Math worksheet on multiplication properties with problems to classify as grouping, property of one, zero property, or order property.

Let’s go step by step through each problem and match it to the correct property of multiplication.

We are looking for:
- Grouping Property → Also called Associative Property. It means you can group numbers differently when multiplying, and the answer stays the same. Example: (a × b) × c = a × (b × c)
- Property of One → Also called Identity Property. Any number multiplied by 1 equals itself. Example: a × 1 = a or 1 × a = a
- Zero Property → Any number multiplied by 0 equals 0. Example: a × 0 = 0 or 0 × a = 0
- Order Property → Also called Commutative Property. You can switch the order of the numbers being multiplied, and the answer stays the same. Example: a × b = b × a

Now let’s solve each one:

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1) 10 × 7 = 7 × 10
→ The numbers switched places. That’s the order property.

2) 1 × (10 × 5) = (1 × 10) × 5
→ The grouping changed — parentheses moved around the 1 and the (10×5). But note: this also involves the “1”, so we must check if it’s really about grouping or about the 1.
Actually, both sides equal 50. Left: 1 × 50 = 50; Right: 10 × 5 = 50.
But the structure shows regrouping — even though 1 is involved, the main change is in grouping. However, since 1 is always there, some might argue it’s identity + associative. But the question asks which property the *problem shows*. The equation is showing that changing the grouping doesn’t change the result — so grouping property is best fit here.

Wait — actually, let’s think again. The left side is 1 × (10×5), right side is (1×10) × 5. This is exactly the associative property applied with 1 included. So yes — grouping property.

3) (4 × 10) × 2 = 4 × (10 × 2)
→ Parentheses moved — classic grouping property.

4) 2 × 1 = 1 × 2
→ Numbers switched AND 1 is involved. But switching order is commutative (order property), and multiplying by 1 is identity. Which one does the problem show?
The equation is showing that 2×1 equals 1×2 — which is true because of BOTH properties. But the key action here is swapping the order — so order property is what’s being demonstrated. Also, note that 2×1=2 and 1×2=2 — so it’s also identity, but the form emphasizes the swap. Let’s stick with order property as primary.

Actually — wait! In many curricula, if 1 is involved and the product equals the other number, they call it “property of one”. But here, both sides have 1, and they’re swapped. Hmm.

Let me compare to #5: 1 × 6 = 6 → clearly property of one.

#4: 2 × 1 = 1 × 2 → this is showing that multiplication is commutative EVEN when 1 is involved. So I think order property is still correct.

But let’s see official definitions:

- Order/Commutative: a × b = b × a → regardless of values.
- Property of One: a × 1 = a → focuses on the result being the same as the non-one factor.

In #4, the equation is written as an equality between two expressions that are reverses — so it’s demonstrating commutativity. So order property.

5) 1 × 6 = 6
→ Multiplying by 1 gives the same number → property of one

6) 4 × 10 = 10 × 4
→ Swapped order → order property

7) 7 × 0 = 0 × 3
→ Wait — this is tricky. Left side: 7×0=0. Right side: 0×3=0. So both sides equal 0. But the equation is saying 0 = 0, which is true, but it’s not showing a standard property directly.
Actually, this seems like a trick. The zero property says any number times 0 is 0. Here, both sides use 0, so it’s illustrating that no matter what you multiply by 0, you get 0 — even if the other numbers are different. So this is showing the zero property, because both products are 0 due to multiplying by 0.

Even though the non-zero numbers are different (7 and 3), the fact that both sides equal 0 because of the 0 multiplier — so zero property.

8) 3 × 10 = 10 × 3
→ Swapped → order property

9) 4 × 1 = 4
→ Multiply by 1 → property of one

10) 1 × 0 = 0 × 1
→ Both sides equal 0. Again, this is showing that multiplying by 0 gives 0, regardless of order. So zero property. Even though 1 is involved, the key is the 0 making the product 0.

11) 1 × 2 = 2
→ Multiply by 1 → property of one

12) 4 × 8 = 8 × 4
→ Swapped → order property

13) 1 × (8 × 5) = (1 × 8) × 5
→ Grouping changed — parentheses moved. Similar to #2. Even though 1 is involved, the structure is about regrouping. So grouping property

14) (10 × 7) × 3 = (3 × 7) × 10
→ Let’s compute both sides to verify:

Left: (10×7)=70, then 70×3=210
Right: (3×7)=21, then 21×10=210 → same.

But how did we get from left to right? We didn’t just regroup — we also reordered the numbers. Original factors: 10, 7, 3. On right: 3, 7, 10. So we used both commutative (order) and associative (grouping) properties together.

But the question says: “Decide if the problems show grouping property, property of one, zero property, or order property.” — implying pick ONE.

Looking at the transformation:
(10 × 7) × 3 → became (3 × 7) × 10

This required:
- Moving 3 to front (commutative/order)
- Keeping 7 in middle
- Moving 10 to end (another commutative)
- And regrouping? Actually, the grouping is still two numbers first, then multiply by third — so grouping structure is similar.

Actually, the main change is reordering the factors. The parentheses are still around the first two numbers — so grouping hasn't changed in structure, only the order of the numbers inside.

Wait — original: (10 × 7) × 3 → group 10 and 7 first
New: (3 × 7) × 10 → group 3 and 7 first

So the grouping changed — now 3 and 7 are grouped instead of 10 and 7. Plus the order of the outer number changed.

This is a combination, but perhaps the most prominent feature is that the entire set of factors was rearranged — so order property? Or maybe grouping?

Actually, in some contexts, this would be considered using both, but since we have to choose one, and the problem likely expects us to recognize that the factors were reordered (not just regrouped), I’d say order property is dominant here.

But let’s think: if it were only grouping, it would be like (10×7)×3 = 10×(7×3). Here, it’s (10×7)×3 = (3×7)×10 — which requires commuting 3 with (10×7), and then commuting 10 with 7? Not straightforward.

Perhaps it's better to say this demonstrates the commutative property because the factors are permuted.

I recall that in elementary math, sometimes they accept "order property" for any rearrangement of factors.

Alternatively, maybe it's intended to be grouping property because the parentheses are still there, just with different numbers inside.

Wait — let's look back at #2 and #13: those had 1 and showed grouping.

For #14, since all numbers are different and none are 0 or 1, and the equation holds due to commutativity and associativity combined, but the question forces one choice.

Given that the final arrangement has the same three numbers but in different order and different grouping, but the key insight is that multiplication allows us to reorder freely — I think order property is acceptable.

However, upon second thought, many textbooks would classify this under the associative and commutative properties together, but since we must choose one, and the problem might expect "grouping" because of the parentheses, I'm conflicted.

Let me calculate numerically:

Left: (10*7)*3 = 70*3 = 210
Right: (3*7)*10 = 21*10 = 210

To go from left to right, you could do:
Step 1: Use commutative on the outer part: (10×7)×3 = 3×(10×7) [but that's not standard]
Actually, standard way:
(10×7)×3 = 10×(7×3) [associative]
Then 10×(7×3) = 10×(3×7) [commutative inside]
Then 10×(3×7) = (3×7)×10 [commutative overall]

So multiple steps.

But perhaps for this level, they consider any rearrangement of factors as "order property".

I found a similar example online: sometimes (a*b)*c = (c*b)*a is attributed to commutative property.

I'll go with order property for #14.

But let's double-check with a reliable source in my knowledge: In Common Core and similar standards, the commutative property is for order, associative for grouping.

In #14, the grouping is technically the same format: (first two) times third. But the "first two" are different numbers. So the grouping operation is applied to different pairs, which implies reordering happened first.

Perhaps it's safer to say grouping property is not purely shown because the factors themselves changed position.

Another approach: count how many properties are violated if we remove one.

If we assume only commutative property exists, can we derive it? Yes, by swapping factors.

If we assume only associative, can we? No, because associative doesn't allow changing order, only grouping.

So commutative (order) is necessary.

Therefore, order property is the best single answer.

Final decisions:

1) order
2) grouping
3) grouping
4) order
5) property of one
6) order
7) zero
8) order
9) property of one
10) zero
11) property of one
12) order
13) grouping
14) order

Wait, for #10: 1 × 0 = 0 × 1 — both sides are 0, so it's zero property, even though 1 is there. The presence of 0 makes the product 0, so zero property dominates.

Similarly, #7: 7×0=0×3 — both 0, so zero property.

Now, let's write them down.

Final Answer:

1) order property
2) grouping property
3) grouping property
4) order property
5) property of one
6) order property
7) zero property
8) order property
9) property of one
10) zero property
11) property of one
12) order property
13) grouping property
14) order property