15.2Grade 5 Math Lab
0 of 10 missions complete0%
Chapter 15 · Divisibility

Split Large Numbers into Blocks

Lesson 15.2 · 大数分组判断整除

Replace a huge number with a much smaller block calculation. Learn when three-digit blocks alternate, when two-digit blocks add, and how Hua Luogeng turned a seven-digit car-plate number into 119.

Group carefully. Use the right sign. Keep the same divisibility conclusion.
10 missionsoriginal examples includedHua Luogeng enrichment storyEverything needed is on this page.
Autosaves in this browser
Mission 1

Why three-digit blocks can replace a large number

The shortcut begins with one unusually useful number: 1001.

Not complete

The shared doorway

7 × 11 × 13=1001

Because 1001 is divisible by 7, 11, and 13, adding or subtracting any whole-number multiple of 1001 cannot change whether a number is divisible by any of those three divisors.

1000 = 1001 − 1

Split after the last three digits

Write a number as:

N = 1000q + r

Now replace 1000 by 1001 − 1:

N = 1001q + (r − q)

The first part is divisible by 7, 11, and 13. So only r − q matters.

Build the decomposition yourself

Original N96,928
r − q832
Same divisibility result?Yes

Check the idea

Mission 2

Correct the arithmetic, then test the worked examples

A correct shortcut still needs accurate arithmetic. Check the subtraction in these examples before using the reduced value to test divisibility.

Not complete
original correction: the book prints 928 − 96 = 823. The correct subtraction is 928 − 96 = 832. Since 832 = 13 × 64, the worked example’s divisibility conclusion remains correct.

Example A · 96,928

96|928
|928 − 96| = 832 = 13 × 64

Therefore 96,928 is divisible by 13.

Example B · 690,578

690|578
|578 − 690| = 112 = 7 × 16

Therefore 690,578 is divisible by 7. A negative difference would be equally valid; divisibility is unchanged by changing the sign.

Audit both examples

Mission 3

Use alternating three-digit blocks for 7, 11, and 13

Group digits from the right in threes. The leftmost group may have one or two digits. When testing divisibility by 7, 11, or 13, successive powers of 1000 can be replaced by alternating −1 and +1 because the difference is divisible by the chosen divisor.

Not complete
Starting at the rightmost block: +, −, +, −, … (the right-positive convention)For 3 | 103 | 219, calculate 3 − 103 + 219.

Three-digit block laboratory

Presets
3 − 103 + 219 = 119
Reduced value119
Original remainder0
VerdictDivisible

Why the entire block state matters

Do not simply subtract every block from the first one. The signs must alternate according to place value. For three blocks:

a | b | c → a − b + c

For four blocks:

a | b | c | d → −a + b − c + d

Use the plate-number blocks

Mission 4

For 99, add two-digit blocks

Group digits from the right in twos, retaining a one-digit leading group if needed, then add. The symbol ≡ below means that the two sides differ by a multiple of the stated divisor; it does not mean they are equal.

Not complete

For 7, 11, or 13

1000 ≡ −1 (for the chosen divisor 7, 11, or 13)

Use three-digit blocks and alternate signs.

For 99

100 ≡ 1 (for divisor 99)

Use two-digit blocks and add all of them.

Two-digit block laboratory

Presets
52 + 00 + 47 = 99
Block sum99
Remainder mod 990
VerdictDivisible

Choose the correct rule

Mission 5

Worked example: fill □2004□ to make a multiple of 99

The two boxes may contain different digits. The notation a2004b is a six-digit number, not multiplication. Let the first missing digit be a and the last missing digit be b. The first digit cannot be zero.

Not complete

Complete problem

Fill the two boxes in the six-digit number a2004b so the completed number is divisible by 99.

a2+00+4b
(10a+2)+0+(40+b)=99
10a+b=57

Search all digit pairs

The search tests all 90 possibilities with a from 1–9 and b from 0–9. It accepts a pair only when the completed number is divisible by 99.

Run the search and inspect its candidate ledger before checking this mission.

Record the unique solution

Mission 6

Guided practice: finish 2003□□

The final two boxes form one two-digit block from 00 through 99. Here xy means the two-digit number 10x+y, not x multiplied by y.

Not complete

Build the equation

20+03+xy
20+3+xy must be a multiple of 99

The possible sum ranges from 23 through 122. The only multiple of 99 in that range is 99.

Explore the final block

00
Block sum23
Completed number200300
Multiple of 99?No

Complete the guided practice

Mission 7 · Enrichment

Hua Luogeng’s “large-scale disarmament”

The worked example closes the lesson with a story showing that divisibility can be discovered in ordinary surroundings.

Not complete
original story · May 9, 1978

A car plate becomes a divisibility puzzle

After dinner, mathematician Hua Luogeng was walking with graduate students and several middle-school teachers in the courtyard of a guesthouse near Xiangshan. He pointed to his car plate:

3103219

Ignoring the dash gives the number 3,103,219. Hua asked why it is divisible by 7, but not by 11 or 13. Later that evening, he revealed the “small doorway”:

1001 = 7 × 11 × 13

He described his reduction as a kind of “large-scale disarmament”: remove huge multiples of 1001 until only a tiny number remains.

1

Write the plate as a number.
31-03219 → 3,103,219

2

Remove 3,000 copies of 1001.
3,103,219 − 3,003,000 = 100,219

3

Remove another 100 copies of 1001.
100,219 − 100,100 = 119

4

Inspect the small result.
119 = 7 × 17, while 119 is not divisible by 11 or 13.

The worked example uses the story to encourage teachers and students to notice interesting mathematics nearby, think about it, and use a “mathematical eye” to understand everyday things.

The block shortcut says the same thing even faster

3 | 103 | 219→3 − 103 + 219→119

Reveal all four steps, then explain the mathematical core

Mission 8

Choose the block size and operation before calculating

The most common mistake is using a correct block idea with the wrong divisor.

Not complete
Target divisorWhyBlock sizeOperation
7, 11, or 131000 ≡ −1 (for the chosen divisor 7, 11, or 13)3 digitsAlternate signs; for just two blocks, subtract
99100 ≡ 1 (for divisor 99)2 digitsAdd all blocks

Error detective

Claim: “Because 690−578 is positive, only that order works.”

Repair: either order is acceptable. The two differences are opposites, and a number is divisible by a divisor exactly when its opposite is.

Smallest block warning

Keep the leading block even if it has fewer digits. In:

3 | 103 | 219

the first block is 3. Writing it as 003 would still mean 3; neither version may be discarded.

Match each situation to its method

Mission 9

Block-divisibility workshop

Solve all eight. Correct all eight responses to complete the workshop.

Not complete
Worked explanation for question 1

35 | 035 gives |35 − 35| = 0, which passes all three divisors.

Worked explanation for question 2

2 − 15 + 27 = 14.

Worked explanation for question 3

12 + 34 + 56 = 102; it does not pass 99.

Worked explanation for question 4

12 + 34 + tail must be 99, so tail = 53.

Worked explanation for question 5

45 + 54 = 99.

Worked explanation for question 6

|84 − 42| = 42 = 7 × 6.

Worked explanation for question 7

100 differs from 99 by 1, so use blocks of two digits.

Worked explanation for question 8

8 − 2 + 15 = 21.

Workshop score0 / 8
Mission 10

Objective exit ticket

Complete all ten missions, including a perfect exit score, to earn the certificate. Every item has one objective response.

Not complete
Worked explanation for question 1

|114 − 57| = 57.

Worked explanation for question 2

67 + 32 = 99, so 6732 passes 99.

Worked explanation for question 3

56 + 00 + 43 = 99.

Worked explanation for question 4

4 − 9 + 26 = 21.

Worked explanation for question 5

26 = 13 × 2; it is not a multiple of 7 or 11.

Exit score0 / 5

Certificate of mathematical thinking

Large-Number Block Navigator

Grade 5 Mathematician

has completed Lesson 15.2 by choosing the correct block size, operation, and divisibility conclusion.

Optional reflection

What stayed unchanged?

Explain in your own words why replacing 3,103,219 by 119 does not change whether the number is divisible by 7, 11, or 13.

Learning notes

About this lesson

The student page explicitly corrects the worked example’s printed subtraction 928−96=823 to 928−96=832. The worked example’s intended divisibility conclusion is unchanged because 832 is divisible by 13.

The interactive laboratories, complete candidate search, error detective, workshop, and exit ticket are added instructional scaffolds.