New product is $24\times27$. Using the distributive property, $(a+1)b = ab+b$, with $a=23,\ b=27$:

New product is $23\times28$. Using $a(b+1) = ab + a$, with $a=23,\ b=27$:

New product is $24\times28 = (23+1)(27+1)$. Expanding using distributivity:

Pattern noticed: increasing one number by 1 increases the product by the value of the other number; increasing both by 1 increases the product by the sum of both numbers plus 1.

No — the product does not always increase. Recall the identity:

The increase is $b – a – 1$. This is negative (i.e. the product decreases) whenever $b \lt a+1$, i.e. when $a \ge b$.

The distributive property holds for all integers, positive or negative, since $x(y+z)=xy+xz$ is true for any integers $x,y,z$. Let’s verify with the identity $(a+1)(b-1)=ab+b-a-1$.

Treat $(b+1)$ as one block and distribute over $(a+1)$:

• $(a+1)(b+1) = a(b+1) + 1(b+1)$
• $= ab + a + b + 1$

Write as $(a + (-2))(b+3)$, so $m=-2,\ n=3$:

Write as $(a+(-3))(b+(-4))$, so $m=-3,\ n=-4$:

Both results match what you’d get by direct distribution, confirming Identity 1 works for decreases too — just treat the decrease as adding a negative number.

We know $(a+b)^2 = a^2+2ab+b^2$. So:

• If $ab \gt 0$ (both same sign, both positive or both negative) → $(a+b)^2 \gt a^2+b^2$.
• If $ab \lt 0$ (opposite signs) → $(a+b)^2 \lt a^2+b^2$.
• If $a=0$ or $b=0$ → $(a+b)^2 = a^2+b^2$.

Decompose $104 = 100+4$:

Decompose $37 = 30+7$:

Two terms can only be combined (added/subtracted into one term) if they are like terms — meaning they have exactly the same letter-numbers (variables) raised to the same powers.

To multiply any number by 11: write down the last (units) digit as it is. Then, moving leftward, add each pair of neighbouring digits and write down that sum (carrying over if the sum is 10 or more). Finally, write down the first digit (plus any carry).

For a 4-digit number $dcba$: $\ dcba \times 11 = d\ (c+d)\ (b+c)\ (a+b)\ a$  (handling carries as needed).

Using $dcba \times 101 = dcba\times100 + dcba$, for digits $d{=}3, c{=}8, b{=}7, a{=}4$:

For 101: leave the last two digits as-is, then for each pair moving left, add digits two places apart, carrying as needed, finishing with the leading digit(s). This generalises: multiplying by $10^k+1$ means “shift and add” — write the number, then add the same number shifted left by $k$ places.

• By $101 = 100+1$: shift left by 2 places and add.
• By $1001 = 1000+1$: shift left by 3 places and add.
• By $10001 = 10000+1$: shift left by 4 places and add.

Write $99 = 100-1$:

Write $999 = 1000-1$:

The identity $(a+b)^2 = a^2+2ab+b^2$ works for any split of 65 into two parts — the total area of the big square stays the same regardless of how we partition the side length.

Using $(a+b)^2=a^2+2ab+b^2$ with $a=m, b=3$:

With $a=6, b=p$:

With $a=3j, b=2k$:

Draw a square of side $a$, and remove a square of side $b$ from one corner (where $b \lt a$), leaving an L-shaped region with side $a-b$.

The square of side $(a-b)$ is obtained by taking the big square ($a^2$), and removing the two rectangle strips of dimensions $a\times b$ — but this removes the small corner square of side $b$ twice, so we add it back once.

Write $99 = 100-1$, so $a=100, b=1$:

Write $58 = 60-2$, so $a=60, b=2$:

Take $a’=-2a,\ b’=3$ (or equivalently use $(3-2a)^2$):

Take $a=4, b=7$. Sum of squares: $4^2+7^2 = 16+49=65$. Twice this sum: $2\times65=130$.

Add the two identities:

Write as $(100-2)(100+2)$, so $a=100, b=2$:

Write as $(50-5)(50+5)$, so $a=50, b=5$:

Start with a large square of side $a$, with a smaller square of side $b$ cut from one corner. The remaining L-shaped region has area $a^2-b^2$.

Cut this L-shape into two rectangles as shown, then re-arrange (slide the smaller rectangle) to form a single rectangle of dimensions $(a+b)$ by $(a-b)$.

This is just Identity 1C, $(a+b)(a-b)=a^2-b^2$, rearranged by adding $b^2$ to both sides:

The pattern shows hollow square frames: Step 1 is a $3\times3$ frame with a $1\times1$ hole, Step 2 is a $4\times4$ frame with a $2\times2$ hole, Step 3 is a $5\times5$ frame with a $3\times3$ hole.

Number of tiles = (area of outer square) − (area of inner hole) = $(n+2)^2 – n^2$.

Expanding $(n+2)^2-n^2$ using the difference-of-squares idea:

The full bounding rectangle has dimensions $s \times p$. The missing piece (top-left, not part of the L) is $(s-r)\times r$ — wait, let’s set it up carefully using the figure: the overall width is $s$, overall height is $p$; the small square cut out (bottom-left corner notch) has side $r$.

This comes from: take rectangle $p \times s$, subtract the two rectangle strips of width $r$ along two adjoining sides ($p\times r$ and $s\times r$), then add back the doubly-subtracted corner square $r^2$ — exactly like the $(a-b)^2$ identity logic.

The grid is a multiplication table where the entry at row $i$, column $j$ is $i\times j$. If the centre cell of a $3\times3$ block is $p\times q = pq$ (row $p$, column $q$), then the row above is $(p-1)$, row below is $(p+1)$; column to the left is $(q-1)$, column to the right is $(q+1)$.

Let the numbers be $a$ and $b$. We need $(a+2)(b-4) = ab$.

$(a-b)$ multiplied by a sum of $n$ terms (powers of $a$ decreasing, powers of $b$ increasing) always telescopes down to a simple difference of powers: $a^2-b^2,\ a^3-b^3,\ a^4-b^4,\ \ldots$

Verification by expansion: $a^5+a^4b+a^3b^2+a^2b^3+ab^4 -a^4b-a^3b^2-a^2b^3-ab^4-b^5 = a^5-b^5$ ✓ (all middle terms cancel out).

Note that $(b-a) = -(a-b)$. So:

Verification with numbers: Let $a=7, b=3$. Then $(a-b)^2=4^2=16$ and $(b-a)^2=(-4)^2=16$.

We want $a^2-b^2 = 100$, i.e. $(a+b)(a-b)=100$. Try $a-b=4$ and $a+b=25$… but those need to give integer $a,b$. Easier: try $a-b=2,\ a+b=50 \Rightarrow a=26, b=24$.

Both patterns are derived purely from the algebraic identities $(a+b)^2=a^2+2ab+b^2$ and $(a-b)^2=a^2-2ab+b^2$, which are true for any real numbers $a$ and $b$ — there is no restriction requiring them to be counting numbers.

Let $a=-3, b=2$:

Let $a=\tfrac32, b=\tfrac12$:

The student treated $-3p$ as if it were just $-3$, forgetting to multiply the variable $p$ into each term. Multiplication, not addition, must be distributed across the bracket.

In the first step, $2$ was multiplied only into the $x$ term of $(x-1)$, but not into $-1$. (It should be $2 \times x = 2x$ and $2\times(-1)=-2$, not just $-1$.)

This treats $y+2(y+2)$ as if it were $(y+2)\times(y+2)$ — but the expression has $y$ added to $2(y+2)$, not multiplied. The mistake is converting an addition into a squaring operation.

The middle “cross term” $2ab$ from the identity $(a+b)^2=a^2+2ab+b^2$ was forgotten entirely. Squaring a sum is not the same as squaring each term separately.

The sign of the cross term was computed incorrectly. With $a=-q, b=2$: the middle term is $2ab = 2(-q)(2) = -4q$ — that part is actually right — but $a^2=(-q)^2=q^2$ is correct too. Let’s verify carefully: actually $(-q+2)^2 = (-q)^2+2(-q)(2)+2^2 = q^2-4q+4$. This is correct!

The expression $3a(2b\times3c)$ has only one multiplication needed — first simplify inside the bracket, then multiply by $3a$ once. Instead, the student incorrectly distributed $3a$ separately into $2b$ and $3c$ as if it were $3a(2b+3c)$, creating two separate products and multiplying them together — a double mistake.

Actually, this one is computed correctly! $\tfrac12(10s-6) = 5s-3$, and adding $3$ gives $5s-3+3=5s$.

$5w^2$ and $6w$ are not like terms — one has $w^2$, the other has $w^1$. Only like terms (same variable, same power) can be combined into a single term. The student incorrectly added the coefficients while ignoring that the powers of $w$ are different.

Since the terms are unlike, the expression is already in its simplest form:

Two separate mistakes: (1) $2a^3+3a^3=5a^3$ is correct, since these are like terms. (2) But $6a^2b$ and $6ab^2$ are not like terms (different powers of $a$ and $b$ — $a^2b$ vs $ab^2$), so they cannot be combined into $12a^2b^2$.

Nothing — this expansion is done correctly, step by step, using the distributive property properly.

Only the “outer” terms ($a\times b$) and the product of the two constants ($2\times4=8$) were multiplied; the two “cross” terms ($a\times4$ and $2\times b$) were completely skipped. All four pairwise products are needed when expanding $(a+2)(b+4)$.

Let’s verify by factoring out the common factor $ab$ from each term: $ab^2 = ab\times b$; $a^2b = ab\times a$; $a^2b^2 = ab\times ab$. So:

“A square number” means some number squared, say $s^2$. “Two more than” it means add 2:

If one number is $m$, the next consecutive number is $m+1$. Sum of their squares:

Label the 2×2 block as: top-left $=a$, top-right $=a+1$; bottom-left $=a+7$ (one row below, same weekday column), bottom-right $=a+8$.

This depends on $k$ (e.g. $k=1$ gives $1+2-1=2$, but $k=2$ gives $4+4-1=7\ne2$).

$4q^2-4q$ is a multiple of 4, but subtracting 3 means the whole expression leaves remainder $-3 \equiv 1 \pmod 4$ — not a multiple of 4. Check $q=1$: $(3)(-1)=-3$, not a multiple of 4.

Even number: $2n$. $(2n)^2=4n^2$ — clearly a multiple of 4. ✓

Odd number: $2n+1$. $(2n+1)^2 = 4n^2+4n+1 = 4n(n+1)+1$. Since $n(n+1)$ is always even (product of consecutive integers), $4n(n+1)$ is a multiple of 8, so $(2n+1)^2$ is 1 more than a multiple of 8. ✓

For this to be “5 less than a square number,” we’d need $20n^2-5 = (\text{some square})-5$, i.e. $20n^2$ itself a perfect square for all $n$ — but $20n^2$ is a perfect square only when $20$ is a perfect square times something, which fails generally (e.g., $n=1$: $20$ is not a perfect square).

Let $n_1 = 7a+3$ and $n_2=7b+5$ for some integers $a, b$.

Remainder when divided by 7: 1

Remainder when divided by 7: 5  (equivalently, $n_2-n_1$ leaves remainder 2)

Remainder when divided by 7: 1

Take $4, 5, 6$: middle squared $= 5^2=25$; product of outer two $=4\times6=24$. Difference $=25-24=1$.

Try $9,10,11$: $10^2=100$; $9\times11=99$; difference $=1$.

Let consecutive numbers be $(n-1), n, (n+1)$:

Let the two numbers be $a$ and $b$.

• Add the two numbers: $a+b$
• Multiply by half of the sum: $(a+b) \times \tfrac12(a+b)$

Write $14\times26 = (16-2)(24+2)$:

Write $25\times75 = (26-1)(74+1)$:

General insight: for a fixed sum, the product of two numbers is larger when the numbers are closer together — moving them further apart (while keeping the sum fixed) always decreases the product.

Looking at the layout: the path of width $w$ surrounds and separates the two square plots. The full park’s length (left-to-right) $= w + g + 2w + g + w = 2g+4w$. The park’s breadth (top-to-bottom) $= w+g+w = g+2w$.

Counting units: Step 1 has $9=3^2$ units, Step 2 has $16=4^2$, Step 3 has $25=5^2$ — each step is a perfect square of (Step number + 2).

Step 1 has $5$ squares (a $2\times2$ block of 4, plus 1 extra) $=2^2+1$. Step 2 has $11 = 3^2+2$. Step 3 has $19=4^2+3$. Pattern: Step $y$ has $(y+1)^2+y$.

For a triangle with $n$ rows, the total number of coins is the triangular number $T_n = \tfrac{n(n+1)}{2}$.

The minimum number of moves needed follows a known result for “reversing a triangular array”: for the cases above, the minimums $1, 2, 3, 5$ for $n=2,3,4,5$ rows match a quadratic-style growth pattern related to $n^2$.

More simply, by direct construction: for the 15-coin (5-row) triangle, the minimum is 5 moves — found by identifying which corner coins lie outside the “flipped” footprint and relocating just those.