Recover Hidden Dividends, Divisors, and Quotients
第16讲 余数问题 · 隐藏的被除数、除数与商
Use one division equation to connect every clue. Bound the unknowns, compare totals, and verify that the remainder is smaller than the divisor.
Learn it on your own
Work through one or two missions at a time. First predict an answer on paper, then use a laboratory to test it. If stuck, open the local hint and then the reasoning. Try the workshop yourself before choosing a worked review after an attempted check. Every explanation you need is on this page; external chat is optional.
Always check: whole-number quantities, positive divisor, and 0≤r<d. A zero quotient is possible when N<d; a zero remainder means exact division. If a question asks for positive numbers, exclude zero.
Put every clue into one division story
A hidden quantity becomes manageable when all four parts of the division are written in the same equation.
Live division-story builder
Translate the rules
Need a hint? Try this first
Use one set of letters throughout.
Read the reasoning, then explain it yourself
N is the dividend, d the divisor, q the quotient and r the remainder. Write N=dq+r, then translate the extra clue separately.
Worked example 2: use the sum of several remainders
The same unknown divisor is used for 70, 110, and 160. Their three remainders total 50. Find the divisor.
1. Build a lower bound
Each remainder is at most d − 1. Therefore:
So the divisor must satisfy:
2. Build an upper bound
If d > 70, then 70 itself would be the first remainder—already larger than the total 50. At d = 70, the remainders are 0, 40, and 20, whose sum is 60, so d = 70 also fails.
3. Make the divisor divide a known total
Removing all three remainders leaves complete multiples of d, so:
Within the range 18 through 69, the factor candidates are:
Test a candidate divisor
Here “70 mod d” means the remainder when 70 is divided by d; the other two labels work the same way.
Complete the worked example reasoning
Need a hint? Try this first
Add the three division equations.
Read the reasoning, then explain it yourself
70+110+160−50=290 is divisible by d. Three remainders each below d total 50, so d≥18. If d≥70, the remainder of 70 is 70 (or 0 at 70, where the other two total 60), so no solution. Between 18 and 69, factors 29 and 58 remain: their remainder totals are 50 and 108. Thus d=29.
Use a remainder-sum candidate laboratory
A factor condition creates a short candidate list; the original remainder total decides which candidates truly work.
Enter several dividends and one target remainder sum
Necessary factor candidates
Exact solutions after checking remainders
Read the worked example preset
Need a hint? Try this first
Test candidates, not just factors.
Read the reasoning, then explain it yourself
For any candidate d, calculate each actual remainder and add them. A divisor of the adjusted total is necessary but not sufficient. When the adjusted total is positive, use its factors. When it is zero, the laboratory lists the allowed zero-quotient family.
The factor-list method uses a positive sum of quotients. The laboratory also handles the zero-quotient case separately. If all quotients are zero, each remainder equals its dividend: the total remainders equal the total dividends, and every divisor above the largest dividend works within the stated range.
Guided Practice 2: recover the dividend from the four-part total
A nonnegative whole-number dividend is divided by a positive whole-number divisor. The quotient is 8, the remainder is 16, and the dividend, divisor, quotient, and remainder total 463.
Write the division equation
The quotient and remainder are already known.
Use the total clue
Substitute the first equation into the second.
Move the divisor and watch the total
Complete the guided practice
Need a hint? Try this first
Substitute N=8d+16 into the total.
Read the reasoning, then explain it yourself
463=(8d+16)+d+8+16=9d+40. Thus d=47 and N=392. Check all four quantities add to 463 and 16<47.
Use a general four-quantity total solver
When q, r, and the total T = N + d + q + r are known, solve for d first and then reconstruct N.
Derive a reusable formula
Four-quantity solver
Use the two Practice presets
Need a hint? Try this first
Collect all copies of d.
Read the reasoning, then explain it yourself
Generally T=(qd+r)+d+q+r=(q+1)d+q+2r. So d=(T−q−2r)/(q+1). With q=4,r=8,T=415, d=79 and N=324; with q=17,r=13,T=2113, d=115 and N=1968. Reject a fractional divisor.
Combine the sum of two numbers with a division clue
original Exercise 4 says A + B = 1,088 and A divided by B gives quotient 11 and remainder 32.
Translate the division
B is the divisor, so the remainder check will be 32 < B.
Use the total
Substitute A:
Two-number total solver
Complete the practice problem
Need a hint? Try this first
Replace A by its division story.
Read the reasoning, then explain it yourself
A=11B+32 and A+B=1088 give 12B+32=1088. B=88 and A=1000; 32<88.
Compare two calculations that keep the same remainder
original Exercise 7: 171 was copied as 117. The quotient became 3 smaller, but the remainder did not change. Find the original remainder.
Same-remainder comparison laboratory
Read the worked example comparison
Need a hint? Try this first
Subtract to cancel the equal remainders.
Read the reasoning, then explain it yourself
171−117=3d=54, so d=18. Now 171=18×9+9 and 117=18×6+9. The original remainder is 9.
Build families when quotient and remainder are linked
The remainder bound makes some quotient–remainder relationships produce only a short finite family.
original Exercise 9: quotient equals remainder when dividing by 7
Let q = r. Then:
Because 1 ≤ r < 7, the positive solutions are:
Quotient–remainder family generator
This generator lists positive dividends only, with positive whole-number k. The rule is q = k × r. Then N = (dk + 1)r for r = 1, 2, …, d−1. If zero were allowed, N=q=r=0 would be another solution.
Practice Question 10: find a positive whole number satisfying both division rules
So N belongs to the family:
So N belongs to the family:
Complete the family reasoning
Need a hint? Try this first
Express everything using the remainder.
Read the reasoning, then explain it yourself
If q=r when dividing by 7, N=8r. For positive N, r=1,…,6, giving 8,16,24,32,40,48. If zero is allowed, N=0 is another answer. For division by 11 with q=r, N=12r; for division by 9 with q=3r, N=28r. Their positive bounded lists intersect at 84.
Hidden-quantity workshop
Correct all eight responses to complete the workshop.
Workshop strategy reminder
Write the division equation or identify the repeated change first. Use the nearby mission hints for the matching method. For “all possible” answers, list the candidates systematically and explain why no other candidate can pass. Check every answer in the original story.
Exit ticket
Try these questions with the examples covered. Complete Missions 1–8, earn 8/8 in the workshop, and earn 5/5 here to unlock your certificate. Reflection is optional and ungraded.
Optional reflection — not automatically graded
Certificate of completion
Hidden Division-Quantity Solver
has completed Lesson 16.3 and can combine N = d×q+r with remainder totals, quantity totals, comparison clues, and quotient–remainder relationships.
Worked explanation for the exit ticket
93=(3d+5)+d+3+5=4d+13, so d=20 and N=65. For division by 5 with q=r, N=6r and r≤4, giving 24. The worked example copying error has divisor 18 and remainder 9. For the new error, 205−145=4d, so d=15; both divisions leave 10.
If you used this explanation, close it and solve the questions again on paper. Explain why each remainder is smaller than its divisor.
Further practice
The common-divisor remainder-sum problem, the quotient-8 remainder-16 guided practice, and the chapter exercises involving a two-number total, a copying error, and quotient equal to remainder follow Chapter 16 on. The additional four-part-total examples and the two simultaneous quotient–remainder conditions come from Test 16 on Practice .