Number Play · Lesson 4 of 8
Mental Arithmetic and Possible Results
“Build useful calculations and explain which sums and differences can exist.”
• Compose target amounts by reusing available numbers and grouping repeated additions. • Use addition and subtraction to adjust an amount efficiently. • Find smallest and largest possible results from specified digit lengths. • Construct examples for possible results and justify impossible requests. • Classify numerical claims as always, sometimes, or never true using sound reasoning.
Choose a useful route to the target
Mental arithmetic becomes easier when you notice structure in the numbers. To make 38,800, you could count hundreds for a long time, but starting with 25,000 and 13,000 gets you to 38,000 immediately. You then need only 800 more. Breaking a target into convenient parts gives a calculation you can remember and explain.
Suppose the available amounts are 25,000, 400, 13,000, 1500, and 60,000. In the first puzzle, you may add any of these amounts as many times as you like. You may not subtract, split an amount into smaller pieces, or invent an extra amount. These conditions define what a valid solution is. The question is whether a target can be written as a sum of allowed amounts.
Problem
Use only the available amounts to make 38,800 and 3400.
- 1.25,000 + 13,000 = 38,000. The remaining amount is 38,800 − 38,000 = 800.
- 2.800 is two copies of 400, so 38,800 = 25,000 + 13,000 + 400 × 2.
- 3.For 3400, use two copies of 1500 to make 3000, then one copy of 400: 3400 = 1500 + 1500 + 400.
- 4.Check each expression against the target and confirm that every addend is allowed.
| Target | One valid construction | Mental route |
|---|---|---|
| 28,000 | 25,000 + 1500 × 2 | Start at 25,000; add 3000. |
| 63,000 | 60,000 + 1500 × 2 | Start at 60,000; add 3000. |
| 61,600 | 60,000 + 400 × 4 | Four 400s make 1600. |
| 19,500 | 13,000 + 1500 × 3 + 400 × 5 | Add 4500 and 2000 to 13,000. |
| 31,000 | 25,000 + 1500 × 4 | Four 1500s make 6000. |
| 20,900 | 13,000 + 1500 × 5 + 400 | 13,000 + 7500 + 400. |
A multiplication such as 400 × 4 is a compact way to record four equal addends. It does not change the rules. You can check a result by grouping those addends differently: two 400s make 800, so four make two 800s, or 1600. Comparing routes is useful because one student may find a convenient grouping that another has missed.
Problem
Can these amounts make 1000? Which positive whole thousands can be made?
- 1.Every available amount except 400 already exceeds 1000. Since subtraction is forbidden, using any of them would overshoot.
- 2.Only repeated 400s remain. Two make 800 and three make 1200, so none makes 1000.
- 3.2000 = 400 × 5 and 3000 = 1500 × 2. Every even number of thousands is a sum of copies of 2000.
- 4.Every odd number of thousands from 3000 upward is 3000 plus copies of 2000. Therefore all positive whole thousands except 1000 can be made. For example, 14,000 = 7 × 2000, 15,000 = 3000 + 6 × 2000, and 16,000 = 8 × 2000.
Failing to find a construction does not show that none exists. For 1000, the argument rules out every large addend and then every possible number of 400s. Explaining that limit is stronger than saying that you tried several sums.
Addition and subtraction give different building options
Now use the amounts 40,000, 7000, 300, 1500, 12,000, and 800. This time both addition and subtraction are allowed, and amounts may be reused. A convenient route can start above the target and remove the excess, or start below it and add the missing amount. State what you are adjusting and why the adjustment has the right size.
Problem
Make 39,800 from the new set of amounts.
- 1.40,000 is near the target, but it is 200 too high.
- 2.The available 800 is too large to subtract by itself. Adding two copies of 300 restores 600 of that subtraction, giving a net decrease of 200.
- 3.Thus 40,000 − 800 + 300 + 300 = 39,800. Evaluate in order: 39,200 + 600 = 39,800.
| Target | One expression | Check |
|---|---|---|
| 45,000 | 40,000 + 7000 − 800 − 300 − 300 − 300 − 300 | 47,000 − 800 − 1200 = 45,000. |
| 5900 | 7000 − 800 − 300 | 7000 − 1100 = 5900. |
| 17,500 | 12,000 + 7000 − 1500 | 19,000 − 1500 = 17,500. |
| 21,400 | 12,000 + 7000 + 1500 + 300 + 300 + 300 | 20,500 + 900 = 21,400. |
Evaluate each complete route and compare with its target. For 45,000, the total subtracted from 47,000 is 800 + 1200 = 2000. A calculation can use allowed amounts and still miss the target, so checking the final amount is part of solving the puzzle.
Digit lengths put limits on results
A five-digit positive number can be as small as 10,000 or as large as 99,999. Knowing these bounds helps before you try any examples. The smallest sum of two five-digit numbers is 20,000, and the largest is 199,998. Some sums have five digits and others six, but none can have fewer than five or more than six.
A limit that a value cannot go below or above under stated conditions.
Problem
Can a five-digit number plus a three-digit number give a six-digit sum?
- 1.To reach six digits, the result must be at least 100,000. Most five-digit numbers are too far below that boundary for a three-digit addend to cross it.
- 2.Choose a five-digit number close to the boundary: 99,999. Then add the three-digit number 999.
- 3.99,999 + 999 = 100,998, which has six digits. One valid example proves that the result is possible; it does not mean every such sum has six digits.
Problem
Can two four-digit numbers have a six-digit sum? Can two five-digit numbers sum to 18,500?
- 1.The greatest four-digit addends are 9999 and 9999. Even their sum is only 19,998, below 100,000. Every other pair gives a sum no larger, so six digits are impossible.
- 2.The smallest five-digit addends are 10,000 and 10,000. Their sum is 20,000, already above 18,500. Every other pair gives a sum at least as large, so 18,500 is impossible.
- 3.The first argument uses an upper bound; the second uses a lower bound. Choose the bound that addresses the requested target.
Differences need the same care. When discussing their digit lengths here, subtract the smaller positive number from the larger and allow a result of zero when the numbers are equal. Two five-digit numbers can be almost equal, giving a short difference, or very far apart, giving a five-digit difference. The maximum possible nonnegative difference is 99,999 − 10,000 = 89,999.
Problem
Can the difference of two five-digit numbers be 91,500?
- 1.Make the difference as large as possible by using the largest first number and smallest second number.
- 2.99,999 − 10,000 = 89,999. This is the greatest possible difference for two five-digit positive numbers.
- 3.91,500 exceeds 89,999, so no qualifying pair exists. Trying additional pairs cannot change this bound.
| Requested result | Possible? | Example or reason |
|---|---|---|
| Two five-digit numbers: five-digit sum above 90,250 | Yes | 45,000 + 45,400 = 90,400. |
| Two five-digit numbers: six-digit sum | Yes | 60,000 + 40,000 = 100,000. |
| Two five-digit numbers: difference below 56,503 | Yes | 80,000 − 50,000 = 30,000. |
| Five-digit minus three-digit: four-digit difference | Yes | 10,000 − 999 = 9001. |
| Five-digit minus four-digit: four-digit difference | Yes | 12,000 − 2500 = 9500. |
| Five-digit minus five-digit: three-digit difference | Yes | 50,999 − 50,000 = 999. |
Always, sometimes, or never?
A claim is always true only if it covers every allowed case. It is sometimes true if a working example and a failing example both exist. It is never true when no allowed case can satisfy it. To disprove always, one counterexample is enough. To establish never, use a reason that covers all choices, such as a bound.
| Claim | Classification | Reason |
|---|---|---|
| Five-digit + five-digit gives five digits | Sometimes | 10,000 + 10,000 = 20,000; 60,000 + 40,000 = 100,000. |
| Four-digit + two-digit gives four digits | Sometimes | 1000 + 10 = 1010; 9999 + 99 = 10,098. |
| Four-digit + two-digit gives six digits | Never | The maximum sum is 10,098. |
| Five-digit − five-digit gives five digits | Sometimes | 30,000 − 10,000 = 20,000; 10,001 − 10,000 = 1. |
| Five-digit − two-digit gives three digits | Never | The minimum is 10,000 − 99 = 9901, already four digits. |
Quiz
Using addition only with 400, 1500, 13,000, 25,000, and 60,000, which target is impossible?
Which expression equals 5900?
What is the smallest sum of two positive five-digit numbers?
Why can two four-digit numbers not have a six-digit sum?
Which classification fits “five-digit plus five-digit gives a five-digit sum”?
What is the greatest nonnegative difference of two five-digit positive numbers?
Practice Problems
- Use the first set of available amounts to make 19,500 in a different valid way.
- Make 14,000, 15,000, and 16,000 using only copies of 400 and 1500.
- Explain why 1000 is impossible when addition alone is allowed.
- Use the second set of amounts to make 45,000 and check each step.
- Construct a five-digit plus three-digit sum with six digits and another with only five digits.
- Construct a five-digit minus five-digit difference with three digits and another with five digits.
- Classify “four-digit plus two-digit gives five digits” as always, sometimes, or never, with evidence.
- Create one impossible digit-length request and prove its impossibility using a bound.
19,500 = 1500 × 13. For 14,000 use 1500 × 8 + 400 × 5; for 15,000 use 1500 × 10; for 16,000 use 400 × 40. Every amount used comes from the allowed set.
Key Takeaways
• Break a target into convenient allowed amounts and check the completed expression. • Repeated addition can be recorded by multiplication. • Whether subtraction is allowed can change which targets are possible. • Digit lengths give lower and upper bounds that can prove impossibility. • Sometimes needs contrasting examples; always and never need reasoning that covers all allowed cases.