Math Education · Chapter 17
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Chapter 17 · Lesson 17.2

Use Parity and Remainders to Pin Down Primes

利用奇偶性与余数锁定质数

Use the fact that 2 is the only even prime, then organize numbers by their remainders to eliminate whole families of impossible cases before testing individual primes.

Restrict first. Search second. Verify every survivor.
10 interactive missionsGrade 5No this lesson requiredoriginal examples fully restated
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Mission 1

Use the one exceptional even prime

Every prime except 2 is odd. This single exception turns many long prime searches into short parity arguments.

Not complete

The special prime

2is the only even prime

Every prime greater than 2 is odd.

Two-prime sums

odd + odd = even
2 + odd = odd

So an odd sum of two primes forces one prime to be 2.

Several odd primes

odd count of odd numbers → odd sum
even count of odd numbers → even sum

Parity composer

Odd-prime count5
Total primes5
Sum parityOdd
Does adding 2 change parity?No

Check the parity toolkit

Need a hint? Start here

Parity means whether a number is odd or even.

How to check my reasoning

An odd count of odd addends gives an odd sum; adding 2 does not change it. 2; odd; 2.

Mission 2

Worked example 1: seven consecutive primes

Seven consecutive primes are named a, b, c, d, e, f, g from greatest to least. Their total is even. Determine c.

Not complete
Example 1

Why the block must contain 2

Consecutive primes occupy neighboring positions in the prime list, with no prime skipped. They need not be consecutive whole numbers.

1

If the seven primes were all odd, their sum would be odd.

2

The total is even, so the block must include the only even prime, 2.

3

Because the primes are consecutive, the block is exactly the first seven primes.

2, 3, 5, 7, 11, 13, 17

Consecutive-prime block explorer

Smallest prime7
Block sum
Sum parityOdd
Contains 2?No

Move the slider until the only even-sum block appears.

Descending labels

The third entry from the greatest end is c.

Finish the worked example deduction

Need a hint? Start here

Seven odd primes would have an odd sum.

How to check my reasoning

The consecutive block must include 2, so it is the first seven primes. Read labels from greatest to least. 2; c=11; first seven primes.

Mission 3

Guided Practice 1: three distinct primes total 40

The word distinct means all three primes must be different. Parity first forces one prime; then a complete pair search handles the remaining total.

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Guided Practice 1

Force the even prime

Three odd primes would have an odd sum. Since 40 is even, one of the three primes must be:

2

The remaining two primes must total:

40 − 2 = 38

Search prime pairs totaling 38

Do not stop at 19 + 19. It uses the same prime twice, but the worked example requires three different primes.

Enter the three primes from least to greatest

Run the complete pair search before checking your answers.

Need a hint? Start here

All three primes must be different. Run the pair search.

How to check my reasoning

Parity forces 2. The remaining pair totals 38; 19+19 repeats a prime. 2,7,31 in increasing order.

Mission 4

Worked example 2: prove a prime sum is impossible

Can two primes add to the 20-digit number made entirely of 1s? The worked example combines parity with a digit-sum divisibility test.

Not complete
Example 2

The target number

This is a 20-digit odd number.

Reveal the proof one step at a time

Why the digit sum finishes the proof

11111111111111111109

Its digit sum is:

18 × 1 + 0 + 9 = 27

So it is divisible by 9 and is greater than 9. It is composite, not prime.

Audit the impossibility proof

Reveal all four proof steps before checking your answers.

Need a hint? Start here

Reveal all four steps. The odd sum forces one addend to be 2.

How to check my reasoning

Subtracting 2 changes the final 11 into 09; the digit sum is 27. No; forced prime 2; one-digit composite divisor 9.

Mission 5

Guided Practice 3: a weighted prime equation

Two primes p and q satisfy 3p + 2q = 2000. Use parity before doing any primality search.

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Guided Practice 3

Parity forces p

3p+2q=2000

2q is even and 2000 is even, so 3p must be even. Since 3 is odd, p must be even.

The only even prime is 2.

Candidate explorer

3p parityOdd
Computed q
q statusNot whole
p + q

Complete the equation

Need a hint? Start here

Both 2q and 2000 are even.

How to check my reasoning

Therefore 3p is even; since 3 is odd, p must be the only even prime. p=2; q=997; p+q=999.

Mission 6

Guided Practice 4: opposite cube faces

The sum of the two numbers in each pair of opposite faces is the same for all three pairs. The faces opposite 18, 14, and 35 contain primes a, b, and c.

Not complete
181435opposite: aopposite: bopposite: c

Common-sum explorer

Parity shortcut

If the common sum were even, then a = S − 18 and b = S − 14 would both be even. The only even prime is 2, but the two equations cannot both work with the same even sum.

So the common sum is odd. Then:

c = odd − 35 = even prime = 2

Complete the cube

Run the common-sum search before checking. The parity argument proves uniqueness for all sums; the tool checks only the displayed range 20 through 100.

Need a hint? Start here

Opposite-pair sums are equal: 18+a=14+b=35+c.

How to check my reasoning

An even common sum would force both a and b to be 2, contradicting the different fixed faces. Thus the sum is odd and c=2. Common sum 37; c=2; a+b+c=19+23+2=44. Run the displayed search to complete the check.

Mission 7

Use remainder classes to pin down prime chains

Three expressions can cover all three possible remainders after division by 3. One must then be divisible by 3—and if it is prime, it must equal 3.

Not complete

Prime-chain explorer

exercise

p, p + 10, p + 14

Because:

10 leaves remainder 1 after division by 3
14 leaves remainder 2 after division by 3

the three terms occupy remainder classes 0, 1, and 2 in some order. The divisible-by-3 term must itself be the prime 3.

p = 3 → 3, 13, 17

17

p, p + 20, p + 40

p = 3 → 3, 23, 43

Use the remainder-class argument

Run a search through p≤200 before checking. The remainder argument proves the result for every prime p; the search confirms cases only in its stated range.

Need a hint? Start here

Reduce the offsets modulo 3 and run the comparison search.

How to check my reasoning

One member is divisible by 3. The larger members exceed 3, so the first member must be 3. p=3 for both chains; the three terms occupy all three remainder classes.

Mission 8

Explore prime sums systematically

Parity predicts whether 2 is necessary. A complete search counts each group of prime terms once, in increasing order (or nondecreasing order if repeats are allowed). Rearrangements such as 2+7+31 and 31+2+7 count as the same representation.

Not complete

Prime-sum enumerator

Representations0
Parity forecast
Contains 2 in every result?
Search statusReady

Two transfer conclusions

exercise

If two primes total 2001, the odd sum forces one prime to be 2. The other is 1999, which is prime.

2 × 1999 = 3998
17

If primes satisfy x + y = z, then the two addends cannot both be odd, because their sum would be an even prime greater than 2. Therefore one addend is 2.

The smallest of x, y, z is 2.

Transfer the method

Need a hint? Start here

For two primes with an odd sum, one must be 2.

How to check my reasoning

2001−2=1999. For x+y=z, two odd addends would make an even number greater than 2. 2; 1999; product 3998; smallest prime 2.

Mission 9

Parity-and-remainder workshop

Correct all eight questions to complete the workshop.

Not complete

1. What is the only even prime?

2. The sum of seven odd primes is:

3. Seven consecutive primes a>b>c>d>e>f>g have an even total. What is c?

4. What is the largest of the three distinct primes totaling 40?

5. Can the 20-digit all-ones number be written as a sum of two primes?

6. If 3p + 2q = 2000 for primes p and q, what is p + q?

7. The primes a,b,c satisfy 18+a=14+b=35+c. What is a+b+c?

8. If p, p+20, and p+40 are all prime, what is p?

Workshop score0 / 8
Need a hint? Start here

Use parity before testing large candidates.

How to check my reasoning

Check distinctness, descending labels, and which prime is forced to equal 2 or 3. 2; odd; 11; 31; no; 999; 44; 3.

Mission 10

Exit ticket

Complete Missions 1–8, earn 8/8 in the workshop, and earn 5/5 here to unlock your certificate. The optional reflection is not automatically graded.

Not complete

1. If two primes have an odd sum, one of them must be:

2. Seven consecutive primes have an even total. Their block must begin with which smallest prime?

3. Three distinct primes total 40. What is the middle prime?

4. After subtracting 2 from the 20-digit all-ones number, what is the digit sum?

5. Two primes total 45. What is the larger prime?

Exit score0 / 5

Optional reflection — not automatically graded

Certificate of completion

Prime Parity & Remainder Navigator

Grade 5 Mathematician

has completed Lesson 17.2 and demonstrated accurate use of parity, divisibility, remainder classes, complete candidate searches, and prime verification.

Teaching notes

Need a hint? Start here

Every prime except 2 is odd.

How to check my reasoning

Explain your method in the reflection box. The checker grades the answer; your explanation is for you to review.