21.5Math Education • Chapter 21
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Lesson 21.5 • Digit Sums

Track Digit Sums through Carries and Products

数字和:进位与乘积

Learn exactly why a carry changes a digit sum by 9, use column evidence to recover hidden digit totals, and distinguish a true multiplication pattern from a mere modulo-9 clue.

Carry units subtract 9. Products preserve a remainder—not necessarily the exact digit sum.

originalCompanion transfer: Test 21, page 5010 missionsCarry and product laboratories
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Your toolkit: s(n) means the sum of the decimal digits of n. For example, s(203)=2+0+3=5. Try each checkpoint yourself, then open its on-page hint if you get stuck. No lesson or external chat is needed.
Mission 1

See exactly what one carry does

Ten ones are renamed as one ten. The number stays the same, but the visible digit sum drops by 9.

Not complete

The place-value exchange

8 + 7 = 15

Before regrouping, the ones digits contribute 15. After regrouping, the written digits are 1 ten and 5 ones:

15 → 1 + 5 = 6
The digit sum changed by 15 − 6 = 9.

Two-number carry laboratory

Checkpoint

Need a hint for Mission 1?

Use the fixed addition38+47, even if the laboratory changes.

Check your method

Input digit sums total3+8+4+7=22. The ones column carries1;85 has digit sum13. 22;one carry;13;drop9.

Why the long product pattern continues

A row of n nines is 10ⁿ−1. Multiplying by the row of n eights therefore shifts that row left n places and subtracts the original row. For n=3: 888×999=888000−888=887112. The borrowing leaves n−1 eights, then 7, then n−1 ones, then 2. Its digit sum is 8(n−1)+7+(n−1)+2=9n. Predict the n=4 pattern before using the tool.

Mission 2

Use the carry formula on the worked example exercises

For two whole numbers, count every column that sends a carry to the next column.

Not complete
Exercises 2–3

One carry

If s(A)=8, s(B)=7, and the addition has one carry, then:

s(A+B)=8+7−9=6

Three carries

If s(A)=35, s(B)=26, and the addition has three carries, then:

s(A+B)=35+26−3×9=34

Carry-formula laboratory

Input digit sums61
9 per carry27
=
Result digit sum34
This formula gives a possible nonnegative digit sum only when the stated carry count is consistent with an actual addition.

Checkpoint

Need a hint for Mission 2?

Subtract9 for each carry unit.

Check your method

First case8+7−9=6. Second case35+26−27=34. 15;6;61;34.

Mission 3

Read digit totals from a two-digit sum

Place value can reveal the sum of all four digits without identifying either two-digit number.

Not complete

Here xy and zw are two-digit numerals: xy=10x+y and zw=10z+w. The leading digits x,z are 1–9; y,w are 0–9. Digits may repeat. They are not products of letters.

Exercise 4

If xy + zw = 139, find x+y+z+w.

Ones columny + w = 9

The largest possible sum of two digits is 18, so a ones digit of 9 cannot come from 19. There is no ones carry.

Tens columnx + z = 13

The tens digits make 13, writing 3 and carrying 1 hundred.

x+y+z+w=(x+z)+(y+w)=13+9=22

Slide through actual pairs that total 139

Checkpoint

Need a hint for Mission 3?

The ones total is9 because it cannot reach19.

Check your method

The tens total is13, creating the only carry. Sum all four digits as13+9. 9;13;one carry;22.

Mission 4

Count carry units in a many-addend original sum

With several addends, one column may send a carry of 2 or 3. Count each carried ten as one carry unit.

Not complete
Exercise 1
11 + 192 + 1,993 + 19,994 + 199,995 = 222,185

The addends’ digit sums total 110. The columns send 1, 3, 3, 2, 1 carry units, for 10 altogether:

110 − 9×10 = 20 = s(222185)

General column-carry ledger

PlaceAddend digitsIncoming carryColumn totalWritten digitOutgoing carry units

Checkpoint

Need a hint for Mission 4?

A carry of3 counts as three units, not one event.

Check your method

For the fixed five-addend original sum, outgoing carries1,3,3,2,1 total10. 222185;input digit sums110;10 carry units;result digit sum20.

Mission 5

Use modulo 9 as a filter for products

A number and its digit sum have the same remainder modulo 9, but matching remainders do not guarantee matching digit sums.

Not complete

Why modulo 9 works

Every decimal place value is one more than a multiple of 9:

10 ≡ 1, 100 ≡ 1, 1000 ≡ 1 (mod 9)

Therefore:

n ≡ s(n) (mod 9)

For multiplication:

s(kn) ≡ k·s(n) (mod 9)

A filter is not a proof

For n=27, both s(27)=9 and s(7×27)=s(189)=18 are divisible by 9.

They have the same remainder modulo 9, but the exact digit sums are different.

Product digit-sum audit

Checkpoint

Need a hint for Mission 5?

Compare exact sums as well as remainders.

Check your method

27 has digit sum9;7×27=189 has digit sum18. Both leave remainder0 modulo9. 9;9;18;no.

Mission 6

Find every two-digit number whose digit sum never changes

The worked example requires the digit sum to stay exactly the same after multiplication by every digit from 1 through 9.

Not complete
Exercise 6

If s(2n)=s(n), then modulo 9:

2s(n) ≡ s(n) ⟹ s(n) ≡ 0 (mod 9)

A two-digit number has digit sum at most 18, so only digit sums 9 or 18 remain. This leaves the ten two-digit multiples of 9 to test completely.

Complete candidate audit

Checkpoint

Need a hint for Mission 6?

The answer must pass all nine multipliers, not just those currently displayed.

Check your method

The ten two-digit multiples of9 are candidates; only18,45,90,99 pass the full test. Ten candidates;four pass;list18,45,90,99 in any order; s(99)=18.

Mission 7

Discover the repeated-digit product pattern

The worked example multiplies a 2,000-digit string of 9s by a 2,000-digit string of 8s. A short pattern replaces impossible long multiplication.

Not complete
Exercise 5
9×8=72
99×88=8712
999×888=887112

The pattern for n digits is:

(n−1) eights, then 7, then (n−1) ones, then 2
digit sum = 8(n−1)+7+(n−1)+2 = 9n

Repeated-digit product laboratory

Checkpoint

Need a hint for Mission 7?

Use the fixed n=2000 original product.

Check your method

The product has n−1 copies of8, then7, then n−1 copies of1, then2. Its digit sum is9n. 1999 eights;1999 ones;last two digits12;sum18000.

Mission 8

Transfer the carry method to two companion-test sums

The target numeral determines the carry pattern. A distinct-digit restriction may reduce the actual addend pairs, but the digit-total argument still comes from place value.

Not complete
21 • Question 4

ab + cd = 149

The ones digits total 9, while the tens digits total 14. Therefore:

a+b+c+d=14+9=23
21 • Question 10

ab + cd = 182, with four distinct digits

The ones digits could total 2 or 12. If they totaled 2, the tens digits would have to total 18, forcing both tens digits to be 9 and violating the distinct-digit condition. So the ones column must total 12 and carry 1. The tens digits then total 17. Therefore:

a+b+c+d=12+17=29

A complete unordered-pair audit gives 85+97 and 87+95.

General two-digit-pair explorer

Checkpoint

Need a hint for Mission 8?

ab and cd are two-digit numerals. In the182 case all four digits must differ.

Check your method

149 gives ones total9 and tens total14. For182, distinctness forces ones12 and tens17. The two unordered pairs are85+97 and87+95. 23;one carry;29;two pairs.

Try a new problem before the workshop

Find the digit sum of 58+67 using carries, then verify it directly.

Compare your reasoning after trying

The input digit sums total 26. There are two carry units, so 26−18=8. Directly, 58+67=125 and 1+2+5=8.

Mission 9

Independent workshop

Correct all eight answers to complete the workshop.

Not complete
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Optional reflection — not automatically graded

Need a hint for Mission 9?

Distinguish carry units, result digit sums and complete number lists.

Check your method

Use the original original values, independent of the current laboratory settings. 9;6;34;22;20;list18,45,90,99;18000;29.

Mission 10

Objective exit ticket

Answer all five correctly and complete the other nine missions to unlock the certificate.

Not complete
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21.5

Carry & Product Digit-Sum Architect

This certifies that Student completed all ten missions and can track digit sums through carries, use modulo 9 responsibly, and prove structured multiplication patterns.

Chapter 21 • Digit Sums

Need a hint for Mission 10?

Subtract nine times the carry count.

Check your method

27 first fails at multiplier7, since7×27=189 has digit sum18. Repeated-digit products have digit sum9n. 9;10;7;225;9.