Putnam 2018 A1

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62861

3564 posts

62861

#1 h Dec 2, 2018, 10:50 PM • 2 Y

Y by Davi-8191, Adventure10

Find all ordered pairs $(a, b)$ of positive integers for which
$\frac{1}{a} + \frac{1}{b} = \frac{3}{2018}.$

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greenturtle3141

3544 posts

greenturtle3141

#2 h Dec 2, 2018, 11:25 PM • 4 Y

Y by Demin, centslordm, Adventure10, Mango247

Solution

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jeff10

1117 posts

jeff10

#3 h Dec 2, 2018, 11:30 PM • 3 Y

Y by Binomial-theorem, Adventure10, Mango247

greenturtle3141 wrote:

(Is this the easiest Putnam yet?)

I was going to make a post about how we should keep in mind that AoPSers learn Blank earlier than people who don't use AoPS and that it could have been harder for others. But then I went back to look at some previous years' problems... And I can't help but agree.

This post has been edited 1 time. Last edited by jeff10, Dec 2, 2018, 11:31 PM

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62861

3564 posts

62861

#4 h Dec 2, 2018, 11:49 PM • 5 Y

Y by Madhavi, Adventure10, Mango247, Mango247, Mango247

Solution

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acegikmoqsuwy2000

767 posts

acegikmoqsuwy2000

#5 h Dec 2, 2018, 11:50 PM • 3 Y

Y by removablesingularity, Adventure10, Mango247

I forgot that $1$ is a factor of $2018^2$ ... We'll see whether this merits a zero or if it'll be a positive score

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Stormersyle

2785 posts

Stormersyle

#7 h Dec 3, 2018, 3:19 AM • 2 Y

Y by Binomial-theorem, Adventure10

wait a second how is this a putnam even i can solve this

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Anzoteh

126 posts

Anzoteh

#8 h Dec 3, 2018, 3:30 AM • 3 Y

Y by Adventure10, Mango247, Mango247

Small aside: how do we justify that 1009 is a prime in Putnam? I just did 1009/p has this remainder for p= prime between 2 and 31, inclusive

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Stormersyle

2785 posts

Stormersyle

#9 h Dec 3, 2018, 3:31 AM • 5 Y

Y by Anzoteh, Binomial-theorem, Cpi2728, Adventure10, Mango247

Anzoteh wrote:

Small aside: how do we justify that 1009 is a prime in Putnam? I just did 1009/p has this remainder for p= prime between 2 and 31, inclusive

i'm pretty sure you can just assert that 1009 is a prime because that's a well-known fact

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TomCalc

1635 posts

TomCalc

#10 h Dec 3, 2018, 4:37 AM • 2 Y

Y by Adventure10, Mango247

$1009$ is an odd prime of the form $4k+1$ .Then...

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whatRthose

1792 posts

whatRthose

#11 h Dec 3, 2018, 12:18 PM • 2 Y

Y by Adventure10, Mango247

If you want to check if an integer $k$ is prime, check if it has any factors less than or equal to $\sqrt{k}$ , therefore your justification for why 1009 is prime is correct, since $\sqrt{1009} \approx 31$

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Mathlete2017

3233 posts

Mathlete2017

#12 h Dec 3, 2018, 3:12 PM • 2 Y

Y by Adventure10, Mango247

We can use SFFT to write the equation as $(3a-2018)(3b-2018) = 2018^2.$ Finding the pairs $(a,b)$ is relatively simple from here. (How is this a Putnam problem? I'm a middle schooler.)

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scrabbler94

7551 posts

scrabbler94

#13 h Dec 3, 2018, 3:39 PM • 2 Y

Y by integrated_JRC, Adventure10

Mathlete2017 wrote:

(How is this a Putnam problem? I'm a middle schooler.)

I'm actually not sure - A1 is usually almost a "freebie" in the sense that not much clever insight is needed (in terms of content, some A1's do require calculus) but this one seems really bland and easy.

You still have to write your solution very carefully and rigorously, as with all Putnam problems.

This post has been edited 1 time. Last edited by scrabbler94, Dec 3, 2018, 3:40 PM

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Mathlete2017

3233 posts

Mathlete2017

#14 h Dec 3, 2018, 4:18 PM • 2 Y

Y by Adventure10, Mango247

whatRthose wrote:

If you want to check if an integer $k$ is prime, check if it has any factors less than or equal to $\sqrt{k}$ , therefore your justification for why 1009 is prime is correct, since $\sqrt{1009} \approx 31$

Rather than checking if it has any factors less than $\sqrt{k},$ you could just check if it has any factors that are prime less than $\sqrt{k}.$

This post has been edited 1 time. Last edited by Mathlete2017, Dec 3, 2018, 4:19 PM

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hydrohelium

245 posts

hydrohelium

#15 h Feb 4, 2019, 4:22 AM • 3 Y

Y by denery, Adventure10, Mango247

Finally solved a Putnam Problem

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Aniv

1141 posts

Aniv

#16 h Feb 4, 2019, 3:16 PM • 3 Y

Y by Adventure10, Mango247, Mango247

we already have $(3a-2018)(3b-2018)=4*1009^2$

now if $a,b>\frac{4036}{3}$ then $\frac{1}{a}+\frac{1}{b}<\frac{3}{2018}$

again if $a,b<\frac{4036}{3}$ then $\frac{1}{a}+\frac{1}{b}>\frac{3}{2018}$

so as we know that $a\neq b$ so WLOG let $a<b$ then we have $a<\frac{4036}{3}=1345+\frac{1}{3}$
so $a\le 1345$ or $3a-2018\le 2017$
similarly $b>\frac{4036}{3}=1346-\frac{2}{3}$ which implies $b\geq 1346$ or $3b-2018=2020$

so $3a-2018$ can only be equal to 1,2,4,1009 and hence $a$ can be found and similarly b can also be deducted

This post has been edited 2 times. Last edited by Aniv, Feb 4, 2019, 3:17 PM

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AopsUser101

1750 posts

AopsUser101

#17 h Apr 6, 2020, 4:03 AM • 1 Y

Y by v4913

$\frac{a+b}{ab}=\frac{3}{2018} \implies 3ab = 2018a + 2018b \implies 3\left(a-\frac{2018}{3} \right)\left(b-\frac{2018}{3}\right) = \frac{2018^2}{3} \implies (3a-2018)(3b-2018) = 2018^2$ Let $3a - 2018 = x$ and $3b-2018 = y$ . We want $xy = 2018^2$ . Assume WLOG that $|x| \le |y|$ . Then, we have following solutions for $(x,y)$ :
$\left(1,2018^2 \right), \left(2, \frac{2018^2}{2}\right), \left(1009,\frac{2018^2}{1009} \right), \left(2018,2018 \right),\left(-1,-2018^2 \right), \left(-2, -\frac{2018^2}{2}\right), \left(-1009,-\frac{2018^2}{1009} \right), \left(-2018,-2018 \right)$ However, note that $x \mod 3 \equiv 1$ and $y \mod 3 \equiv 1$ and $a > 0, b > 0$ . Therefore, we can narrow them down some more:
$\left(1,2018^2 \right), \left(1009,\frac{2018^2}{1009} \right), \left(-2, -\frac{2018^2}{2}\right)$ After some computation, we get that the only results for $(a,b)$ are:
$(673, 1358114),(674, 340033),(1009, 2018)$ and symmetric counterparts.

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dikhendzab

108 posts

dikhendzab

#18 h Apr 12, 2020, 3:17 PM

Y by

Let some $d$ be the greatest common divisor of $m$ and $n$ , so $a=dx, b=dy$ . Equations turns to be:
$\frac{1}{dx}+\frac{1}{dy}=\frac{3}{2018}$ or $3dxy=2018(x+y)$ . Since $2018$ is not divisible by $3$ , $x+y$ has to be divisible by $3$ , so:
$(x,y)=(1,2);(2,1009);(1,2018)$ . We will now find $d=\frac{2018(x+y)}{3xy}$ and $d=(1009);(337);(673)$ . Finally, we have these pairs:
$(a,b)=(1009,2018);(674,340033);(673,1358114)$

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OlympusHero

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OlympusHero

#19 h Apr 14, 2021, 6:37 PM

Y by

First Putnam solve!

Solution

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BrainiacVR

225 posts

BrainiacVR

#20 h Dec 31, 2021, 7:11 AM

Y by

OMG !! Did I just solve a Putnam problem ?!! :oops:

Easy Soln To Easy Putnam :P :first:

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pog

4917 posts

pog

#21 h Jan 3, 2022, 9:22 AM

Y by

Obviously, first Putnam solve :roll:

Solution

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megarnie

5579 posts

megarnie

#22 h Apr 3, 2022, 1:12 PM

Y by

We have $\frac{a+b}{ab}=\frac{3}{2018}$ , so $2018a+2018b=3ab\implies 3ab-2018a-2018b=0$
Now, $9ab-6054a-6054b=0\implies (3a-2018)(3b-2018)=2018^2=4072324$
Note that both $3a-2018$ and $3b-2018$ are $1\pmod 3$ . So they can be any $1\pmod 3$ factors that multiply to $2018^2$ .

If we WLOG $a\le b$ , the possible ways are $\begin{align*} 3a-2018=1, 3b-2018=4072324 \\ 3a-2018=4, 3b-2018=1018081 \\ 3a-2018=1009, 3b-2018=2018 \\ \end{align*}$
These give the solutions $\boxed{(a,b)=(673,1358114), (674,340033), (1009,2018), \text{ and permutations}}$

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ZETA_in_olympiad

2211 posts

ZETA_in_olympiad

#23 h Apr 5, 2022, 9:24 AM

Y by

CantonMathGuy wrote:

Find all ordered pairs $(a, b)$ of positive integers for which
$\frac{1}{a} + \frac{1}{b} = \frac{3}{2018}.$

The equation is equivalent to $(3a-2018)(3b-2018)=2018^2,$ where all factors equiv to $1$ in mod $3$ . But there are six factors of $2018$ equiv to $1$ in mod $3: 1, 2^2, 1009, 1009^2, 2^2\cdot 1009, 2^2\cdot 1009^2.$

So our required ordered pairs: $\boxed{(a,b)=(673, 1358114), (674, 340033), (1009, 2018), (2018, 1009), (340033, 674), (1358114, 674)}$ .

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RedFireTruck

4220 posts

RedFireTruck

#24 h Aug 6, 2022, 1:22 AM

Y by

We can multiply by $6054ab$ to get $6054a+6054b=9ab$ . This factors as $(3a-2018)(3b-2018)=2018^2=2^2\cdot1009^2$ . Clearly, $3a-2018\equiv 3b-2018\equiv 1\pmod{3}$ . This gives us that $(3a-2018, 3b-2018)=(1,2018^2), (2^2, 1009^2), (1009, 2^2\cdot1009), (2018^2,1), (1009^2,2^2),(1009,2^2\cdot1009)$ . Each of these gives us the pairs $(a,b)=(673, 1358114), (674, 340033), (1009, 2018), (1358114,673),(340033,674),(2018,1009)$ .

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mathstudent5

374 posts

mathstudent5

#25 h Jul 24, 2023, 8:05 PM

Y by

Is this a Putnam problem? Can't believe.

How easy is this !!

Just play with Simon's favourite factoring trick.

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KHOMNYO2

97 posts

KHOMNYO2

#26 h Jan 20, 2024, 2:26 PM

Y by

pretty common in hs olympiad

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pie854

243 posts

pie854

#27 h Apr 15, 2024, 5:02 PM

Y by

Note that $(3b-2018)a=2018b$ . Write $3b-2018=t$ where $t\equiv 1\pmod 3$ and $t>0$ . So $2018\cdot \frac{t+2018}3\equiv 0\pmod t$ thus $2018^2\equiv 0\pmod t$ . So $t$ is any divisor of $2018^2$ that is $1\pmod 3$ . Now we can check and find $(a,b)$ . They are $(a,b)=(1358114,673),(340033,674),(2018,1009)$ and symmetric pairs.