BTT Mathematics / Primary Mathematics Learning Hub / Multiple Constraints
A simultaneous-condition problem asks for a value, arrangement or pair that satisfies every stated condition at the same time. One condition may create many candidates. The next condition removes some. A complete solution keeps intersecting the conditions until the valid set is identified.
Suppose a whole number is greater than twenty, less than fifty, divisible by six and divisible by four. Multiples of both six and four are multiples of twelve. In the required interval, the candidates are twenty-four, thirty-six and forty-eight. All three satisfy every stated condition. The correct conclusion is therefore three possible values, not one guessed value.
This guide develops condition tables, sums and differences, multiples, parity, digit restrictions, ranges, totals and before-and-after requirements. It also asks an important question: do the conditions determine exactly one answer, several answers or none?
The method connects to Systematic Listing and Logical Counting, but the emphasis here is not always on counting every arrangement. It is on treating each sentence as a mathematical constraint and checking that no condition disappears during the solution.
Use the learner’s current school scope and the MOE Primary curriculum and syllabus page as reference points. The examples and 24 questions below are original teaching material.
Turn sentences into conditions · Build candidate sets · Sums and differences · Number constraints · Context problems · Unique, multiple or impossible? · 24 questions · Worked solutions
1. Every condition is a gate the answer must pass
Read a problem once for the situation, then again for constraints. A statement such as “the two numbers total forty” gives one condition. “The larger is eight more than the smaller” gives a second. “Both are whole numbers” gives a third.
A pair that satisfies the total but not the difference is not partly correct. It is not a solution to the full problem.
Worked example A: Sum and difference
Two whole numbers total forty and differ by eight. Remove the difference from the total: 40−8=32. Split the remaining thirty-two equally to get sixteen for the smaller number. The larger is twenty-four.
Check both conditions: 16+24=40 and 24−16=8.
Worked example B: Total and multiplication relationship
Two whole numbers total seventy-two and one is three times the other. Think of four equal units altogether. One unit is 72÷4=18. The numbers are eighteen and fifty-four.
The total condition and “three times” condition are both used. A pair such as twenty and fifty-two has the correct total but fails the multiplication relationship.
Write conditions before operations
For a more complicated problem, list short statements such as:
• total=84
• A=2×B
• A is even
• A and B are positive whole numbers
Some conditions may be automatic once others are satisfied. For example, if A=2×B and B is a whole number, A is automatically even. Recognising redundancy is useful, but do not discard a condition until its role is understood.
2. Build candidates from the strongest easy condition
Suppose a two-digit whole number is between thirty and seventy, is even and has digit sum nine. Start with the range and parity: 30,32,34,…,68. Then test digit sums. The valid numbers are thirty-six and fifty-four.
Starting with all two-digit numbers from ten to ninety-nine would also work, but it creates unnecessary cases. A useful first condition sharply reduces the search without risking omission.
Worked example C: Multiples in a range
Find whole numbers greater than forty and less than one hundred that are divisible by six and eight. Common multiples of six and eight are multiples of twenty-four. In the interval, the candidates are forty-eight, seventy-two and ninety-six.
All three satisfy the conditions. Do not stop at forty-eight merely because it is the first success.
Worked example D: A final digit restriction
Find two-digit multiples of nine ending in five. Two-digit numbers ending in five are 15,25,35,45,55,65,75,85,95. Only forty-five is divisible by nine because 4+5=9.
The two conditions produce a unique answer.
Use a table for candidate testing
Create one row per candidate and one column per condition. Mark pass or fail. This is especially useful when the conditions are of different types, such as range, parity and digit sum.
A table makes it visible if one candidate has not been tested against every condition. It also prevents rejecting a candidate for a condition the problem never stated.
3. Sum, difference and ratio constraints can be solved structurally
Worked example E: Sum and difference
Two quantities total ninety and differ by twenty-six. The smaller is (90−26)÷2=32. The larger is fifty-eight.
This works because removing the extra twenty-six from the larger leaves two equal copies of the smaller.
Worked example F: Sum and ratio
A:B=5:4 and A+B=117. Nine ratio units represent 117, so one unit is thirteen. A=65 and B=52.
Worked example G: Difference and ratio
A:B=7:3 and A−B=36. Four ratio units represent thirty-six, so one unit is nine. A=63 and B=27.
Worked example H: Sum, difference and parity
Two whole numbers total fifty and differ by ten. The structural solution gives twenty and thirty. Both are even, so the additional condition “both even” is satisfied.
If the problem instead required both to be odd, the same first two conditions would force twenty and thirty, which fail the third condition. The correct conclusion would be that no pair satisfies all three conditions.
Worked example I: Three numbers with relationships
A, B and C total eighty-four. B is twice A, and C is three times A. The total contains 1+2+3=6 equal A-units. One unit is fourteen. Therefore A=14, B=28, C=42.
Check all relationships as well as the total. A triple whose total is eighty-four but whose multiples are wrong is not valid.
4. Divisibility, parity, digits and bounds can work together
Worked example J: Divisible by two conditions
A positive whole number below eighty is divisible by both nine and six. Common multiples are multiples of eighteen. The candidates are eighteen, thirty-six, fifty-four and seventy-two.
If another condition says the number is greater than fifty, the candidates reduce to fifty-four and seventy-two. If its digit sum is nine, only fifty-four remains.
Worked example K: Consecutive whole numbers
Two consecutive positive whole numbers total forty-one. Write them as n and n+1. Then 2n+1=41, so 2n=40 and n=20. The numbers are twenty and twenty-one.
A Primary route avoids formal algebra: remove one from the total to make the two numbers equal, giving forty; split equally into twenty and twenty, then restore the one to the larger.
Worked example L: Consecutive even numbers
Two consecutive even numbers differ by two. If their total is seventy-four, subtract two to obtain seventy-two and split equally: thirty-six and thirty-six. Restore the two to the larger: 36 and 38.
Worked example M: Digit sum and reversal
A two-digit number has digits summing to eleven, and the tens digit is three more than the units digit. Let the units digit be four; then tens is seven and the number seventy-four. More systematically, two digit-values total eleven and differ by three, giving four and seven.
The difference condition identifies which digit is larger. Without it, both forty-seven and seventy-four would use the same pair of digits.
Worked example N: Divisibility and digit conditions
A two-digit number is divisible by five, greater than sixty and has digit sum ten. Candidates ending zero or five between sixty and ninety-nine are 65,70,75,80,85,90,95. Digit sums are 11,7,12,8,13,9,14. None equals ten.
The correct conclusion is no solution. Do not change a condition to manufacture a number.
5. Context problems add units and physical constraints
Worked example O: Tickets and total cost
A family buys eight tickets. Adult tickets cost twelve dollars and child tickets cost seven. The total is seventy-six dollars. Let a trial of eight child tickets cost fifty-six. The twenty-dollar extra comes from replacing child tickets with adult tickets, five dollars per replacement.
Four replacements are needed, so there are four adult and four child tickets. Check the ticket count and cost. This is an assumption-method structure inside a multiple-constraint problem.
Worked example P: Rectangles
A rectangle has whole-number side lengths, area forty-eight square units and perimeter twenty-eight units. Factor pairs of forty-eight are 1×48, 2×24, 3×16, 4×12 and 6×8.
Perimeters are 98,52,38,32 and 28 respectively. Only 6×8 satisfies both conditions. The side lengths are six and eight units.
Worked example Q: Equal groups and a remainder
A positive whole number is between fifty and eighty. Dividing it into groups of six leaves remainder two, and dividing it into groups of four leaves remainder two. Subtract two from the number. The result must be divisible by both six and four, hence by twelve.
Multiples of twelve between forty-eight and seventy-eight are sixty and seventy-two. Add two to get 62 and 74. Both are valid; the conditions do not give a unique answer.
Worked example R: Measurement condition
A rectangle has perimeter thirty centimetres and one side is three centimetres longer than the other. The two adjacent sides sum to fifteen. Remove the three-centimetre difference: twelve remains for two equal shorter sides, giving six and nine centimetres.
Check perimeter=2×(6+9)=30 and difference=3.
Keep units compatible
A condition stated in centimetres cannot be combined directly with another stated in metres without conversion. A solution can satisfy the numerical equations and still be physically wrong if the units were mixed.
6. Ask whether the conditions determine one answer, several or none
Finding one candidate proves existence. It does not prove uniqueness. To claim a unique answer, explain why every other candidate fails or why the structure forces only one value.
Unique answer
A two-digit number ends in five and is divisible by nine. The candidates ending in five are finite, and only forty-five passes the divisibility test. The answer is unique.
Several answers
A number between twenty and fifty divisible by four and six may be twenty-four, thirty-six or forty-eight. Every one satisfies the conditions. Unless another condition is stated, all three belong in the answer set.
No answer
Two whole numbers total fifty and differ by ten, while both are required to be odd. The first two conditions force twenty and thirty, so the oddness condition cannot also hold. State that the conditions are inconsistent.
Insufficient information
Two positive numbers total thirty. There are many possible pairs. The information is not contradictory; it is simply insufficient to identify one pair. Distinguish “many solutions” from “no solution.”
Redundant conditions
If a number is stated to be divisible by twelve, it is automatically divisible by three and four. Adding “divisible by three” may not narrow the candidate set. A condition can be true and still add no new information.
Final checking is intersection checking
Return to every sentence in the problem and mark it against the candidate. Do not stop after checking the calculation that produced the candidate. A complete solution belongs to the intersection of all stated conditions.
7. Practice: 24 questions
For each question, list the conditions before solving. State whether the answer is unique, multiple or impossible when relevant.
Questions 1–8: Sums, differences and ratios
1. Two positive whole numbers total 64 and differ by 12. Find them.
2. Two positive whole numbers total 91 and one is six times the other. Find them.
3. A:B=4:7 and A+B=99. Find A and B.
4. P:Q=9:5 and P−Q=28. Find P and Q.
5. A, B and C total 120. B=2A and C=3A. Find all three.
6. Two consecutive positive whole numbers total 67. Find them.
7. Two consecutive even numbers total 106. Find them.
8. Two whole numbers total 48 and differ by 8. Can both be odd? Explain.
Questions 9–16: Number candidates and restrictions
9. List all whole numbers greater than 20 and less than 70 divisible by both 4 and 6.
10. List all whole numbers from 30 to 100 inclusive divisible by both 8 and 12.
11. Find the two-digit number ending in 5 that is divisible by 9.
12. Find all two-digit even numbers with digit sum 7 and tens digit greater than units digit.
13. A two-digit number has digit sum 13 and tens digit five more than units digit. Find the number.
14. Find all whole numbers greater than 50 and less than 100 that are divisible by 9 and have digit sum 9.
15. Find all whole numbers between 40 and 80 that leave remainder 2 when divided by both 4 and 6.
16. A number is divisible by 12 and less than 50. It is also divisible by 3. Does the second condition narrow the candidates? List the positive candidates.
Questions 17–24: Context and completeness
17. A rectangle has whole-number side lengths, area 48 square units and perimeter 28 units. Find the side lengths.
18. A rectangle has perimeter 34 cm and one side is 5 cm longer than the other. Find its sides.
19. Ten tickets cost either $4 or $7. Total cost is $55. Find each ticket count.
20. Twelve full packets hold either 3 or 6 cards each. There are 54 cards altogether. Find each packet count.
21. Two positive whole numbers total 30. Is the pair uniquely determined? Give three valid pairs.
22. Two whole numbers total 40, differ by 6 and are both even. Can all conditions hold?
23. A positive whole number is below 60, divisible by 8 and divisible by 6. List all possible values.
24. A learner finds one number satisfying “between 20 and 60, even, digit sum 8” and stops. Explain why one success does not prove uniqueness, then list all valid numbers.
8. Worked solutions
Solutions 1–8
1. 26 and 38. Remove the twelve difference from sixty-four to get fifty-two, split equally to get twenty-six, then add twelve.
2. 13 and 78. Seven equal units total ninety-one. One unit is thirteen.
3. A=36, B=63. Eleven units represent ninety-nine, so one unit is nine.
4. P=63, Q=35. Four ratio units equal twenty-eight, so one unit is seven.
5. A=20, B=40, C=60. Six A-units total 120.
6. 33 and 34. Remove one from sixty-seven, split sixty-six equally, then restore one to the larger.
7. 52 and 54. Consecutive even numbers differ by two. Remove two from 106 to get 104, split equally to fifty-two, then add two.
8. No. The sum-and-difference conditions force the pair 20 and 28: (48−8)÷2=20 and 20+8=28. Both are even, so the additional condition that both are odd cannot hold.
Solutions 9–16
9. 24, 36, 48, 60. Common multiples of four and six are multiples of twelve. Keep those strictly inside the interval.
10. 48, 72, 96. Common multiples of eight and twelve are multiples of twenty-four. These lie from thirty to one hundred inclusive.
11. 45. It ends in five and 4+5=9, so it is divisible by nine. No other two-digit number ending in five has digit sum a multiple of nine.
12. 52 and 70. Even units digits and digit sum seven give 16, 34, 52 and 70 among the two-digit possibilities; only fifty-two and seventy have tens digit greater than units digit.
13. 94. The digits total thirteen and differ by five, with tens larger. Smaller digit=(13−5)÷2=4; larger digit=9.
14. 54, 63, 72, 81, 90. These multiples of nine lie in the interval and each has digit sum nine.
15. 50, 62, 74. Subtract two; the result must be divisible by both four and six, hence by twelve. The relevant multiples are forty-eight, sixty and seventy-two. Add two.
16. No; candidates 12, 24, 36, 48. Divisibility by twelve already guarantees divisibility by three. The second condition is redundant.
Solutions 17–24
17. 6 by 8. Factor pairs of forty-eight are checked against perimeter. Only six and eight give 2×(6+8)=28.
18. 6 cm and 11 cm. Adjacent sides sum to seventeen because half the perimeter is seventeen. Their difference is five. Smaller=(17−5)÷2=6.
19. Five of each. Ten four-dollar tickets cost forty. Fifteen extra dollars are needed; each upgrade to seven dollars adds three, so five upgrades.
20. Six packets of each size. Twelve three-card packets would contain thirty-six. Eighteen additional cards require six upgrades of three cards each.
21. No. Examples include 1 and 29, 10 and 20, and 14 and 16. The total alone allows many positive pairs.
22. No. The sum-and-difference conditions force seventeen and twenty-three: (40−6)÷2=17 and 23. They are both odd, so the evenness condition fails.
23. 24 and 48. Common multiples of eight and six are multiples of twenty-four below sixty.
24. 26 and 44. One successful candidate proves only that the valid set is non-empty. Check all even two-digit numbers in the interval whose digits sum to eight. Twenty-six and forty-four work; sixty-two would also have digit sum eight but lies outside the stated interval.
9. Teaching and diagnosis
If a learner uses one condition and forgets another, ask them to write a short condition list before calculating. After finding a candidate, tick every condition only when it has been checked explicitly.
When the first valid candidate is treated as the only answer
Ask where another answer would have to appear. For a range-and-divisibility problem, continue through the remaining multiples until the upper bound is reached. For digit problems, organise by tens digit or units digit.
When no-solution and insufficient-information cases are confused
Show two examples: “two numbers total thirty” has many solutions; “two numbers total thirty, differ by ten and are both even” may be checked for consistency. One lacks enough information, while the other may contain incompatible information.
When a redundant condition distracts
Ask whether satisfying one condition automatically satisfies another. Divisibility by twelve implies divisibility by three. Recognising this can simplify working without ignoring any requirement.
Use independent representations
A ratio solution can be checked with actual totals. A factor-pair solution can be checked with both area and perimeter. A digit candidate can be checked numerically against divisibility and digit conditions.
Continue through the structural-reasoning collection
For finding one equal unit, use Unitary Method and Equal-Unit Reasoning. For comparing two complete sharing plans, use Excess and Shortage Problems. For minimum, maximum and guaranteed conclusions, continue to Bounds, Extremes and Guaranteed Conclusions.
Return to the BTT Primary Mathematics Learning Hub.
Original explanations and 24 original practice questions with separate worked solutions. Contexts are hypothetical mathematical models. No examination prediction or performance guarantee is implied.