of nearby candidates, not every tested row. Decimal outputs may be rounded to eight places for display; a match requires exact equality.
Test 25 transfer: for natural numbers with b>a, define a△b=a+(a+1)+⋯+(a+b−1). If x△10=65, recover x.
Need a hint?

Decide which input is unknown, and check every candidate against all domain restrictions.

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Test 25 transfer studio

Use several conditions to recover linked parameters and inputs

Not complete

Two useful backward steps

Worked explanation: reverse a triangular total

Let n=ab, meaning the product a × b, not joined digits. Pair the first and last terms in 1+2+…+n: twice the total is n(n+1). So n(n+1)=1190. Since 34 × 35=1190, n=34. The total strictly increases as positive n increases, so no other positive n works. With a > b > 1, the factor pair is 17 and 2.

Worked explanation: reverse a nested consecutive product

The earlier consecutive product is positive, so y must be positive. Since 60 × 61=3660, y=60. The negative algebraic possibility y=−61 is outside the domain. Then 3 × 4 × 5=60 gives x=3; positive consecutive triple-products strictly increase as x increases.

A · One hidden constant and a telescoping result

x*y=1/(xy)+1/[(x+1)(y+A)]

2*1=2/3

Recover A, then find 1998*1999. All inputs are positive and all denominators must be nonzero.

The known equation forces A=1. Then:

1998*1999=1/(1998·1999)+1/(1999·2000)=1/1998000.

B · Three linked positive-integer parameters

x*y=mx+ny,   x△y=kxy

1*2=5,   (2*3)△4=64

Recover the positive integer parameters and find (1△2)*3.

The condition m+2n=5 leaves positive-integer branches (m,n)=(3,1) or (1,2). Since k(2m+3n)=16, the first gives k=16/9 and is rejected. The second gives k=2. Then1△2=4 and4*3=1×4+2×3=10.

C · Output is a triangular number

a*b=1+2+⋯+ab

Here ab means a×b. Find (4*5)−(2*3), then find positive integers a>b>1 with a*b=595.

(4*5)−(2*3)=T₂₀−T₆=210−21=189.

a*b=595⇒ab=34. With a>b>1, (a,b)=(17,2).

D · Output becomes the next input

a*b=a(a+1)⋯(a+b−1)

(x*3)*2=3660

Let y=x*3. Then y(y+1)=3660=60·61, so y=60. Now x(x+1)(x+2)=60, giving x=3.

Transfer note. These are companion-test extensions.
Enter the key recovered values.
Need a hint?

Name an intermediate result when a product becomes the input of another operation.

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Independent practice

Backward-operation workshop

Not complete
Current rule reminders

For the questions below: a∧b=lcm(a,b)+gcd(a,b); X@Y=6XY/(mX+2Y), with fixed positive parameter m and 1@2=1; a★b=2a/b+3; and the consecutive-product rule is a(a+1)…(a+b−1). Here [t] is the integer part and {t}=t−[t] is the fractional part of a nonnegative number. Unknown inputs are positive integers unless the question uses integer and fractional parts.

1. What must happen before ordinary equation solving begins?

2. If 6∧x=33, what is x?

3. If X@Y=6XY/(mX+2Y) and 1@2=1, find the fixed parameter m.

4. Under a*b=ab-a-b+1, solve (2*x)*2=0.

5. Under a★b=2a÷b+3, solve 256★x=19.

6. From a+[b]=15.3 and {a}+b=7.8, what is a?

7. If x△10 adds ten consecutive integers and equals 65, what is x?

8. Let x*y=mx+ny and x△y=kxy, with positive integers m,n,k. If 1*2=5 and (2*3)△4=64, find (1△2)*3.

Need a hint?

Correct all eight answers: replace the symbol;15;8;2;32;8.3;2;10. Substitute the recovered values in each original rule. In the linked-parameter problem, m=1,n=2,k=2, so (1△2)*3=4*3=10.

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Objective assessment

Exit ticket and certificate

Not complete
Current rule reminders

For the questions below: a∧b=lcm(a,b)+gcd(a,b); X@Y=6XY/(mX+2Y), with fixed positive parameter m and 1@2=1; a★b=2a/b+3; and the consecutive-product rule is a(a+1)…(a+b−1). Here [t] is the integer part and {t}=t−[t] is the fractional part of a nonnegative number. Unknown inputs are positive integers unless the question uses integer and fractional parts.

Try a new equation

For this question only, a◇b=2a/b+3, with positive whole-number inputs. Solve 48◇x=11 and verify your answer in the original rule.

1. Solve 6∧x=33.

2. After recovering the worked example parameter, find 2@8.

3. Solve 256★x=19 under a★b=2a÷b+3.

4. If a+[b]=15.3 and {a}+b=7.8 for nonnegative a,b, find b.

5. If (x*3)*2=3660 under the consecutive-product rule, what is x?

Need a hint?

Answers:15;3;32;7.5;3. Check gcd(6,15)+lcm(6,15)=33;2@8=96/32=3;512/32+3=19. For the parts problem,0.3+7.5=7.8. For the nested product,x=3 gives60, then60×61=3660. The separate new equation has x=12.

Optional reflection · not automatically graded

Chapter 25 achievement

Defined-Operation Equation Detective

This certifies that the learner decoded rules, recovered hidden inputs and parameters, checked domains, and verified every result.

Completed