Two and Three-digit numbers | AIME I, 1997 | Question 3

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1997 based on Two and Three-digit numbers.

Two and Three-digit numbers - AIME I, 1997

Sarah intended to multiply a two digit number and a three digit number, but she left out the multiplication sign and simply placed the two digit number to the left of the three digit number, thereby forming a five digit number. This number is exactly nine times the product Sarah should have obtained, find the sum of the two digit number and the three digit number.

is 107
is 126
is 840
cannot be determined from the given information

Key Concepts

Twodigit Number

Threedigit Number

Factors

Check the Answer

Answer: is 126.

AIME I, 1997, Question 3

Elementary Number Theory by David Burton

Try with Hints

Let p be a two digit number and q be a three digit number

here 1000p+q=9pq

$\Rightarrow 9pq-1000p-q=0$

$(9p-1)(q-\frac{1000}{9})$=$\frac{1000}{9}$

$\Rightarrow(9p-1)(9q-1000)$=1000

from factors of 1000 gives 9p-1=125

$\Rightarrow p=14,q=112$

$\Rightarrow 112+14=126$.

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

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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=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"]

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Two and Three-digit numbers | AIME I, 1997 | Question 3