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 3rd,6th,10th3^{\text{rd}},6^{\text{th}},10^{\text{th}}. I could not solve the 7th7^{\text{th}} one also, but he provided me a solution, so there is no need for it. Any help is appreciated! Thank you!

Note by Vinayak Srivastava
4 months, 2 weeks ago

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

Vinayak Srivastava - 4 months, 2 weeks ago

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

Zakir Husain - 4 months, 2 weeks ago

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Also I have copied some of my questions from here.

Zakir Husain - 4 months, 2 weeks ago

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Do you have more papers?

Vinayak Srivastava - 4 months, 2 weeks ago

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

Zakir Husain - 4 months, 2 weeks ago

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

Vinayak Srivastava - 4 months, 2 weeks ago

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Question 6th6_{th}

Zakir Husain - 4 months, 2 weeks ago

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Oh, I forgot this question! Thank you!

Vinayak Srivastava - 4 months, 2 weeks ago

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Question 7th7_{th}

Zakir Husain - 4 months, 2 weeks ago

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

Siddharth Chakravarty - 4 months, 2 weeks ago

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@Siddharth Chakravarty Thank you in advance!

Vinayak Srivastava - 4 months, 2 weeks ago

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Problem 3's solution:

Let the polynomial P(x) = a0+a1x1+a2x2+...+akxk{ a }_{ 0 }+{ a }_{ 1 }{ x }^{ 1 }+{ a }_{ 2 }{ x }^{ 2 }+...+{ a }_{ k }{ x }^{ k } where a0,a1,a2,...,akxk{ a }_{ 0 },{ a }_{ 1 },{ a }_{ 2 },...,{ a }_{ k }{ x }^{ k } are the respective coefficients and xx is the variable.

So for P(n), x = n and thus P(n) = a0+a1n1+a2n2+...+aknk{ a }_{ 0 }+{ a }_{ 1 }{ n }^{ 1 }+{ a }_{ 2 }{ n }^{ 2 }+...+{ a }_{ k }{ n }^{ k }. Now we can factor out nn as common from the part of the expression leaving a0{ a }_{ 0 } as n(a1+a2n1+...+aknk1)n({ a }_{ 1 }+{ a }_{ 2 }{ n }^{ 1 }+...+{ a }_{ k }{ n }^{ k-1 }). Now obviously, this part is divisble by nn as nn is factored out.

But it is given that the whole expression P(n) is divisible by the positive integer nn, which means a0{ a }_{ 0 } also has to be divisible by any positive integer nn for the whole expression to be divisible. Now, a0{ 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 a0{ a }_{ 0 } has to be 0.

Hence, now P(x) = a1x1+a2x2+...+akxk{ a }_{ 1 }{ x }^{ 1 }+{ a }_{ 2 }{ x }^{ 2 }+...+{ a }_{ k }{ x }^{ k } as a0{ 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 - 4 months, 2 weeks ago

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

Siddharth Chakravarty - 4 months, 2 weeks ago

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

Vinayak Srivastava - 4 months, 2 weeks ago

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Thank you for your solution! However, please write except 0 in the last line, as 000^0 is not defined! :)

Vinayak Srivastava - 4 months, 2 weeks ago

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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 - 4 months, 2 weeks ago

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

Vinayak Srivastava - 4 months, 2 weeks ago

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

Vinayak Srivastava - 4 months, 2 weeks ago

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

Mahdi Raza - 4 months, 2 weeks ago

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Check the Millennium Prize problems: Link

Siddharth Chakravarty - 4 months, 2 weeks ago

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lol, really 😂

Mahdi Raza - 4 months, 2 weeks ago

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

f(x)×f(x)=1f''(x)\times f(x)=1

Note:

  • f(x)f''(x) is the second derivative of f(x)f(x)

  • Clue : f(x)=2ln(f(x))f'(x)=\sqrt{2\ln(f(x))}

@Mahdi Raza , @Siddharth Chakravarty , @Aryan Sanghi

Zakir Husain - 4 months, 2 weeks ago

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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 - 4 months, 2 weeks ago

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@Siddharth Chakravarty I can't understand.

Zakir Husain - 4 months, 2 weeks ago

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@Zakir Husain Check out now.

Siddharth Chakravarty - 4 months, 2 weeks ago

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@Siddharth Chakravarty e2x2×erf1(???)πe^{\dfrac{2x^2\times erf^{-1}(\red{???})}{\pi}}

Zakir Husain - 4 months, 2 weeks ago

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@Zakir Husain Oh sorry

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

Siddharth Chakravarty - 4 months, 2 weeks ago

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

Siddharth Chakravarty - 4 months, 2 weeks ago

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@Siddharth Chakravarty The first one is very close, I've rechecked my calculations. e(erf1(2π×x))2\large{e^{-({erf^{-1}(\sqrt{\frac{2}{\pi}}\times x)})^2}}, clue : multiply a constant

Zakir Husain - 4 months, 2 weeks ago

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@Siddharth Chakravarty is it e(erf1(2π×x))2\large{e^{-({erf^{-1}(\sqrt{\frac{2}{\pi}}\times x)})^2}} or it is eerf1((2π×x)2)\large{e^{-{erf^{-1}((\sqrt{{\frac{2}{\pi}}}\times x)^2)}}}

Zakir Husain - 4 months, 2 weeks ago

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@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 - 4 months, 2 weeks ago

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@Siddharth Chakravarty The first one is very close, I've rechecked my calculations. e(erf1(2π×x))2\large{e^{-({erf^{-1}(\sqrt{\frac{2}{\pi}}\times x)})^2}}, clue : multiply a constant

Zakir Husain - 4 months, 2 weeks ago

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

Siddharth Chakravarty - 4 months, 2 weeks ago

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@Siddharth Chakravarty Yes, but it will be more better if it will be i×e(erf1(2π×x))2\large{i\times e^{-({erf^{-1}(\sqrt{\frac{2}{\pi}}\times x)})^2}}

Zakir Husain - 4 months, 2 weeks ago

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@Siddharth Chakravarty Check the new problem

Zakir Husain - 4 months, 2 weeks ago

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@Siddharth Chakravarty If it's e(erf1(2π×x))2\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 - 4 months, 2 weeks ago

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

Jeff Giff - 4 months, 2 weeks ago

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

Siddharth Chakravarty - 4 months, 2 weeks ago

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

1+3i323+13i323\frac{\sqrt[3]{1+\sqrt{3}i}}{\sqrt[3]{2}}+\frac{\sqrt[3]{1-\sqrt{3}i}}{\sqrt[3]{2}}

Where ii is the imaginary unit

Zakir Husain - 4 months, 2 weeks ago

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2, since this is w,w2w,w^2, the solutions to x3=1x^3=1.

Jeff Giff - 4 months, 2 weeks ago

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@Jeff Giff What w,w2w,w^2 and what x3x^3, I don't understand what are you saying. (this is cube root)

Zakir Husain - 4 months, 2 weeks ago

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

Jeff Giff - 4 months, 2 weeks ago

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

Jeff Giff - 4 months, 2 weeks ago

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#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 12212^2 to 71+7371+73, then since 71 is the 36th odd number, 122+(361)2=(361+2)212^2+(36-1)^2=(36-1+2)^2.)

So the answer is 84.

Jeff Giff - 4 months, 2 weeks ago

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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 - 4 months, 2 weeks ago

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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 - 4 months, 2 weeks ago

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

Jeff Giff - 4 months, 2 weeks ago

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

Siddharth Chakravarty - 4 months, 2 weeks ago

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

Jeff Giff - 4 months, 2 weeks ago

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#10 The ‘1’ and ‘2’ is derived from the fact that 3Pw=PΩ3\overline{Pw} =\overline{P\Omega}, derived from the fact that PwZPΩY\triangle PwZ\sim \triangle P\Omega Y. Last bit was studying wTΩ\triangle wT\Omega.
@Vinayak Srivastava

Jeff Giff - 4 months, 2 weeks ago

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@Vinayak Srivastava, try this:

OEEOEE
×EE\times EE

EOEEEOEE
EOEEOE

OOEEOOEE
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 - 4 months, 2 weeks ago

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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 - 4 months, 2 weeks ago

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oh. Nice translator!

Jeff Giff - 4 months, 2 weeks ago

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@Vinayak Srivastava which grade are you?

SRIJAN Singh - 4 months ago

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