AMC 10A 2002 Problem 15 | Prime Number

Try this beautiful Problem based on Number theory from AMC 10A, 2002 Problem 15.

Prime Number | AMC 10A 2021, Problem 15

Using the digits $1,2,3,4,5,6,7$, and 9 , form 4 two-digit prime numbers, using each digit only once. What is the sum of the 4 prime numbers?

Key Concepts

Arithmetic

Divisibility

Prime Number

Suggested Book | Source | Answer

Elementary Number Theory by David M. Burton.

AMC 10A 2002 Problem 15

190

Try with Hints

First try to find the probable digits for the unit place of the prime number.

The two digit prime number should end with $1, 3, 7, 9$ since it is prime and should not divisible by $2$ or $5$.

So now try to find which two digit primes will work here.

So, the primes should be $23, 41, 59, 67$.

Now find the sum of them.

AMC 8 2019 Problem 20 | Fundamental Theorem of Algebra

Try out this beautiful algebra problem number 2 from AMC 8 2019 based on the Fundamental Theorem of Algebra.

AMC 8 2019 Problem 20:

How many different real numbers $x$ satisfy the equation $(x^{2}-5)^{2}=16?$

$\textbf{(A) }0$
$\textbf{(B) }1$
$\textbf{(C) }2$
$\textbf{(D) }4$
$\textbf{(E) }8$

Key Concepts

Algebra

Value

Telescoping

Check the Answer

Answer: is (D) 4

AMC 8, 2019, Problem 20

Try with Hints

The given equation is

$(x^2-5)^2 = 16$

and that means

$x^2-5 = \pm 4$

Among both cases, if

$x^2-5 = 4$

then,

$x^2 = 9 \implies x = \pm 3$

and that means we have 2 different real numbers that satisfy the equation.

and if we take another case, then

$x^2-5 = -4$

and so,

$x^2 = 1 \implies x = \pm 1$

and that means we have 2 different real numbers in this `case too that satisfy the equation. So total 2+2=4 real numbers that satisfy the equation.

AMC 8 2019 Problem 16 | Algebra Problem

Try this beautiful Number Theory problem from the AMC 2019 Problem 16. You may use sequential hints to solve the problem.

Algebra Question - AMC 8, 2019 Problem 16

Qiang drives 15 miles at an average speed of 30 miles per hour. How many additional miles will he have to drive at 55 miles per hour to average 50 miles per hour for the entire trip?
(A) 45
(B) 62
(C) 90
(D) 110
(E) 135

Key Concepts

Algebra

Value

Average Speed

Check the Answer

Answer: is (D) 110

AMC 8, 2019, Problem 16

Try with Hints

Among the options, there is only one option which is divisible by 55 and that is 110.

That option tells the travel hour is 2.

Qiang drives 15 miles at an average speed of 30 miles per hour.

And we know, Average speed = Total Distance/Total Time

So, by the formula,

$\frac{15}{30} + \frac{110}{55} = \frac{5}{2}$

In this case, If we consider the whole journey, Total Distance is (110+15)=125

And as Qiang has to drive at 50 miles per hour for the entire trip, and as Average speed = Total Distance/Total Time ,

$\frac{125}{50} = \frac{5}{2}$

As both are same , our answer 110 is established.

AMC 8 2019 Problem 17 | Value of Product

Try out this beautiful algebra problem from AMC 8, 2019 based on finding the value of the product. You may use sequential hints to solve the problem.

AMC 8 2019: Problem 17

What is the value of the product

$\left(\frac{1 \cdot 3}{2 \cdot 2}\right)\left(\frac{2 \cdot 4}{3 \cdot 3}\right)\left(\frac{3 \cdot 5}{4 \cdot 4}\right) \cdots\left(\frac{97 \cdot 99}{98 \cdot 98}\right)\left(\frac{98 \cdot 100}{99 \cdot 99}\right) ?$

(A) $\frac{1}{2}$

(B) $\frac{50}{99}$

(D) $\frac{100}{99}$

(E) $50$

Key Concepts

Algebra

Value

Telescoping

Check the Answer

Answer: is $\frac{50}{99}$

AMC 8, 2019, Problem 17

Try with Hints

We write

$\left(\frac{1.3}{2.2}\right)\left(\frac{2.4}{3.3}\right)\left(\frac{3.5}{4.4}\right) \ldots\left(\frac{97.99}{98.98}\right)\left(\frac{98.100}{99.99}\right)$

in a different form like

$\frac{1}{2} \cdot\left(\frac{3.2}{2.3}\right) \cdot\left(\frac{4.3}{3.4}\right) \cdots \cdots \left(\frac{99.98}{98.99}\right) \cdot \frac{100}{99}$

All of the middle terms eliminate each other, and only the first and last term remains i.e.

$\frac{1}{2} \cdot \frac{100}{99}$

$\frac{1}{2} \cdot \frac{100}{99}=\frac{50}{99}$

and that is the final answer.

Digits and Rationals | AIME I, 1992 | Question 5

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on Digits and Rationals.

Digits and Rationals - AIME I, 1992

Let S be the set of all rational numbers r, 0<r<1, that have a repeating decimal expression in the form 0.abcabcabcabc.... where the digits a,b and c are not necessarily distinct. To write the elements of S as fractions in lowest terms find number of different numerators required.

is 107
is 660
is 840
cannot be determined from the given information

Key Concepts

Integers

Digits

Prime

Check the Answer

Answer: is 660.

AIME I, 1992, Question 5

Elementary Number Theory by David Burton

Try with Hints

Let x=0.abcabcabcabc.....

$\Rightarrow 1000x=abc.\overline{abc}$

$\Rightarrow 999x=1000x-x=abc$

$\Rightarrow x=\frac{abc}{999}$

numbers relatively prime to 999 gives us the numerators

$\Rightarrow 999(1-\frac{1}{3})(1-\frac{1}{111})$=660

=660.

Smallest Positive Integer | PRMO 2019 | Question 14

Try this beautiful problem from the PRMO, 2019 based on Smallest Positive Integer.

Smallest Positive Integer - PRMO 2019

Find the smallest positive integer n$\geq$10 such that n+6 is a prime and 9n+7 is a perfect square.

is 107
is 53
is 840
cannot be determined from the given information

Key Concepts

Integers

Primes

Perfect Square

Check the Answer

Answer: is 53.

PRMO, 2019, Question 14

Elementary Number Theory by David Burton

Try with Hints

Let 9n+7=$m^{2}$ n+6 prime then n+6 odd then n is odd then n=2k+1 then 9(2k+1)+7=$m^{2}$ then 18k=$m^{2}$-16=(m+4)(m-4) then 18k even m is even then m=2p

18k=(2p+4)(2p-4)=4(p+2)(p-2) then 9k=2(p+2)(p-2)then k even then k=2d then 18d=2(p+2)(p-2) then 9d=(p+2)(p-2) then p of form 9q+2,9q-2

for p=9q-2 then m=2(9q-2) for q=1 then$m^{2}$=196then n=21 then n+6=27 non prime, for p=9q+2 then m=2(9q+2) for q=1 $m^{2}$=484 then n=53 then n+6=59 prime then n=53.

Problem based on divisibility - CMI 2015 -problem 3

Problem based on divisibility

The problem is based upon the divisibility and prime factorization of a numbers. Also we have to deal with the number divisible by either one prime number or more than one prime numbers.

Try the problem

A positive integer n is called a magic number if it has the following property: if a and b
are two positive numbers that are not coprime to n then a + b is also not coprime to n.
For example, 2 is a magic number, because sum of any two even numbers is also even.
Which of the following are magic numbers? Write your answers as a sequence of four
letters (Y for Yes and N for No) in correct order.
(i) 129 (ii) 128 (iii) 127 (iv) 100.

I.S.I. Entrance 2015 for B. sc. program at CMI Sub problem 3

Divisibility and Prime factorisation

6 out of 10

Secrets in Inequalities.

Knowledge Graph

Use some hints

Take the LCM, and point out all the numbers that divides the given number, now select any two of them or any two of the prime factors out of calculated ones.

Now we can say those two prime factors a and b, and then we can easily calculate a+b. now check weather a+b and the number itself has any co prime factor or not other than 1.

Number Theory Problem | AMC 10B 2019| Problem 19

[et_pb_section fb_built="1" _builder_version="4.0"][et_pb_row _builder_version="4.2.2" width="100%"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_text _builder_version="4.2.2" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="10px|10px|10px|10px|false|false" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

What are we learning ?

Competency in Focus: Number Theory

This problem from American Mathematics Contest 10B (AMC 10B, 2019) is based on calculation of number theory. It is Question no. 19 of the AMC 10B 2019 Problem series.

[/et_pb_text][et_pb_text _builder_version="4.2.2" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="10px||10px||false|false" custom_padding="10px|10px|10px|10px|false|false" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

First look at the knowledge graph:-

[/et_pb_text][et_pb_image src="https://cheenta.com/wp-content/uploads/2020/02/p19.png" alt="calculation of mean and median- AMC 8 2013 Problem" title_text=" mean and median- AMC 8 2013 Problem" align="center" force_fullwidth="on" _builder_version="4.2.2" min_height="429px" height="189px" max_height="198px" custom_padding="10px|10px|10px|10px|false|false"][/et_pb_image][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Next understand the problem

[/et_pb_text][et_pb_text _builder_version="4.2.2" text_font="Raleway||||||||" text_font_size="20px" text_letter_spacing="1px" text_line_height="1.5em" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]Let $S$ be the set of all positive integer divisors of $100,000.$ How many numbers are the product of two distinct elements of $S?$ $\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121$[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="4.0"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.2.2" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" open="off" _builder_version="4.2.2"]American Mathematical Contest 2019, AMC 10B Problem 19[/et_pb_accordion_item][et_pb_accordion_item title="Key Competency" _builder_version="4.2.2" inline_fonts="Abhaya Libre" open="off"]

Number Theory

[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.2.2" open="off"]4/10[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.2.2" hover_enabled="0" open="on"]

Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics [/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="4.0.9" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|0px|20px||" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Start with hints

[/et_pb_text][et_pb_tabs _builder_version="4.2.2"][et_pb_tab title="HINT 0" _builder_version="4.0.9"]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title="HINT 1" _builder_version="4.2.2"]Any number is divisible by all of its factors. For eaxmple 50 is divisible by $2,5,10$ and $25$ out of these their are some prime numbers called Prime factors. [/et_pb_tab][et_pb_tab title="HINT 2" _builder_version="4.2.2"]The prime factor of 100,000 are only 2 and 5, the rest of them are not the prime factor, they are composite factor. Also The prime factorization of $100,000$ is $2^5 \cdot 5^5$.[/et_pb_tab][et_pb_tab title="HINT 3" _builder_version="4.2.2"]Any Number which divides 100,000 must be multiple of 2 and (or) 5. So it can be 10=5x2 or $200=2^{3} 5^{2}$. [/et_pb_tab][et_pb_tab title="HINT 4" _builder_version="4.2.2"]Since prime factorization of $100,000$ is $2^5 \cdot 5^5$. Thus We can find possible value of a,b,c and d being between 0 and 5.[/et_pb_tab][/et_pb_tabs][/et_pb_column][/et_pb_row][/et_pb_section][et_pb_section fb_built="1" fullwidth="on" _builder_version="4.2.2" global_module="50833"][et_pb_fullwidth_header title="AMC - AIME Program" button_one_text="Learn More" button_one_url="https://cheenta.com/amc-aime-usamo-math-olympiad-program/" header_image_url="https://cheenta.com/wp-content/uploads/2018/03/matholympiad.png" _builder_version="4.2.2" title_level="h2" background_color="#00457a" custom_button_one="on" button_one_text_color="#44580e" button_one_bg_color="#ffffff" button_one_border_color="#ffffff" button_one_border_radius="5px"]

AMC - AIME - USAMO Boot Camp for brilliant students. Use our exclusive one-on-one plus group class system to prepare for Math Olympiad

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Number Theory, Greece MO 2019, Problem 3

[et_pb_section fb_built="1" _builder_version="3.22.4"][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Understand the problem

[/et_pb_text][et_pb_text _builder_version="4.0" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]Find all positive rational $(x,y)$ that satisfy the equation : $yx^y=y+1$ [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.22.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.0"]Greece MO 2019, Problem 3[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0" open="off"]Number Theory[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0" open="off"]7/10[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0" open="off"]Challenges and Thrills of Pre College Mathematics[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.22.4" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.0"]Let us the given equation in terms of rational numbers and simplify the equation. Let $y=\frac{p}{q}$ and $x=\frac{r}{s}$ where $\gcd(p,q)=\gcd(r,s)=1$ . We have $p\cdot \left(\frac{r}{s}\right)^{\frac{p}{q}}=p+q \Longleftrightarrow p^q \cdot r^p=(p+q)^q \cdot s^p$ And since $\gcd(p+q,p)=\gcd(r,s)=1$ we must have $p^q=s^p$ and $r^p=(p+q)^q$ . Now, we have to find the solutions from these equations.[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0"]Observe both the equations are of the form $x^a = y^b$. The idea is that due to the prime factorization theorem, we can specify that $x^a = y^b$ leads to a special form of the x and y. Claim: If $x^a=y^b$ for some $x,y,a,b$ naturals , then there exists a natural $z$ such that $x=z^m$ and $y=z^n$ where $m=\frac{b}{\gcd(a,b)}$ and $n=\frac{a}{\gcd(a,b)}$ .[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0"]Proof of Claim: Consider the prime factorization theorem of the x and y in $x^a = y^b$. Observe that it implies x and y must have same set of primes by the prime factorization theorem. Let x and y contain the primes $p_1, p_2, ..., p_k$. Let $ x = \prod_{i=1}^{k} x_i$ and $ y = \prod_{i=1}^{k} y_i$. The above equation implies that $ a.x_i = b.y_i $. This implies that $ y_i = n.c $ and $ x_i = m.c $, where c is a natural number. Hence $ x = z^m, y = z^n$.

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0"]Using the intuitive claim, $p^q=s^p$ , there exists a $z$ such that $p=z^p$ , if $z \neq 1$ we have $p=z^p=z^{z^p}$ and continuing like this , $p$ is unbounded ,contradiction. So , we must have $z=1$ wich means $p=s=1$ and from $r^p=(p+q)^q$ we have $r=(q+1)^q$ So, the solutions come out to be $x=(q+1)^q$ and $y=\frac{1}{q}$ , where $q$ is any positive integer.[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

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Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/matholympiad/" url_new_window="on" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" background_layout="dark" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3"][/et_pb_button][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Prime Number | AMC 10A 2021, Problem 15

Key Concepts

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AMC 8 2019 Problem 20:

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Algebra Question - AMC 8, 2019 Problem 16

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AMC 8 2019: Problem 17

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Digits and Rationals - AIME I, 1992

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Smallest Positive Integer - PRMO 2019

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Problem based on divisibility

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Knowledge Graph

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What are we learning ?

Competency in Focus: Number Theory

First look at the knowledge graph:-

Next understand the problem

Number Theory

Start with hints

Understand the problem

Start with hints

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