Course›Chapter 27 bonus studio›Lesson 27.3

Puzzle studio · digit placement · systematic search

Place Digits to Make Arithmetic Statements True

Use every permitted digit exactly once, respect place value, and turn a huge trial-and-error search into a short chain of arithmetic filters. You will solve both original puzzles and see why one printed filling does not necessarily mean one unique solution.

Digit inventoryPlace valueExact divisionWork backwardMultiple solutions

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Study a filling, then build your own

Your goal: fill every digit slot so the equation is true and each allowed digit is used exactly once. Any filling that meets all the rules is accepted.

Learn with the 2005 puzzle: read the slots in Missions 1–2, verify a worked filling in Mission 3, then learn the backwards method in Mission 4.
Make an attempt: use the 2005 construction board in Mission 6. After trying it, return to Mission 5 to compare alternatives. Its complete computer search is optional.
Transfer the method: Mission 7 prepares you for the different 2008 equation; Mission 8 is your own 2008 construction.

Checking an example shows that it works. Building a filling lets you practise choosing the digits yourself.

Puzzle habit

Do not shuffle all digits blindly

A strong solver chooses the positions that control the arithmetic first. The divisor and the final two-digit term determine a required quotient. That quotient then determines almost the whole large numeral.

Core route
Read the slots → protect the digit inventory → work backward from the target → test divisibility → verify every digit and every operation.

Two puzzle structures

(ABCDE-F)÷GH+IJ=2005

(ABCD+E)÷F+GH-I=2008

Letters are placeholders for digits, not multiplication. Adjacent letters form one multi-digit numeral. The 2005 puzzle uses digits 0–9 exactly once; the 2008 puzzle uses digits 1–9 exactly once, with no zero.

Read before placing

Decode the rules of a digit-placement equation

Not complete

①

Use each allowed digit onceA digit cannot be repeated or omitted.

□

Respect place valueABCDE is one five-digit numeral.

≠0

Protect leading positionsA multi-digit numeral cannot begin with zero.

Division must be exactThe other terms are integers, so the quotient must be an integer too.

Operation order: calculate inside the parentheses, divide, and then combine the remaining addition or subtraction.

Each permitted digit is used

May a five-digit numeral begin with 0?

What is evaluated first?

A legal completed division has

Worked example 2 · slot anatomy

Give every circle a job

Not complete

The ten original slots

(ABCDE−F) ÷GH+IJ=2005

Select a letter.

The page will identify its role and place value.

0123456789

The ten circles receive the ten digits 0,1,2,3,4,5,6,7,8,9, one digit in each circle.

Number of digit slots

Digits in the large numeral

Digits in the divisor

Digits in the final addend

Worked example · supplied filling

Verify one worked 2005 construction

Not complete

Worked example, target 2005. This filling is already supplied. Your task is to check its arithmetic and digit inventory, so you know what a valid answer must satisfy.

(25046-8)÷13+79=2005

1Subtract inside the parenthesesHidden

2Check exact divisibilityHidden

3DivideHidden

4Add the final two-digit numberHidden

Digit audit: 25046 · 8 · 13 · 79 together use every digit from 0 through 9 exactly once.

25046-8

25038÷13

Final value after adding 79

Work backward instead of permuting 10 digits

Let the divisor and addend determine the missing numeral

Not complete

How could we find a filling? Work backwards from 2005 instead of guessing every slot. Change the divisor and addend to explore; the checkpoint below names the fixed pair it asks about.

Q=2005-IJ, then ABCDE-F=GH×Q. Once GH and IJ are chosen, test each unused single digit F; then ABCDE is forced to equal GH×Q+F. Reject it if its five digits break any rule.

Two-digit divisor GH

Two-digit addend IJ

required quotient—

required adjusted numerator—

valid subtraction digits—

recovered five-digit numeral—

With 13 and 79, required quotient

Required ABCDE−F for GH=13, IJ=79

F in the worked 25046 filling

Recovered ABCDE in the worked 25046 filling

Added exhaustive audit

Find every filling of the worked example equation

Not complete

Compare after trying. First try your own 2005 filling in Mission 6, then return here. A different valid filling shows that an answer need not match the worked example. You can answer this checkpoint using the example below; running the complete search is optional.

More than one answer can work. One counterexample to uniqueness is (26703−5)÷14+98=2005, since 26698÷14=1907 and 1907+98=2005; it also uses digits 0–9 exactly once.

Optional: explore all valid 2005 fillings after your attempt

Filter by divisor

Filter by subtraction digit

If a valid equation uses every digit once, how many copies of digit 7 does it use?

Is the worked 25046 filling the only valid answer?

Interactive practice problem

Build any valid 2005 equation

Not complete

Your construction, target 2005. Choose the digits yourself, using the backwards method from Mission 4. Aim for a true equation using 0–9 once each. Use a hint if you get stuck, and check both the arithmetic and the inventory before validating.

digit inventoryIncomplete

adjusted numerator—

quotient—

final result—

original Practice 2 · reverse reconstruction

Use the target 2008 to rebuild the numerator

Not complete

A new equation, target 2008. This time use digits 1–9 once each. The signs and slot sizes have changed. Rebuild the backwards steps for this equation before using the construction board in Mission 8.

(ABCD+E)÷F+GH-I=2008

Q=2008-GH+I, so ABCD+E=F×Q. The final product may split into the four-digit block and the one-digit addend in more than one way.

One-digit divisor F

Two-digit addend GH

Final subtraction digit I

required quotient—

required numerator total—

valid splits ABCD+E—

reference split9681 + 4

Reference answer: the book’s answer section gives (9681+4)÷5+73-2=2008. The live audit also finds 9684+1=9685 for the same chosen divisor and final terms.

Required quotient for F=5, GH=73, I=2

Required ABCD+E for that reference trio

Number of valid splits for 5, 73, 2

Interactive Practice 2 and complete audit

Build any valid 2008 equation

Not complete

Your construction, target 2008. Apply the method you just explored. Your answer may differ from the worked filling: it passes when every digit rule and the full equation are satisfied.

digit inventoryIncomplete

numerator total—

quotient—

final result—

Filter solutions by divisor

In the practice puzzle, how many copies of digit 6 may be used?

Is the reference filling the only answer?

Independent practice

Digit-equation workshop

Not complete

Correct all eight answers to complete the workshop. Questions 4 and 7 are short arithmetic exercises, not full use-every-digit puzzles.

1. Target of Worked example 2

2. 25046-8

3. 25038÷13

4. If the target is 100 and the final addend is 23, what quotient is required?

5. For Practice 2 with F=5, GH=73, I=2, required quotient

6. For F=5, GH=73, I=2, required value of ABCD+E

7. A required quotient is 24 and the divisor is 3. What numerator total is required?

8. Best first move when the target and final addend are known

Mastery check

Exit ticket and certificate

Not complete

1. How many times is each permitted digit used?

2. Worked example 2 final result

3. If the target is 90 and the final addend is 14, what quotient is required?

4. Practice reference numerator total 9681+4

5. Working backward is useful because it

Study a filling, then build your own

Do not shuffle all digits blindly

The ten original slots

Digit-Equation Constraint Solver