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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Odd and Even integers | AIME I, 1997 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1997 based on Odd and Even integers.

Odd and Even Integers – AIME I, 1997


Find the number of integers between 1 and 1000 that can be expressed as the difference of squares of two non-negative integers.

  • is 107
  • is 750
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Divisibility

Difference of squares

Check the Answer


Answer: is 750.

AIME I, 1997, Question 1

Elementary Number Theory by David Burton

Try with Hints


Let x be a non-negetive integer \((x+1)^{2}-x^{2}=2x+1\)

Let y be a non-negetive integer \((y+1)^{2}-(y-1)^{2}=4y\)

Numbers 2(mod 4) cannot be obtained as difference of squares then number of such numbers =500+250=750.

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AMC 8 Math Olympiad USA Math Olympiad

Divisibility Problem from AMC 10A, 2003 | Problem 25

Try this beautiful problem from Number theory based on divisibility from AMC 10A, 2003.

Number theory in Divisibility – AMC-10A, 2003- Problem 25


Let \(n\) be a \(5\)-digit number, and let \(q\) and \(r\) be the quotient and the remainder, respectively, when \(n\) is divided by \(100\). For how many values of \(n\) is \(q+r\) divisible by \(11\)?

  • \(8180\)
  • \(8181\)
  • \(8182\)
  • \(9190\)
  • \(9000\)

Key Concepts


Number system

Probability

divisibility

Check the Answer


Answer: \(8181\)

AMC-10A (2003) Problem 25

Pre College Mathematics

Try with Hints


Since \(11\) divides \(q+r\) so may say that \(11\) divides \(100 q+r\). Since \(n\) is a \(5\) digit number …soTherefore, \(q\) can be any integer from \(100\) to \(999\) inclusive, and \(r\) can be any integer from \(0\) to \(99\) inclusive.

can you finish the problem……..

Since \(n\) is a five digit number then and \(11 | 100q+r\) then \(n\) must start from \(10010\) and count up to \(99990\)

can you finish the problem……..

Therefore, the number of possible values of \(n\) such that \(900 \times 9 +81 \times 1=8181\)

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AMC 8 Math Olympiad USA Math Olympiad

Problem based on LCM | AMC 8, 2016 | Problem 20

Try this beautiful problem from Algebra based on LCM from AMC-8, 2016.

Problem based on LCM – AMC 8, 2016


The least common multiple of $a$ and $b$ is $12$, and the least common multiple of $b$ and $c$ is $15$. What is the least possible value of the least common multiple of $a$ and $c$?

  • \(26\)
  • \(20\)
  • \(28\)

Key Concepts


Algebra

Divisor

multiplication

Check the Answer


Answer:20

AMC-8, 2016 problem 20

Challenges and Thrills of Pre College Mathematics

Try with Hints


We have to find out the least common multiple of $a$ and $c$.if you know the value of \(a\) and \(c\) then you can easily find out the required LCM. Can you find out the value of \(a\) and \(c\)?

Can you now finish the problem ……….

Given that the least common multiple of $a$ and $b$ is $12$, and the least common multiple of $b$ and $c$ is $15$ .then b must divide 12 and 15. There is only one possibility that b=3 which divide 12 and 15. therefore \(a\)=\(\frac{12}{3}=4\)

can you finish the problem……..

so\(b\)=3. Given that LCM of \(b\) and \(c\) is 15. Therefore c=5

Now lcm of \(a\) and \(c\) that is lcm of 4 and 5=20

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AMC 8 Math Olympiad

Linear Equations | AMC 8, 2007 | Problem 20

Try this beautiful problem from Algebra based on Linear equations from AMC-8, 2007.

Linear equations – AMC 8, 2007


Before the district play, the Unicorns had won $45$% of their basketball games. During district play, they won six more games and lost two, to finish the season having won half their games. How many games did the Unicorns play in all?

  • \(40\)
  • \(48\)
  • \(58\)

Key Concepts


Algebra

linear equation

multiplication

Check the Answer


Answer:48

AMC-8, 2007 problem 20

Challenges and Thrills of Pre College Mathematics

Try with Hints


At first, we have to Calculate the number of won games and lost games. Unicorns had won $45$% of their basketball game.so we may assume that out of 20 unicorns woned 9.

Can you now finish the problem ……….

Next unicorns won six more games and lost two.so find out the total numbers of won game and total numbers of games i.e won=\(9x+6\) and the total number of games become \(20x+8\)

can you finish the problem……..

Given that Unicorns had won \(45\)% of their basketball games i.e \(\frac{45}{100}=\frac{9}{20}\)

During district play, they won six more games and lost two,

Therefore they won\(9x+6\) and the total number of games becomes \(20x+8\)

According to the question, Unicorns finish the season having won half their games. …

Therefore,\(\frac{9x+6}{20x+8}=\frac{1}{2}\)

\(\Rightarrow 18x+12=20x+8\)

\(\Rightarrow 2x=4\)

\(\Rightarrow x=2\)

Total number of games becomes \(20x+8\) =\((20 \times 2) +8=48\)

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Algebra AMC 8 Math Olympiad

Divisibility | AMC 8, 2014 |Problem 21

Try this beautiful problem from Algebra based on multiplication and divisibility of two given numbers.

Multiplication and Divisibility- AMC 8, 2014


The  7-digit numbers 74A52B1 ana 326AB4C are each multiples of 3.which of the following could be the value of c ?

  • 1
  • 2
  • 3

Key Concepts


Algebra

Division algorithm

Integer

Check the Answer


Answer:1

AMC-8, 2014 problem 21

Challenges and Thrills of Pre College Mathematics

Try with Hints


Use the rules of Divisibility ……..

Can you now finish the problem ……….

If both numbers are divisible by 3 then the sum of their digits has to be divisible by 3……

can you finish the problem……..

Since both numbers are divisible by 3, the sum of their digits has to be divisible by three. 7 + 4 + 5 + 2 + 1 = 19. In order to be a multiple of 3, A + B has to be either 2 or 5 or 8… and so on. We add up the numerical digits in the second number; 3 + 2 + 6 + 4 = 15. We then add two of the selected values, 5 to 15, to get 20. We then see that C = 1, 4 or 7, 10… and so on, otherwise the number will not be divisible by three. We then add 8 to 15, to get 23, which shows us that C = 1 or 4 or 7… and so on. In order to be a multiple of three, we select a few of the common numbers we got from both these equations, which could be 1, 4, and 7. However, in the answer choices, there is no 7 or 4 or anything greater than 7, but there is a 1. so the answer is 1

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Algebra AMC 8 Math Olympiad Number Theory

Least common multiple | AMC 8, 2016 – Problem 20

LCM – AMC 8, 2016 – Problem 20


The least common multiple of a and b is 12 .and the lest common multiple of b and c is 15.what is the least possible value of the least common multiple of a and c?

  • 30
  • 60
  • 20

Key Concepts


Algebra

Division algorithm

Integer

Check the Answer


Answer:20

AMC-8, 2016 problem 20

Challenges and Thrills of Pre College Mathematics

Try with Hints


Find greatest common factors

Can you now finish the problem ……….

Find Least common multiple….

can you finish the problem……..

we wish to find possible values of a,b and c .By finding the greatest common factor 12 and 15, algebrically ,it’s some multiple of b and from looking at the numbers ,we are sure that it is 3.Moving on to a and c ,in order to minimize them,we wish to find the least such that the LCM of a and 3 is 12,$\to 4$.similarly with 3 and c,we obtain 5.the LCM of 4 and 5 is 20 .

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AMC 8 USA Math Olympiad

Divisibility – Understanding the Rule : AMC 8, 2016 Problem 24

What is Divisibility?


Divisibility is the study of finding remainder when a number is divided by another number.

Try the problem


The digits $1$, $2$, $3$, $4$, and $5$ are each used once to write a five-digit number $PQRST$. The three-digit number $PQR$ is divisible by $4$, the three-digit number $QRS$ is divisible by $5$, and the three-digit number $RST$ is divisible by $3$. What is $P$?

$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$

AMC 8, 2016 Problem number 24

Divisibility

6 out of 10

Mathematical Circles

Knowledge Graph


Divisibility Rule- Knowledge graph

Use some hints


See this image:

Divisibility rule step 1

So we have $5$ blank space to fill with the numbers $1,2,3,4,5$ following the given restriction.

See the given conditions care fully :

$1.\quad PQR$ is divisible by $4$,

$2.\quad QRS$ is divisible by $5$,

$3. \quad RST$ is divisible by $3$

Can you make a definite choice for any of the places ???

$1.\quad PQR$ is divisible by $4 \Rightarrow$ $QR$ is divisible by $4$

$2.\quad QRS$ is divisible by $5 \Rightarrow $, $S$ is either $5$ or $0$. [But we don’t have $0$ so $S$ must be $5$]

$3. \quad RST$ is divisible by $3 \Rightarrow $ $R+S+T$ is divisible by 3.

Divisibility Rule- Step 2

Try to find out other choices, other blanks !!

$QR$ is divisible by $4$ so $R$ needs to be $2$ or $4$.

Let $R=2$

$\Rightarrow R+S+T=2+5+T=7+T$

As, $R+S+T$ is divisible by 3 then possible values of $T$ are $2, 5, 8$

We have already used $2$ and $5$. And $8$ is not in the given set of numbers. (Contradiction)

Therefore $R \ne 2 \Rightarrow R = 4$

Divisibility Rule - Step 3

Now the things are easy, try to drive it from here !!!!!!!!!

Again, $RST$ is divisible by $3$

$\Rightarrow R+S+T$ is divisible by $3$

$\Rightarrow 4+5+T$ is divisible by $3$

Then the possibilities for $T$ are $3, 6$

$6$ is not in the given number, hence $T=3$

 Step 4

We have two numbers left and two blank places left.

if $Q=1$

then $QR=14$ is not divisible by $4$

Then $Q=2$ [$24$ is divisible by $4$]

And certainly $P=1$

The number is :

 Step 5

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AMC 8 USA Math Olympiad

Probability of an event- AMC 8 2009 Problem 13

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What are we learning ?

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Competency in Focus: Probability of an event

This problem from American Mathematics Contest 8 (AMC 8, 2019) is based on calculation of probability of an event using the concept of divisibility. It is Question no. 13 of the AMC 8 2009 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://www.cheenta.com/wp-content/uploads/2020/02/Amc8_1102.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=”4.3.1″ 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″]A three-digit integer contains one of each of the digits $1$, $3$, and $5$. What is the probability that the integer is divisible by $5$? $\textbf{(A)}\ \frac{1}{6} \qquad \textbf{(B)}\ \frac{1}{3} \qquad \textbf{(C)}\ \frac{1}{2} \qquad \textbf{(D)}\ \frac{2}{3} \qquad \textbf{(E)}\ \frac{5}{6}$[/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=”on” _builder_version=”4.3.1″ hover_enabled=”0″]

American Mathematical Contest 2009, AMC 8 Problem 13 [/et_pb_accordion_item][et_pb_accordion_item title=”Key Competency” _builder_version=”4.3.1″ hover_enabled=”0″ inline_fonts=”Abhaya Libre” open=”off”]

Finding probability of an event using the concept of divisibility

[/et_pb_accordion_item][et_pb_accordion_item title=”Difficulty Level” _builder_version=”4.2.2″ open=”off”]5/10[/et_pb_accordion_item][et_pb_accordion_item title=”Suggested Book” _builder_version=”4.0.9″ open=”off”]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″]First note that a number is divisible by $5$ if the unit digit is $5$ or $0$[/et_pb_tab][et_pb_tab title=”HINT 2″ _builder_version=”4.2.2″]Can you find all the three digit numbers that can be made using the digits given  [/et_pb_tab][et_pb_tab title=”HINT 3″ _builder_version=”4.2.2″]So the three digit numbers are $135,153,351,315,513,531$. among these only two numbers are divisible by $5$ [/et_pb_tab][et_pb_tab title=”HINT 4″ _builder_version=”4.2.2″]Probability of an event = $\frac{\textbf{Number points supporting the event}}{Total number of outcomes}$ Ans=$\frac{\textbf{Number of 3 digit numbers divisible by 5}}{\textbf{Number of 3 digits number can be made}}=\frac{2}{6}=\frac{1}{3}$[/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://www.cheenta.com/amc-aime-usamo-math-olympiad-program/” header_image_url=”https://www.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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AMC 8 USA Math Olympiad

Divisibility AMC 8, 2017 problem 7

[et_pb_section fb_built=”1″ _builder_version=”4.0″][et_pb_row _builder_version=”3.25″ hover_enabled=”0″][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″]

What are we learning ?

[/et_pb_text][et_pb_text _builder_version=”4.1″ 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″]Competency in Focus: Divisibility This problem from American Mathematics contest (AMC 8, 2017) is based on the concept of divisibility .[/et_pb_text][et_pb_text _builder_version=”4.3.1″ 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” hover_enabled=”0″ border_radii=”on|5px|5px|5px|5px” box_shadow_style=”preset3″]

First look at the knowledge graph.

[/et_pb_text][et_pb_image src=”https://www.cheenta.com/wp-content/uploads/2020/01/divisibilityamc82017_7.png” align=”center” force_fullwidth=”on” _builder_version=”4.1″ min_height=”388px” height=”198px” max_height=”207px”][/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=”ABeeZee”]

Next understand the problem

[/et_pb_text][et_pb_text _builder_version=”4.1″ 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 \(Z\) be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of \(Z\)? \((A)11\qquad(B)19\qquad(A)101\qquad(A)111\qquad(A)1111\)[/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=”4.3.1″ 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=”4.1″]American Mathematical Contest 2017, AMC 8  Problem 7[/et_pb_accordion_item][et_pb_accordion_item title=”Key Competency” _builder_version=”4.1″ open=”off”]Divisibility[/et_pb_accordion_item][et_pb_accordion_item title=”Difficulty Level” _builder_version=”4.1″ open=”off”]2/10[/et_pb_accordion_item][et_pb_accordion_item title=”Suggested Book” _builder_version=”4.3.1″ open=”off”]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″]

Start with hints 

[/et_pb_text][et_pb_tabs _builder_version=”4.1″][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=”4.1″]Take \(Z\) as \(\overline{\rm ABCABC}\).[/et_pb_tab][et_pb_tab title=”HINT 2″ _builder_version=”4.1″]Calculate the alternating sum of the digits. i.e., \(A-B+C-A+B-C=0\) then it is dibisible by \(11\)[/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″ min_height=”12px” custom_margin=”50px||50px” custom_padding=”20px|20px|20px|20px” border_radii=”on|5px|5px|5px|5px” box_shadow_style=”preset3″]

Connected Program at Cheenta

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Cheenta AMC Training Camp consists of live group and one on one classes, 24/7 doubt clearing and continuous problem solving streams.[/et_pb_blurb][et_pb_button button_url=”https://www.cheenta.com/amc-8-american-mathematics-competition/” url_new_window=”on” button_text=”Learn More” button_alignment=”center” _builder_version=”4.0.9″ 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″]

Similar Problems

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AMC 8 USA Math Olympiad

Number Theory Problem | AMC 10B 2019| Problem 19

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What are we learning ?

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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.

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First look at the knowledge graph:-

[/et_pb_text][et_pb_image src=”https://www.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

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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=5×2 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://www.cheenta.com/amc-aime-usamo-math-olympiad-program/” header_image_url=”https://www.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”]

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