Imagine getting this wrong (2021 10A #1)

Art of Problem Solving

AoPS Online

Contests & Programs AMC and other contests, summer programs, etc.

AMC and other contests, summer programs, etc.

3 M G

BBookmark VNew Topic kLocked

Contests & Programs AMC and other contests, summer programs, etc.

AMC and other contests, summer programs, etc.

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Imagine getting this wrong (2021 10A #1)

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: AMC 10A 2021 #1

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

8791 posts

#1 h Feb 5, 2021, 1:59 PM • 2 Y

Y by centslordm, megarnie

$1.$ What is the value of $(2^2-2)-(3^2-3)+(4^2-4)?$
$\textbf{(A) }1 \qquad \textbf{(B) }2 \qquad \textbf{(C) }5 \qquad \textbf{(D) }8 \qquad \textbf{(E) }12$

This post has been edited 2 times. Last edited by samrocksnature, Feb 5, 2021, 5:52 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1155 posts

#5 h Feb 5, 2021, 5:50 PM

Y by

$2-6+12=8$
Answer: D

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1657 posts

#6 h Feb 5, 2021, 5:50 PM

Y by

You could do difference of squares. It's pretty hard for a 1 though

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

193 posts

#7 h Feb 5, 2021, 5:51 PM

Y by

LOL that would uh... suck
$~~~~~~$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

4917 posts

pog

#8 h Oct 19, 2021, 2:08 AM • 56 Y

Y by ChrisWren, fuzimiao2013, 618173, channing421, hh99754539, RP3.1415, suvamkonar, eagles2018, tigerzhang, DramaticFlossKid, OlympusHero, AOPqghj, centslordm, john0512, Jndd, ChromeRaptor777, sugar_rush, greenturtle3141, mahaler, NoSignOfTheta, tenebrine, HrishiP, fukano_2, Jwenslawski, sargamsujit, kante314, wamofan, thinker123, megarnie, violin21, WhitePhoenix, FIREDRAGONMATH16, judgefan99, danprathab, jason543, CyclicISLscelesTrapezoid, ihatemath123, sanaops9, ChrisalonaLiverspur, Eagle42, peelybonehead, Math4Life7, bobthegod78, michaelwenquan, IMUKAT, qwerty123456asdfgzxcvb, Ultroid999OCPN, eibc, ryanbear, doulai1, EpicBird08, Facejo, mathmax12, zhoujef000, OronSH, Turtwig113

Solved with literally nobody

Lemma 1.1: $x^2-y^2=(x+y)(x-y)$ .

Proof: Expand the right side using FOIL as mandated by the federal government, giving $(x)(x) + (x)(-y) + (x)(y) + (x)(-y) = x^2 - xy + yx - y^2$ . By the commutative property of multiplication, $xy = yx$ , so $a^2 - xy + yx - b^2 = a^2 - xy + xy - b^2$ .

Sublemma 1.2: $\ell - \ell = 0$ for any arbitrary value of $\ell$ .

Proof: Note that since $\ell + 0 = \ell$ by the identity property, subtracting $\ell$ from both sides gives $-\ell + \ell = 0$ , as desired. $\square$

By Sublemma 1.2, we get that $xy - xy = 0$ , and since $-xy + xy = xy - xy$ by the commutative property of addition, we get that $-xy + xy = 0$ . Thus, $a^2 - xy + xy - b^2 = a^2 - 0 - b^2$ , which by the commutative property of addition is equal to $a^2-b^2-0$ . Since $-0 = 0$ , we get $a^2 - b^2 + 0$ . Substitute $a^2-b^2=k$ . Then $k+0 =k$ by the identity property, so thus $a^2 - b^2 + 0 = k + 0 = k = a^2 - b^2$ .

Consequently, as desired, we have proved that $x^2-y^2=(x+y)(x-y)$ .

Looking at the problem, we wish to use Lemma 1.1. If we let $2 = (\sqrt2)^2$ , $3 = (\sqrt3)^2$ , and $4 = (\sqrt4)^2$ , then we can use Lemma 1.1 to give
$(2^2-2)-(3^2-3)+(4^2-4) = (2 - \sqrt2)(2 + \sqrt2) - (3+\sqrt3)(3 - \sqrt3) + (4+\sqrt4)(4 -\sqrt4).$ Expanding gives
$(2 - \sqrt2)(2 + \sqrt2) - (3+\sqrt3)(3 - \sqrt3) + (4+\sqrt4)(4 -\sqrt4) = [(2)(2) + (-\sqrt2)(\sqrt2) + (2)(\sqrt2) + (2)(-\sqrt2)] - [(3)(3) + (-\sqrt3)(\sqrt3) + (3)(\sqrt3) + (3)(-\sqrt3)] + [(4)(4) + (-\sqrt4)(\sqrt4) + (4)(\sqrt4) + (4)(-\sqrt4)].$ We can rearrange the first, second, third, and fourth terms in each set of brackets into their own sets of brackets individually by the commutative property of addition, so hence we get
$[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] + [(2\sqrt2)-(3\sqrt3)+(4\sqrt4)] + [(-2\sqrt2)-(-3\sqrt3)+(-4\sqrt4)].$ Let $a = (2\sqrt2)-(3\sqrt3)+(4\sqrt4)$ and $b = (-2\sqrt2)-(-3\sqrt3)+(-4\sqrt4)$ . Substitution gives that
$[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ]+ [(2\sqrt2)-(3\sqrt3)+(4\sqrt4)] + [(-2\sqrt2)-(-3\sqrt3)+(-4\sqrt4)]$ is equal to $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] + a + b.$ Cleverly, we note that $a = -b$ , so substituting $a = -b$ into $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] + a - b$ gives $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] -b + b.$ By Sublemma 1.2, we get that $-b+b = 0$ , so thus $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] -b + b = [2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ]+ 0.$ Let $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ]= n$ . Then $n + 0 = n$ by the identity property, so $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] + 0 = n + 0 = n = [2^2-3^2+4^2] + [ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ]$ .

We are now ready for the brutal answer extraction. Expanding the squares, we get that $2^2 - 3^2 + 4^2 + [ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] = 2^2 - 3^2 + 4^2 - (\sqrt2)(\sqrt2) + (\sqrt3)(\sqrt3) - (\sqrt4)(\sqrt4)$ . We have that $(\sqrt2)(\sqrt2) = 2$ , $(\sqrt3)(\sqrt3) = 3$ , and $(\sqrt4)(\sqrt4) = 4$ , so consequently
$2^2 - 3^2 + 4^2 - (\sqrt2)(\sqrt2) + (\sqrt3)(\sqrt3) - (\sqrt4)(\sqrt4) = 2^2 - 3^2 + 4^2 - 2 + 3 - 4.$
By the commutative property of addition, we rearrange this equation to give $2^2 - 2 - 3^2 + 3 + 4^2 - 4$ . Factoring the common factor of $2$ out of $2^2 - 2$ , we get $2(2 - 1)$ ; factoring the common factor of $3$ out of $-3^2 + 3$ , we get $3(-3 + 1)$ ; finally, factoring the common factor of $4$ out of $4^2 - 4$ , we get $4(4 - 1)$ . Thus, $2^2 - 2 - 3^2 + 3 + 4^2 - 4 = 2(2-1)+3(-3+1)+4(4-1)$ , so by order of operations, the desired answer is thus
$2(1) + 3(-2) + 4(3) = \boxed{\textbf{(D) }8}. \qquad \blacksquare$

This post has been edited 4 times. Last edited by pog, Oct 19, 2021, 3:28 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

3363 posts

#9 h Oct 19, 2021, 2:11 AM • 1 Y

Y by babyfish

pog wrote:

Solved with literally nobody

Lemma 1.1: $x^2-y^2=(x+y)(x-y)$ .

Proof: Expand the right side using FOIL as mandated by the federal government, giving $(x)(x) + (x)(-y) + (x)(y) + (x)(-y) = x^2 - xy + yx - y^2$ . By the commutative property of multiplication, $xy = yx$ , so $a^2 - xy + yx - b^2 = a^2 - xy + xy - b^2$ .

Sublemma 1.2: $\ell - \ell = 0$ for any arbitrary value of $\ell$ .

Proof: Note that since $\ell + 0 = \ell$ by the identity property, subtracting $\ell$ from both sides gives $-\ell + \ell = 0$ , as desired. $\square$

By Sublemma 1.2, we get that $xy - xy = 0$ , and since $-xy + xy = xy - xy$ by the commutative property of addition, we get that $-xy + xy = 0$ . Thus, $a^2 - xy + xy - b^2 = a^2 - 0 - b^2$ , which by the commutative property of addition is equal to $a^2-b^2-0$ . Since $-0 = 0$ , we get $a^2 - b^2 + 0$ . Substitute $a^2-b^2=k$ . Then $k+0 =k$ by the identity property, so thus $a^2 - b^2 + 0 = k + 0 = k = a^2 - b^2$ .

Consequently, as desired, we have proved that $x^2-y^2=(x+y)(x-y)$ .

Looking at the problem, we wish to use Lemma 1.1. If we let $2 = (\sqrt2)^2$ , $3 = (\sqrt3)^2$ , and $4 = (\sqrt4)^2$ , then we can use Lemma 1.1 to give
$(2^2-2)-(3^2-3)+(4^2-4) = (2 - \sqrt2)(2 + \sqrt2) - (3+\sqrt3)(3 - \sqrt3) + (4+\sqrt4)(4 -\sqrt4).$ Expanding gives
$(2 - \sqrt2)(2 + \sqrt2) - (3+\sqrt3)(3 - \sqrt3) + (4+\sqrt4)(4 -\sqrt4) = [(2)(2) + (-\sqrt2)(\sqrt2) + (2)(\sqrt2) + (2)(-\sqrt2)] - [(3)(3) + (-\sqrt3)(\sqrt3) + (3)(\sqrt3) + (3)(-\sqrt3)] + [(4)(4) + (-\sqrt4)(\sqrt4) + (4)(\sqrt4) + (4)(-\sqrt4)].$ We can rearrange the first, second, third, and fourth terms in each set of brackets into their own sets of brackets individually by the commutative property of addition, so hence we get
$[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] + [(2\sqrt2)-(3\sqrt3)+(4\sqrt4)] + [(-2\sqrt2)-(-3\sqrt3)+(-4\sqrt4)].$ Let $a = (2\sqrt2)-(3\sqrt3)+(4\sqrt4)$ and $b = (-2\sqrt2)-(-3\sqrt3)+(-4\sqrt4)$ . Substitution gives that
$[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ]+ [(2\sqrt2)-(3\sqrt3)+(4\sqrt4)] + [(-2\sqrt2)-(-3\sqrt3)+(-4\sqrt4)]$ is equal to $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] + a + b.$ Cleverly, we note that $a = -b$ , so substituting $a = -b$ into $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] + a - b$ gives $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] -b + b.$ By Sublemma 1.2, we get that $-b-b = 0$ , so thus $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] -b + b = [2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ]+ 0.$ Let $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ]= n$ . Then $n + 0 = n$ by the identity property, so $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] + 0 = n + 0 = n = [2^2-3^2+4^2] + [ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ]$ .

We are now ready for the brutal answer extraction. Expanding the squares, we get that $2^2 - 3^2 + 4^2 + [ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] = 2^2 - 3^2 + 4^2 - (\sqrt2)(\sqrt2) + (\sqrt3)(\sqrt3) - (\sqrt4)(\sqrt4)$ . We have that $(\sqrt2)(\sqrt2) = 2$ , $(\sqrt3)(\sqrt3) = 3$ , and $(\sqrt4)(\sqrt4) = 4$ , so consequently
$2^2 - 3^2 + 4^2 - (\sqrt2)(\sqrt2) + (\sqrt3)(\sqrt3) - (\sqrt4)(\sqrt4) = 2^2 - 3^2 + 4^2 - 2 + 3 - 4.$
By the commutative property of addition, we rearrange this equation to give $2^2 - 2 - 3^2 + 3 + 4^2 - 4$ . Factoring the common factor of $2$ out of $2^2 - 2$ , we get $2(2 - 1)$ , factoring the common factor of $3$ out of $-3^2 + 3$ , we get $3(-3 + 1)$ , and factoring the common factor of $4$ out of $4^2 - 4$ , we get $4(4 - 1)$ . Thus, $2^2 - 2 - 3^2 + 3 + 4^2 - 4 = 2(2-1)+3(-3+1)+4(4-1)$ . By order of operations, the desired answer is thus
$2(1) + 3(-2) + 4(3) = \boxed{\textbf{(D) }8}. \qquad \blacksquare$

you cant even lift a finger to common core.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

3302 posts

#10 h Oct 19, 2021, 2:14 AM

Y by

pog wrote:

Solved with literally nobody

Lemma 1.1: $x^2-y^2=(x+y)(x-y)$ .

Proof: Expand the right side using FOIL as mandated by the federal government, giving $(x)(x) + (x)(-y) + (x)(y) + (x)(-y) = x^2 - xy + yx - y^2$ . By the commutative property of multiplication, $xy = yx$ , so $a^2 - xy + yx - b^2 = a^2 - xy + xy - b^2$ .

Sublemma 1.2: $\ell - \ell = 0$ for any arbitrary value of $\ell$ .

Proof: Note that since $\ell + 0 = \ell$ by the identity property, subtracting $\ell$ from both sides gives $-\ell + \ell = 0$ , as desired. $\square$

By Sublemma 1.2, we get that $xy - xy = 0$ , and since $-xy + xy = xy - xy$ by the commutative property of addition, we get that $-xy + xy = 0$ . Thus, $a^2 - xy + xy - b^2 = a^2 - 0 - b^2$ , which by the commutative property of addition is equal to $a^2-b^2-0$ . Since $-0 = 0$ , we get $a^2 - b^2 + 0$ . Substitute $a^2-b^2=k$ . Then $k+0 =k$ by the identity property, so thus $a^2 - b^2 + 0 = k + 0 = k = a^2 - b^2$ .

Consequently, as desired, we have proved that $x^2-y^2=(x+y)(x-y)$ .

Looking at the problem, we wish to use Lemma 1.1. If we let $2 = (\sqrt2)^2$ , $3 = (\sqrt3)^2$ , and $4 = (\sqrt4)^2$ , then we can use Lemma 1.1 to give
$(2^2-2)-(3^2-3)+(4^2-4) = (2 - \sqrt2)(2 + \sqrt2) - (3+\sqrt3)(3 - \sqrt3) + (4+\sqrt4)(4 -\sqrt4).$ Expanding gives
$(2 - \sqrt2)(2 + \sqrt2) - (3+\sqrt3)(3 - \sqrt3) + (4+\sqrt4)(4 -\sqrt4) = [(2)(2) + (-\sqrt2)(\sqrt2) + (2)(\sqrt2) + (2)(-\sqrt2)] - [(3)(3) + (-\sqrt3)(\sqrt3) + (3)(\sqrt3) + (3)(-\sqrt3)] + [(4)(4) + (-\sqrt4)(\sqrt4) + (4)(\sqrt4) + (4)(-\sqrt4)].$ We can rearrange the first, second, third, and fourth terms in each set of brackets into their own sets of brackets individually by the commutative property of addition, so hence we get
$[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] + [(2\sqrt2)-(3\sqrt3)+(4\sqrt4)] + [(-2\sqrt2)-(-3\sqrt3)+(-4\sqrt4)].$ Let $a = (2\sqrt2)-(3\sqrt3)+(4\sqrt4)$ and $b = (-2\sqrt2)-(-3\sqrt3)+(-4\sqrt4)$ . Substitution gives that
$[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ]+ [(2\sqrt2)-(3\sqrt3)+(4\sqrt4)] + [(-2\sqrt2)-(-3\sqrt3)+(-4\sqrt4)]$ is equal to $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] + a + b.$ Cleverly, we note that $a = -b$ , so substituting $a = -b$ into $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] + a - b$ gives $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] -b + b.$ By Sublemma 1.2, we get that $-b-b = 0$ , so thus $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] -b + b = [2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ]+ 0.$ Let $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ]= n$ . Then $n + 0 = n$ by the identity property, so $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] + 0 = n + 0 = n = [2^2-3^2+4^2] + [ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ]$ .

We are now ready for the brutal answer extraction. Expanding the squares, we get that $2^2 - 3^2 + 4^2 + [ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] = 2^2 - 3^2 + 4^2 - (\sqrt2)(\sqrt2) + (\sqrt3)(\sqrt3) - (\sqrt4)(\sqrt4)$ . We have that $(\sqrt2)(\sqrt2) = 2$ , $(\sqrt3)(\sqrt3) = 3$ , and $(\sqrt4)(\sqrt4) = 4$ , so consequently
$2^2 - 3^2 + 4^2 - (\sqrt2)(\sqrt2) + (\sqrt3)(\sqrt3) - (\sqrt4)(\sqrt4) = 2^2 - 3^2 + 4^2 - 2 + 3 - 4.$
By the commutative property of addition, we rearrange this equation to give $2^2 - 2 - 3^2 + 3 + 4^2 - 4$ . Factoring the common factor of $2$ out of $2^2 - 2$ , we get $2(2 - 1)$ , factoring the common factor of $3$ out of $-3^2 + 3$ , we get $3(-3 + 1)$ , and factoring the common factor of $4$ out of $4^2 - 4$ , we get $4(4 - 1)$ . Thus, $2^2 - 2 - 3^2 + 3 + 4^2 - 4 = 2(2-1)+3(-3+1)+4(4-1)$ . By order of operations, the desired answer is thus
$2(1) + 3(-2) + 4(3) = \boxed{\textbf{(D) }8}. \qquad \blacksquare$

I upvoted this not because it was a nice solution or was a troll, but because of the sheer amount of time and energy you had to have put into making this. Respect.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

4917 posts

pog

#11 h Oct 19, 2021, 2:14 AM • 1 Y

Y by ic3paw

Quote:

you cant even lift a finger to common core.

Sorry I couldn't Upload the images of the abacus (corroborated with number line technique) I used for the calculations

This post has been edited 2 times. Last edited by pog, Oct 19, 2021, 2:16 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

323 posts

#12 h Oct 19, 2021, 2:26 AM • 1 Y

Y by ThisUsernameIsTaken

@pog I think you didn't make it hard enough on yourself. Remember to include the proof for $1+1=2$ ; alternatively, you can leave this as an exercise to the reader.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

3363 posts

#13 h Oct 19, 2021, 2:34 AM

Y by

ChrisWren wrote:

@pog I think you didn't make it hard enough on yourself. Remember to include the proof for $1+1=2$ ; alternatively, you can leave this as an exercise to the reader.

include the 379 page proof

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

708 posts

#14 h Oct 19, 2021, 2:34 AM

Y by

Solution

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1002 posts

Afo

#15 h Oct 19, 2021, 2:36 AM

Y by

samrocksnature wrote:

$1.$ What is the value of $(2^2-2)-(3^2-3)+(4^2-4)?$
$\textbf{(A) }1 \qquad \textbf{(B) }2 \qquad \textbf{(C) }5 \qquad \textbf{(D) }8 \qquad \textbf{(E) }12$

Too hard for AMC.. Problems using obscure results should not be on the AMC. The key is the following lemma proved by Alfred North Whitehead and Bertrand Russell.

Principia Mathematica wrote:

$1 + 1 = 2$

Now note that
$(2^2-2)-(3^2-3)+(4^2-4)?$ $= 2 - 6 + 12$ $=2 + 6 ( 2 - 1) = 2 + 6 ( 1) = 1+1+(2 \cdot 3)(1) = 1+1+(1+1+1+1+1+1)(1) = 1+1+1+1+1+1+1+1=8$

This post has been edited 1 time. Last edited by Afo, Oct 19, 2021, 2:37 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

237 posts

#16 h Oct 19, 2021, 3:01 AM

Y by

Solution

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1443 posts

asimov

#17 h Oct 19, 2021, 3:34 AM

Y by

Imagine using difference of squares to solve this.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

558 posts

#18 h Oct 19, 2021, 3:41 AM

Y by

asimov wrote:

Imagine using difference of squares to solve this.

Yeah.

Just brute force it:

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

3084 posts

#19 h Oct 19, 2021, 1:19 PM

Y by

pog wrote:

Solved with literally nobody

Lemma 1.1: $x^2-y^2=(x+y)(x-y)$ .

Proof: Expand the right side using FOIL as mandated by the federal government, giving $(x)(x) + (x)(-y) + (x)(y) + (x)(-y) = x^2 - xy + yx - y^2$ . By the commutative property of multiplication, $xy = yx$ , so $a^2 - xy + yx - b^2 = a^2 - xy + xy - b^2$ .

Sublemma 1.2: $\ell - \ell = 0$ for any arbitrary value of $\ell$ .

Proof: Note that since $\ell + 0 = \ell$ by the identity property, subtracting $\ell$ from both sides gives $-\ell + \ell = 0$ , as desired. $\square$

By Sublemma 1.2, we get that $xy - xy = 0$ , and since $-xy + xy = xy - xy$ by the commutative property of addition, we get that $-xy + xy = 0$ . Thus, $a^2 - xy + xy - b^2 = a^2 - 0 - b^2$ , which by the commutative property of addition is equal to $a^2-b^2-0$ . Since $-0 = 0$ , we get $a^2 - b^2 + 0$ . Substitute $a^2-b^2=k$ . Then $k+0 =k$ by the identity property, so thus $a^2 - b^2 + 0 = k + 0 = k = a^2 - b^2$ .

Consequently, as desired, we have proved that $x^2-y^2=(x+y)(x-y)$ .

Looking at the problem, we wish to use Lemma 1.1. If we let $2 = (\sqrt2)^2$ , $3 = (\sqrt3)^2$ , and $4 = (\sqrt4)^2$ , then we can use Lemma 1.1 to give
$(2^2-2)-(3^2-3)+(4^2-4) = (2 - \sqrt2)(2 + \sqrt2) - (3+\sqrt3)(3 - \sqrt3) + (4+\sqrt4)(4 -\sqrt4).$ Expanding gives
$(2 - \sqrt2)(2 + \sqrt2) - (3+\sqrt3)(3 - \sqrt3) + (4+\sqrt4)(4 -\sqrt4) = [(2)(2) + (-\sqrt2)(\sqrt2) + (2)(\sqrt2) + (2)(-\sqrt2)] - [(3)(3) + (-\sqrt3)(\sqrt3) + (3)(\sqrt3) + (3)(-\sqrt3)] + [(4)(4) + (-\sqrt4)(\sqrt4) + (4)(\sqrt4) + (4)(-\sqrt4)].$ We can rearrange the first, second, third, and fourth terms in each set of brackets into their own sets of brackets individually by the commutative property of addition, so hence we get
$[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] + [(2\sqrt2)-(3\sqrt3)+(4\sqrt4)] + [(-2\sqrt2)-(-3\sqrt3)+(-4\sqrt4)].$ Let $a = (2\sqrt2)-(3\sqrt3)+(4\sqrt4)$ and $b = (-2\sqrt2)-(-3\sqrt3)+(-4\sqrt4)$ . Substitution gives that
$[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ]+ [(2\sqrt2)-(3\sqrt3)+(4\sqrt4)] + [(-2\sqrt2)-(-3\sqrt3)+(-4\sqrt4)]$ is equal to $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] + a + b.$ Cleverly, we note that $a = -b$ , so substituting $a = -b$ into $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] + a - b$ gives $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] -b + b.$ By Sublemma 1.2, we get that $-b+b = 0$ , so thus $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] -b + b = [2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ]+ 0.$ Let $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ]= n$ . Then $n + 0 = n$ by the identity property, so $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] + 0 = n + 0 = n = [2^2-3^2+4^2] + [ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ]$ .

We are now ready for the brutal answer extraction. Expanding the squares, we get that $2^2 - 3^2 + 4^2 + [ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] = 2^2 - 3^2 + 4^2 - (\sqrt2)(\sqrt2) + (\sqrt3)(\sqrt3) - (\sqrt4)(\sqrt4)$ . We have that $(\sqrt2)(\sqrt2) = 2$ , $(\sqrt3)(\sqrt3) = 3$ , and $(\sqrt4)(\sqrt4) = 4$ , so consequently
$2^2 - 3^2 + 4^2 - (\sqrt2)(\sqrt2) + (\sqrt3)(\sqrt3) - (\sqrt4)(\sqrt4) = 2^2 - 3^2 + 4^2 - 2 + 3 - 4.$
By the commutative property of addition, we rearrange this equation to give $2^2 - 2 - 3^2 + 3 + 4^2 - 4$ . Factoring the common factor of $2$ out of $2^2 - 2$ , we get $2(2 - 1)$ , factoring the common factor of $3$ out of $-3^2 + 3$ , we get $3(-3 + 1)$ , and factoring the common factor of $4$ out of $4^2 - 4$ , we get $4(4 - 1)$ . Thus, $2^2 - 2 - 3^2 + 3 + 4^2 - 4 = 2(2-1)+3(-3+1)+4(4-1)$ . By order of operations, the desired answer is thus
$2(1) + 3(-2) + 4(3) = \boxed{\textbf{(D) }8}. \qquad \blacksquare$

Exactly how I would've done it. Great solution. Very concise.

UPVOTED!

This post has been edited 1 time. Last edited by mahaler, Oct 19, 2021, 1:19 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

326 posts

#20 h Oct 19, 2021, 1:26 PM

Y by

i got this wrong.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

237 posts

#21 h Oct 19, 2021, 10:40 PM

Y by

TheAoPSLebron wrote:

i got this wrong.

oof

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

4917 posts

pog

#22 h Oct 19, 2021, 10:44 PM • 7 Y

Y by asimov, kante314, suvamkonar, megarnie, CyclicISLscelesTrapezoid, michaelwenquan, EpicBird08

TheAoPSLebron wrote:

i got this wrong.

I think my solution could have helped

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

3363 posts

#23 h Oct 20, 2021, 12:57 AM

Y by

pog wrote:

TheAoPSLebron wrote:

i got this wrong.

I think my solution could have helped

no common core would have been more helpful

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

808 posts

#24 h Sep 12, 2022, 8:53 PM

Y by

mahaler wrote:

pog wrote:

Solved with literally nobody

Lemma 1.1: $x^2-y^2=(x+y)(x-y)$ .

Proof: Expand the right side using FOIL as mandated by the federal government, giving $(x)(x) + (x)(-y) + (x)(y) + (x)(-y) = x^2 - xy + yx - y^2$ . By the commutative property of multiplication, $xy = yx$ , so $a^2 - xy + yx - b^2 = a^2 - xy + xy - b^2$ .

Sublemma 1.2: $\ell - \ell = 0$ for any arbitrary value of $\ell$ .

Proof: Note that since $\ell + 0 = \ell$ by the identity property, subtracting $\ell$ from both sides gives $-\ell + \ell = 0$ , as desired. $\square$

By Sublemma 1.2, we get that $xy - xy = 0$ , and since $-xy + xy = xy - xy$ by the commutative property of addition, we get that $-xy + xy = 0$ . Thus, $a^2 - xy + xy - b^2 = a^2 - 0 - b^2$ , which by the commutative property of addition is equal to $a^2-b^2-0$ . Since $-0 = 0$ , we get $a^2 - b^2 + 0$ . Substitute $a^2-b^2=k$ . Then $k+0 =k$ by the identity property, so thus $a^2 - b^2 + 0 = k + 0 = k = a^2 - b^2$ .

Consequently, as desired, we have proved that $x^2-y^2=(x+y)(x-y)$ .

Looking at the problem, we wish to use Lemma 1.1. If we let $2 = (\sqrt2)^2$ , $3 = (\sqrt3)^2$ , and $4 = (\sqrt4)^2$ , then we can use Lemma 1.1 to give
$(2^2-2)-(3^2-3)+(4^2-4) = (2 - \sqrt2)(2 + \sqrt2) - (3+\sqrt3)(3 - \sqrt3) + (4+\sqrt4)(4 -\sqrt4).$ Expanding gives
$(2 - \sqrt2)(2 + \sqrt2) - (3+\sqrt3)(3 - \sqrt3) + (4+\sqrt4)(4 -\sqrt4) = [(2)(2) + (-\sqrt2)(\sqrt2) + (2)(\sqrt2) + (2)(-\sqrt2)] - [(3)(3) + (-\sqrt3)(\sqrt3) + (3)(\sqrt3) + (3)(-\sqrt3)] + [(4)(4) + (-\sqrt4)(\sqrt4) + (4)(\sqrt4) + (4)(-\sqrt4)].$ We can rearrange the first, second, third, and fourth terms in each set of brackets into their own sets of brackets individually by the commutative property of addition, so hence we get
$[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] + [(2\sqrt2)-(3\sqrt3)+(4\sqrt4)] + [(-2\sqrt2)-(-3\sqrt3)+(-4\sqrt4)].$ Let $a = (2\sqrt2)-(3\sqrt3)+(4\sqrt4)$ and $b = (-2\sqrt2)-(-3\sqrt3)+(-4\sqrt4)$ . Substitution gives that
$[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ]+ [(2\sqrt2)-(3\sqrt3)+(4\sqrt4)] + [(-2\sqrt2)-(-3\sqrt3)+(-4\sqrt4)]$ is equal to $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] + a + b.$ Cleverly, we note that $a = -b$ , so substituting $a = -b$ into $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] + a - b$ gives $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] -b + b.$ By Sublemma 1.2, we get that $-b+b = 0$ , so thus $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] -b + b = [2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ]+ 0.$ Let $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ]= n$ . Then $n + 0 = n$ by the identity property, so $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] + 0 = n + 0 = n = [2^2-3^2+4^2] + [ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ]$ .

We are now ready for the brutal answer extraction. Expanding the squares, we get that $2^2 - 3^2 + 4^2 + [ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] = 2^2 - 3^2 + 4^2 - (\sqrt2)(\sqrt2) + (\sqrt3)(\sqrt3) - (\sqrt4)(\sqrt4)$ . We have that $(\sqrt2)(\sqrt2) = 2$ , $(\sqrt3)(\sqrt3) = 3$ , and $(\sqrt4)(\sqrt4) = 4$ , so consequently
$2^2 - 3^2 + 4^2 - (\sqrt2)(\sqrt2) + (\sqrt3)(\sqrt3) - (\sqrt4)(\sqrt4) = 2^2 - 3^2 + 4^2 - 2 + 3 - 4.$
By the commutative property of addition, we rearrange this equation to give $2^2 - 2 - 3^2 + 3 + 4^2 - 4$ . Factoring the common factor of $2$ out of $2^2 - 2$ , we get $2(2 - 1)$ , factoring the common factor of $3$ out of $-3^2 + 3$ , we get $3(-3 + 1)$ , and factoring the common factor of $4$ out of $4^2 - 4$ , we get $4(4 - 1)$ . Thus, $2^2 - 2 - 3^2 + 3 + 4^2 - 4 = 2(2-1)+3(-3+1)+4(4-1)$ . By order of operations, the desired answer is thus
$2(1) + 3(-2) + 4(3) = \boxed{\textbf{(D) }8}. \qquad \blacksquare$

Exactly how I would've done it. Great solution. Very concise.

UPVOTED!

I mean, we all do it subconsciously you know?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

6290 posts

#25 h Sep 12, 2022, 10:40 PM

Y by

volcanogobbler wrote:

It's pretty hard for a 1 though

Fax

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

qwerty123456asdfgzxcvb

1077 posts

qwerty123456asdfgzxcvb

#26 h Sep 13, 2022, 5:46 AM

Y by

pog wrote:

Solved with literally nobody

Lemma 1.1: $x^2-y^2=(x+y)(x-y)$ .

Proof: Expand the right side using FOIL as mandated by the federal government, giving $(x)(x) + (x)(-y) + (x)(y) + (x)(-y) = x^2 - xy + yx - y^2$ . By the commutative property of multiplication, $xy = yx$ , so $a^2 - xy + yx - b^2 = a^2 - xy + xy - b^2$ .

Sublemma 1.2: $\ell - \ell = 0$ for any arbitrary value of $\ell$ .

Proof: Note that since $\ell + 0 = \ell$ by the identity property, subtracting $\ell$ from both sides gives $-\ell + \ell = 0$ , as desired. $\square$

By Sublemma 1.2, we get that $xy - xy = 0$ , and since $-xy + xy = xy - xy$ by the commutative property of addition, we get that $-xy + xy = 0$ . Thus, $a^2 - xy + xy - b^2 = a^2 - 0 - b^2$ , which by the commutative property of addition is equal to $a^2-b^2-0$ . Since $-0 = 0$ , we get $a^2 - b^2 + 0$ . Substitute $a^2-b^2=k$ . Then $k+0 =k$ by the identity property, so thus $a^2 - b^2 + 0 = k + 0 = k = a^2 - b^2$ .

Consequently, as desired, we have proved that $x^2-y^2=(x+y)(x-y)$ .

Looking at the problem, we wish to use Lemma 1.1. If we let $2 = (\sqrt2)^2$ , $3 = (\sqrt3)^2$ , and $4 = (\sqrt4)^2$ , then we can use Lemma 1.1 to give
$(2^2-2)-(3^2-3)+(4^2-4) = (2 - \sqrt2)(2 + \sqrt2) - (3+\sqrt3)(3 - \sqrt3) + (4+\sqrt4)(4 -\sqrt4).$ Expanding gives
$(2 - \sqrt2)(2 + \sqrt2) - (3+\sqrt3)(3 - \sqrt3) + (4+\sqrt4)(4 -\sqrt4) = [(2)(2) + (-\sqrt2)(\sqrt2) + (2)(\sqrt2) + (2)(-\sqrt2)] - [(3)(3) + (-\sqrt3)(\sqrt3) + (3)(\sqrt3) + (3)(-\sqrt3)] + [(4)(4) + (-\sqrt4)(\sqrt4) + (4)(\sqrt4) + (4)(-\sqrt4)].$ We can rearrange the first, second, third, and fourth terms in each set of brackets into their own sets of brackets individually by the commutative property of addition, so hence we get
$[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] + [(2\sqrt2)-(3\sqrt3)+(4\sqrt4)] + [(-2\sqrt2)-(-3\sqrt3)+(-4\sqrt4)].$ Let $a = (2\sqrt2)-(3\sqrt3)+(4\sqrt4)$ and $b = (-2\sqrt2)-(-3\sqrt3)+(-4\sqrt4)$ . Substitution gives that
$[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ]+ [(2\sqrt2)-(3\sqrt3)+(4\sqrt4)] + [(-2\sqrt2)-(-3\sqrt3)+(-4\sqrt4)]$ is equal to $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] + a + b.$ Cleverly, we note that $a = -b$ , so substituting $a = -b$ into $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] + a - b$ gives $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] -b + b.$ By Sublemma 1.2, we get that $-b+b = 0$ , so thus $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] -b + b = [2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ]+ 0.$ Let $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ]= n$ . Then $n + 0 = n$ by the identity property, so $[2^2-3^2+4^2] +[ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] + 0 = n + 0 = n = [2^2-3^2+4^2] + [ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ]$ .

We are now ready for the brutal answer extraction. Expanding the squares, we get that $2^2 - 3^2 + 4^2 + [ -(\sqrt2)^2+(\sqrt3)^2-(\sqrt4)^2 ] = 2^2 - 3^2 + 4^2 - (\sqrt2)(\sqrt2) + (\sqrt3)(\sqrt3) - (\sqrt4)(\sqrt4)$ . We have that $(\sqrt2)(\sqrt2) = 2$ , $(\sqrt3)(\sqrt3) = 3$ , and $(\sqrt4)(\sqrt4) = 4$ , so consequently
$2^2 - 3^2 + 4^2 - (\sqrt2)(\sqrt2) + (\sqrt3)(\sqrt3) - (\sqrt4)(\sqrt4) = 2^2 - 3^2 + 4^2 - 2 + 3 - 4.$
By the commutative property of addition, we rearrange this equation to give $2^2 - 2 - 3^2 + 3 + 4^2 - 4$ . Factoring the common factor of $2$ out of $2^2 - 2$ , we get $2(2 - 1)$ ; factoring the common factor of $3$ out of $-3^2 + 3$ , we get $3(-3 + 1)$ ; finally, factoring the common factor of $4$ out of $4^2 - 4$ , we get $4(4 - 1)$ . Thus, $2^2 - 2 - 3^2 + 3 + 4^2 - 4 = 2(2-1)+3(-3+1)+4(4-1)$ , so by order of operations, the desired answer is thus
$2(1) + 3(-2) + 4(3) = \boxed{\textbf{(D) }8}. \qquad \blacksquare$

school math be like

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a