16.2Grade 5 Math Studio
Lesson progress0 / 10 missions
Chapter 16 · Remainder Problems · Lesson 16.2

Find a Hidden Divisor from the Remainder

由余数反求隐藏的除数

Subtract the remainder, factor what remains, and filter the factors by every condition in the story. Then verify each surviving divisor in the original division.

Subtract → Factor → Filter → Verify.
10 interactive missions50–65 minutesComplete original examples includedAutosaves in this browser
Grade 5

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.

Choose a method: missing divisor → subtract the remainder and filter factors; missing quantities → write N=dq+r and the extra clue; several remainder conditions → look for a common gap; repeated changes → track the net change; long sequences → prove a repeating state.
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.
Mission 1

Peel off the remainder

A remainder statement hides a complete multiple of the divisor. Remove the leftover first, and the divisor must divide everything that remains.

Not complete

Start from the division story

A divisor is the number we divide by; a remainder may be left. Saying that a number divides another number means it is a factor, with no remainder. Here d need not be a factor of N, but it is a factor of N − r.

NDividend
=
d × qComplete groups
+
rRemainder
N = d × q + r

Subtract the remainder from both sides:

N − r = d × q
Two facts appear immediately: the divisor d divides N − r, and the remainder rule requires d > r.

Remainder peeler

273 = d × q
37
N − r273
Divisor conditiond ∣ 273
Remainder boundd > 37
Equation310 = d×q + 37

Read the worked example doorway

Need a hint? Try this first

Remove the leftovers.

Read the reasoning, then explain it yourself

310−37=273 must consist of complete groups: d×q. Also d>37.

Mission 2

Build a factor-and-filter sieve

Factors create the candidate pool. The remainder bound, digit length, and any extra story condition remove candidates that cannot be divisors.

Not complete
1

Subtract

Calculate N − r.

2

Factor

List every positive factor of N − r.

3

Filter

Keep factors with d > r and the required size.

4

Verify

Check N = d × q + r in the original problem.

Factor sieve for the worked example

All factors:Every possible divisor must appear here.
Two-digit filter:Keep only factors from 10 through 99.
Remainder filter:Keep only factors greater than 37.

Choose the rules that belong in the sieve

Need a hint? Try this first

Apply every restriction.

Read the reasoning, then explain it yourself

A two-digit divisor is 10 through 99, divides N−r, and is greater than r. A factor alone is not enough.

Mission 3

Solve the worked example completely

310 is divided by an unknown two-digit positive whole number, leaving remainder 37. The worked example asks for every possible two-digit divisor.

Not complete

Step A · Make a multiple

310 − 37 = 273

So the divisor must be a factor of 273.

273 = 3 × 7 × 13

Step B · List the factors

21 is a factor

21 ≤ 37

It fails the remainder rule, so it cannot be the divisor in this division.

39 survives

310 = 39 × 7 + 37

The remainder 37 is smaller than 39.

91 survives

310 = 91 × 3 + 37

The remainder 37 is smaller than 91.

original result: the possible two-digit divisors are 39 and 91.

Rebuild the result yourself

Select every two-digit candidate that gives remainder 37 when dividing 310. Select no other candidate.

Need a hint? Try this first

List factor pairs of 273.

Read the reasoning, then explain it yourself

1×273, 3×91, 7×39, 13×21 exhaust the pairs. Two-digit factors above 37 are 39 and 91; their quotients are 7 and 3.

Mission 4

Verify survivors and reject almost-candidates

A factor can still fail. Always compare the claimed remainder with the divisor and then check the actual division.

Not complete

Candidate verifier: check the remainder

This tool checks the remainder condition. Check size restrictions separately: 273 gives the requested remainder but is not a two-digit answer to the practice problem.

N − r divisible by d?Yes
r < d?No
Actual quotient14
Actual remainder16
21 is not valid.The claimed 37 leftovers would form another group of 21.

Why the tempting equation fails

310 = 21 × 13 + 37

This equality is true, but it is not a finished division statement because:

37 ≥ 21

The actual division

310 = 21 × 14 + 16

The valid remainder is 16 because:

0 ≤ 16 < 21

Audit the candidates

Need a hint? Try this first

Test an actual division.

Read the reasoning, then explain it yourself

310=21×14+16, so 21 does not give remainder 37. But 310=39×7+37=91×3+37.

Mission 5

Guided Practice 1: find every divisor

A two-digit divisor is used to divide 1,477, leaving remainder 49. The worked example asks for every possible two-digit divisor.

Not complete

Subtract the remainder

1477 − 49 = 1428
1428 = 2² × 3 × 7 × 17

The divisor must be a two-digit factor of 1428 that is greater than 49.

Surviving factors

Divisor dQuotient qVerification
51281477 = 51 × 28 + 49
68211477 = 68 × 21 + 49
84171477 = 84 × 17 + 49

Record the complete candidate list

Need a hint? Try this first

First calculate 1477−49.

Read the reasoning, then explain it yourself

1428=2×2×3×7×17. The two-digit factors above 49 are 51, 68 and 84. Rebuild 1477 with each candidate.

Mission 6

Use the hidden-divisor laboratory

Change the dividend, remainder, and allowed divisor range. The laboratory lists every factor, applies every filter, and classifies the result.

Not complete

This search uses a positive quotient, so enter N>r. If q=0, then N=r, and any divisor greater than r works; that is a separate family rather than a factor list of a positive number.

General factor sieve

N − r468
Positive factors18
Valid candidates8
Result typeMultiple
8 candidates12, 13, 18, 26, 36, 39, 52, 78

Read the 474-remainder-6 result with divisors from 10 through 99

Need a hint? Try this first

Keep all two-digit factors greater than 6.

Read the reasoning, then explain it yourself

474−6=468. The complete list is 12,13,18,26,36,39,52,78. There are eight; the greatest is 78.

Mission 7

Let the story add a final filter

Arithmetic may leave several divisors. A story condition—such as “fewer than 60 students”—can select the one that makes sense.

Not complete

16 · Question 1

A class of fewer than 60 students buys 310 workbooks. Each student receives as many whole workbooks as possible, with equal numbers for everyone and 37 workbooks left over. How many students are in the class?

Candidates from the worked example: 39, 91
Class size < 60
Only 39 survives the story filter.

Exercise 1

Find every two-digit divisor that gives remainder 4 when dividing 109.

109 − 4 = 105
15q = 721q = 535q = 3

Apply both story filters

Select every two-digit number that gives remainder 4 when dividing 109.

Need a hint? Try this first

Use the extra story limit last.

Read the reasoning, then explain it yourself

For 310 with remainder 37, the candidates are 39 and 91; fewer than 60 selects 39. For 109 with remainder 4, factor 105: the two-digit candidates are 15,21,35.

Mission 8

Classify unique, multiple, and impossible cases

In each case below, the unknown divisor must be a two-digit positive whole number. The same method may leave one candidate, several candidates, or none. Name exactly what the evidence proves.

Not complete
Unique

5122, remainder 66

5122 − 66 = 5056 = 2⁶ × 79

The two-digit factors are 16, 32, 64, and 79. Only 79 is greater than 66.

5122 = 79 × 64 + 66
Multiple

474, remainder 6

474 − 6 = 468

Eight two-digit factors survive:

1213182636395278
Impossible

100, remainder 60

100 − 60 = 40

Every two-digit factor of 40 is at most 40, but a valid divisor would have to be greater than 60.

Classify the three cases

Need a hint? Try this first

Count the surviving candidates.

Read the reasoning, then explain it yourself

5122−66=5056=64×79; only 79 is two-digit and greater than 66. The 474 problem has eight answers. For 100 with remainder 60, a divisor must divide 40 yet exceed 60: impossible.

Mission 9

Hidden-divisor workshop

Use Subtract → Factor → Filter → Verify. Correct all eight responses to complete the workshop.

Not complete
Workshop score0 / 8
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.

Mission 10

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.

Not complete
Exit score0 / 5

Optional reflection — not automatically graded

Certificate of completion

Hidden-Divisor Factor Detective

Grade 5 Mathematician

has completed Lesson 16.2 and can subtract a remainder, factor the adjusted dividend, apply all divisor restrictions, and verify every survivor.

Worked explanation for the exit ticket

216=2³×3³. Its two-digit factors are 12,18,24,27,36,54,72: seven in all. Each is greater than 5. Only 12 is below 15. For example 221=12×18+5.

If you used this explanation, close it and solve the questions again on paper. Explain why each remainder is smaller than its divisor.

Learning notes

Further practice

The central method and the problems involving 310 with remainder 37 and 1,477 with remainder 49 follow Chapter 16, Example 1 and Guided Practice 1 on this lesson page 97. The exercises involving 109 with remainder 4 and 5,122 with remainder 66 come from the chapter practice on this lesson page 100. The class-size problem and the complete two-digit-divisor problem for 474 with remainder 6 come from Test 16 in this exercise.

The adjustable factor sieve, candidate verifier, unique–multiple–impossible comparison, extra impossible case, workshop, and exit ticket are added teaching scaffolds.