PRMO Sample Questions

Here are some Sample PRMO Questions, which Mahdi Raza gave me today on the Daily Challenges page.

This can also be accessed here

I was able to solve some of them, but I wish for the solutions for the problems $3^{\text{rd}},6^{\text{th}},10^{\text{th}}$ . I could not solve the $7^{\text{th}}$ one also, but he provided me a solution, so there is no need for it. Any help is appreciated! Thank you!

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

@Siddharth Chakravarty, @Mahdi Raza, @David Vreken, @Zakir Husain

Vinayak Srivastava - 10 months, 4 weeks ago

I have them before, I will try to give some solutions tomorrow if I will get time.

Zakir Husain - 10 months, 4 weeks ago

Also I have copied some of my questions from here.

Zakir Husain - 10 months, 4 weeks ago

Do you have more papers?

Vinayak Srivastava - 10 months, 4 weeks ago

@Vinayak Srivastava – Try these questions

Zakir Husain - 10 months, 4 weeks ago

@Vinayak Srivastava – Also try these

Zakir Husain - 10 months, 4 weeks ago

@Vinayak Srivastava – After completing these questions, if you need more then just tell me.

Zakir Husain - 10 months, 4 weeks ago

@Zakir Husain – Thank you! I will surely check these!

Vinayak Srivastava - 10 months, 4 weeks ago

Question $6_{th}$

Zakir Husain - 10 months, 4 weeks ago

Oh, I forgot this question! Thank you!

Vinayak Srivastava - 10 months, 4 weeks ago

Question $7_{th}$

Zakir Husain - 10 months, 4 weeks ago

Ok, then I will post the solutions for the 3rd and 10th as @Zakir Husain has already posted them.

Siddharth Chakravarty - 10 months, 4 weeks ago

@Siddharth Chakravarty – Thank you in advance!

Vinayak Srivastava - 10 months, 4 weeks ago

Problem 3's solution:

Let the polynomial P(x) = ${ a }_{ 0 }+{ a }_{ 1 }{ x }^{ 1 }+{ a }_{ 2 }{ x }^{ 2 }+...+{ a }_{ k }{ x }^{ k }$ where ${ a }_{ 0 },{ a }_{ 1 },{ a }_{ 2 },...,{ a }_{ k }{ x }^{ k }$ are the respective coefficients and $x$ is the variable.

So for P(n), x = n and thus P(n) = ${ a }_{ 0 }+{ a }_{ 1 }{ n }^{ 1 }+{ a }_{ 2 }{ n }^{ 2 }+...+{ a }_{ k }{ n }^{ k }$ . Now we can factor out $n$ as common from the part of the expression leaving ${ a }_{ 0 }$ as $n({ a }_{ 1 }+{ a }_{ 2 }{ n }^{ 1 }+...+{ a }_{ k }{ n }^{ k-1 })$ . Now obviously, this part is divisble by $n$ as $n$ is factored out.

But it is given that the whole expression P(n) is divisible by the positive integer $n$ , which means ${ a }_{ 0 }$ also has to be divisible by any positive integer $n$ for the whole expression to be divisible. Now, ${ a }_{ 0 }$ could either be 0 or 1x2x3x4x...n (n = infinity as we are considering all positive integers) But if it is 1x2x3x4...xn it will be a number which does not have any definite value, as it divergers and infinity is not a number. So ${ a }_{ 0 }$ has to be 0.

Hence, now P(x) = ${ a }_{ 1 }{ x }^{ 1 }+{ a }_{ 2 }{ x }^{ 2 }+...+{ a }_{ k }{ x }^{ k }$ as ${ a }_{ 0 }$ = 0 so adding it to the polynomial would have no effect. Now we need to find P(0) so subtiuting x = 0, all the terms would be reduced to 0 as 0 raise to any power(leaving 0^0 which is still debtable) is 0 and 0 multiplied by any number is 0. Thus P(0) = 0+0+0+..+0 = 0.

Siddharth Chakravarty - 10 months, 4 weeks ago

@Vinayak Srivastava I might post the solution by tomorrow for the 10th as It is night so I might go to sleep.

Siddharth Chakravarty - 10 months, 4 weeks ago

No problem! Do you have more papers of this type?

Vinayak Srivastava - 10 months, 4 weeks ago

Thank you for your solution! However, please write except 0 in the last line, as $0^0$ is not defined! :)

Vinayak Srivastava - 10 months, 4 weeks ago

0^0 is debatable, some say 1 or some say undefined. So I will write the same thing there, you can check the wiki on Brilliant on this.

Siddharth Chakravarty - 10 months, 4 weeks ago

@Siddharth Chakravarty – Yes, that is why I am asking you to add except 0 after "any number".

Vinayak Srivastava - 10 months, 4 weeks ago

@Siddharth Chakravarty – Oh, you have done it. Now it is good! :)

Vinayak Srivastava - 10 months, 4 weeks ago

If at all anyone finds more of such questions, not just PRMO but maybe even harder, then please share :)

Mahdi Raza - 10 months, 4 weeks ago

Check the Millennium Prize problems: Link

Siddharth Chakravarty - 10 months, 4 weeks ago

lol, really 😂

Mahdi Raza - 10 months, 4 weeks ago

Solve the differential equation for $f(x)$ (find $f(x)$ ):

$f''(x)\times f(x)=1$

Note:

$f''(x)$ is the second derivative of $f(x)$
Clue : $f'(x)=\sqrt{2\ln(f(x))}$

@Mahdi Raza , @Siddharth Chakravarty , @Aryan Sanghi

Zakir Husain - 10 months, 4 weeks ago

I worked it out @Zakir Husain I think it will be

if the constants after integrating are 0 as it suggests from your clue and erf is the Error function

Siddharth Chakravarty - 10 months, 4 weeks ago

@Siddharth Chakravarty – I can't understand.

Zakir Husain - 10 months, 4 weeks ago

@Zakir Husain – Check out now.

Siddharth Chakravarty - 10 months, 4 weeks ago

@Siddharth Chakravarty – $e^{\dfrac{2x^2\times erf^{-1}(\red{???})}{\pi}}$

Zakir Husain - 10 months, 4 weeks ago

@Zakir Husain – Oh sorry

It is $\sqrt { \frac { 2 }{ \pi } }$ there.

Siddharth Chakravarty - 10 months, 4 weeks ago

@Siddharth Chakravarty – Is this correct? I had previously forgotten the inputs and simplified falsely.

Siddharth Chakravarty - 10 months, 4 weeks ago

@Siddharth Chakravarty – The first one is very close, I've rechecked my calculations. $\large{e^{-({erf^{-1}(\sqrt{\frac{2}{\pi}}\times x)})^2}}$ , clue : multiply a constant

Zakir Husain - 10 months, 4 weeks ago

@Siddharth Chakravarty – is it $\large{e^{-({erf^{-1}(\sqrt{\frac{2}{\pi}}\times x)})^2}}$ or it is $\large{e^{-{erf^{-1}((\sqrt{{\frac{2}{\pi}}}\times x)^2)}}}$

Zakir Husain - 10 months, 4 weeks ago

@Zakir Husain – The first one where the input is squared, there is also i which I forgot to put because my notebook page has become a mess, so I wrote the answer in a very scribbled way.

Siddharth Chakravarty - 10 months, 4 weeks ago

@Siddharth Chakravarty – The first one is very close, I've rechecked my calculations. $\large{e^{-({erf^{-1}(\sqrt{\frac{2}{\pi}}\times x)})^2}}$ , clue : multiply a constant

Zakir Husain - 10 months, 4 weeks ago

@Zakir Husain – There is a -i (imaginary unit) in the input of the inverse of error function.

Siddharth Chakravarty - 10 months, 4 weeks ago

@Siddharth Chakravarty – Yes, but it will be more better if it will be $\large{i\times e^{-({erf^{-1}(\sqrt{\frac{2}{\pi}}\times x)})^2}}$

Zakir Husain - 10 months, 3 weeks ago

@Siddharth Chakravarty – Check the new problem

Zakir Husain - 10 months, 3 weeks ago

@Siddharth Chakravarty – If it's $\large{e^{-({erf^{-1}(\sqrt{\frac{2}{\pi}}\times x)})^2}}$ then that is not the accurate answer, but it is very close to it!

Zakir Husain - 10 months, 4 weeks ago

@Siddharth Chakravarty – Don’t wrap maths in text :)

Jeff Giff - 10 months, 4 weeks ago

@Jeff Giff – Actually I was doing something else, I was adding a text and separating it.

Siddharth Chakravarty - 10 months, 4 weeks ago

Q. Simplify the complex term below, and don't answer in terms of trigonometric functions:

$\frac{\sqrt[3]{1+\sqrt{3}i}}{\sqrt[3]{2}}+\frac{\sqrt[3]{1-\sqrt{3}i}}{\sqrt[3]{2}}$

Where $i$ is the imaginary unit

Zakir Husain - 10 months, 4 weeks ago

2, since this is $w,w^2$ , the solutions to $x^3=1$ .

Jeff Giff - 10 months, 4 weeks ago

@Jeff Giff – What $w,w^2$ and what $x^3$ , I don't understand what are you saying. (this is cube root)

Zakir Husain - 10 months, 4 weeks ago

@Zakir Husain – Oh! Whups, took it as cube :P

Jeff Giff - 10 months, 4 weeks ago

@Zakir Husain – Yikes, this is $-w$ :P

Jeff Giff - 10 months, 4 weeks ago

#6
Since you want the maximum side length, 12 should be the shortest side in the triangle.
Possible arrangements:12-16-20(3-4-5), 5-12-13(seriously?),

12-35-37(easily missed, but break $12^2$ to $71+73$ , then since 71 is the 36th odd number, $12^2+(36-1)^2=(36-1+2)^2$ .)

So the answer is 84.

Jeff Giff - 10 months, 4 weeks ago

How about this: I have a RadMaths note with loads of question set links. Check out the SAT set and the PRMO set. :)

Jeff Giff - 10 months, 4 weeks ago

I usually do solve the PRMO ones, when I can understand them, but I never solve the SAT ones, as I don't have to give it, and the questions are too tough for me.

Vinayak Srivastava - 10 months, 4 weeks ago

Oh. But I think there is one in a few that’s easy enough :) just check the note ;)

Jeff Giff - 10 months, 4 weeks ago

@Jeff Giff – @Jeff Giff what is your real age, 13?

Siddharth Chakravarty - 10 months, 4 weeks ago

@Siddharth Chakravarty – Yup, and I am dead serious, trust me.

Jeff Giff - 10 months, 4 weeks ago

#10 The ‘1’ and ‘2’ is derived from the fact that $3\overline{Pw} =\overline{P\Omega}$ , derived from the fact that $\triangle PwZ\sim \triangle P\Omega Y$ . Last bit was studying $\triangle wT\Omega$ .
@Vinayak Srivastava

Jeff Giff - 10 months, 4 weeks ago

@Vinayak Srivastava, try this:

$OEE$
$\times EE$
$—$
$EOEE$
$EOE$
$—$
$OOEE$
The multiplication above has been replaced with the parity(even or odd) of its digits, whereas O stands for odd, E stands for even. For example, 84 would become EE, since both 8 and 4 are even, and 1257 would become OEOO according to this pattern.

Jeff Giff - 10 months, 3 weeks ago

请更改符号。太多的细线和曲线使我感到不舒服。我有 OCD

Siddharth Chakravarty - 10 months, 3 weeks ago

oh. Nice translator!

Jeff Giff - 10 months, 3 weeks ago

@Vinayak Srivastava which grade are you?

SRIJAN Singh - 10 months, 2 weeks ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$