Count Prime-Factor Contributions and Product Conditions
Do not calculate an enormous product. Classify each factor by the prime copies it contributes, assemble prime-power packets for an LCM, and merge different card pairs that create the same product.
One classification rule, three powerful uses
In every investigation, replace a large collection of calculations with a small set of case labels:
How many copies of a prime does this factor supply?
Which number receives each prime-power packet of an LCM?
Which different pair selections collapse to the same product?
Prime-copy toolkit
The symbol vp is added notation. The worked example describes the same idea by counting prime factors.
Count prime copies, not just divisible numbers
Count copies, not just multiples
25 = 5 × 5 contributes two copies of 5; 125 contributes three. Count all multiples of 5 once, add another copy for multiples of 25, then another for multiples of 125. Every extra layer counts a different copy.
Prime-copy scanner
Why “multiple of 5” is not enough
105=5×21 contributes one copy of 5, but:
Both are multiples of 5, yet their contributions differ by two prime copies.
Need a hint?
Divide by the chosen prime repeatedly; every successful division removes one copy.
Worked example 5 — classify the factors from 101 through 200
Factors ending in 0
Every value contributes at least one 5. The values 150 and 200 contribute a second copy.
Factors ending in 5
Eight values contribute one copy, 175 contributes two, and 125 contributes three.
| Factor | Prime-5 part | Copies contributed | original case |
|---|
Need a hint?
Multiples of 25 contribute an extra 5 beyond the one already counted for multiples of 5.
Pair 2s with 5s to create trailing zeros
Prime-copy balance for the worked example product
Do not calculate the product
The product contains:
Need a hint?
Every trailing zero uses a pair consisting of one 2 and one 5. The scarcer prime limits the pairs.
Use exponent layers for any consecutive product range
Layers for prime 2
Layers for prime 5
Need a hint?
For an interval, subtract the multiple count below the lower endpoint from the count up to the upper endpoint.
Guided Practice 4 — find the extra contributors
Every multiple of 5 contributes once
There are nine multiples of 5 from 11 through 56.
Which values contribute an extra 5?
Select every value divisible by 5²=25.
Need a hint?
Look for the numbers contributing an extra prime copy beyond the first layer.
original Exercise 7 — distribute the prime packets of an LCM
105 = 3 × 5 × 7 has no repeated prime. Each prime has three choices: in A only, in B only, or in both. That explains 3 × 3 × 3 ordered assignments. Remove disallowed assignments before dividing by two to forget order.
Count ordered assignments
There are three assignment choices for each of the three prime packets.
Apply the printed restrictions
Remove the two ordered assignments containing 1 and the one equal pair (105,105):
Then divide by 2 because swapping A and B does not make a new unordered pair:
Show the complete pair audit
Need a hint?
For each prime of 105, choose A only, B only, or both. Exclude forbidden assignments before forgetting order.
Enumerate all pairs with a chosen LCM
When the LCM has a repeated prime
For LCM 12 = 2² × 3, the maximum exponent of 2 must be 2. The exponent pairs (in A, in B) are (2,0), (2,1), (2,2), (0,2), (1,2): five choices. The smaller exponent need not be zero. For example, 4 and 6 use exponents 2 and 1.
For maximum exponent e, first put e in A and choose B's exponent from 0 to e (e+1 choices). Then put e in B and choose A's exponent from 0 to e−1 (e choices). Total: 2e+1. For 12 this gives 5 × 3 = 15 ordered pairs before restrictions. Remove (1,12), (12,1), and (12,12), then divide by two: six unordered distinct pairs greater than 1.
| Pair | GCD | LCM check | Prime-packet view |
|---|
Need a hint?
List pairs of exponents whose maximum is required. A lower exponent on the other number is allowed.
original Exercise 8 — merge duplicate product values
Need a hint?
Different card pairs can give the same product. Merge equal product values before testing divisibility.
Prime-factor and product-condition workshop
1. How many copies of 5 occur in the prime factorization of 625, written v₅(625)?
2. Which two primes make one factor 10?
3. How many trailing zeros occur in 101×···×200?
4. How many trailing zeros occur in 11×···×56?
5. How many unordered pairs of different positive integers, both greater than 1, have LCM 105?
6. How many unordered pairs of different positive integers, both greater than 1, have LCM 12?
7. Choose two cards from two copies of labels 1–13. How many different product values are possible?
8. How many of those products are divisible by 6?
Need a hint?
Answers: 4; 2 and 5; 25; 11; 12; 6; 72; 21. Count every prime copy; remove forbidden ordered LCM assignments before dividing by two; merge pairs with equal products before counting product values.
Exit ticket and certificate
1. Trailing zeros in 51×52×···×100:
2. What is v₅(175)?
3. How many ordered pairs (A,B) of positive integers have LCM 105? Allow 1 and allow A=B.
4. Unordered distinct pairs greater than 1 with LCM 12:
5. With two copies of labels 1–10, how many different products are divisible by 6?
Need a hint?
Answers: 12; 2; 27; 6; 12. For 51–100, the 5-layers contribute 10+2; enough factors of 2 are available. 175=7×5². For LCM105 there are 3³ assignments before restrictions. For LCM12, (15−2−1)÷2=6 valid unordered pairs. The twelve distinct products divisible by 6 for labels1–10 are 6,12,18,24,30,36,42,48,54,60,72,90.
Prime-Factor Product Classification Architect
This certifies that the learner can count prime-factor contributions, assemble LCM packets, and merge duplicate product cases.
Chapter 24 · Classification · Lesson 24.6