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Algebra Arithmetic Math Olympiad USA Math Olympiad

Algebra and Positive Integer | AIME I, 1987 | Question 8

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1987 based on Algebra and Positive Integer.

Algebra and Positive Integer – AIME I, 1987


What is the largest positive integer n for which there is a unique integer k such that \(\frac{8}{15} <\frac{n}{n+k}<\frac{7}{13}\)?

  • is 107
  • is 112
  • is 840
  • cannot be determined from the given information

Key Concepts


Digits

Algebra

Numbers

Check the Answer


Answer: is 112.

AIME I, 1987, Question 8

Elementary Number Theory by David Burton

Try with Hints


Simplifying the inequality gives, 104(n+k)<195n<105(n+k)

or, 0<91n-104k<n+k

for 91n-104k<n+k, K>\(\frac{6n}{7}\)

and 0<91n-104k gives k<\(\frac{7n}{8}\)

so, 48n<56k<49n for 96<k<98 and k=97

thus largest value of n=112.

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Algebra Arithmetic Math Olympiad USA Math Olympiad

Digits and Numbers | AIME I, 2012 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2012 based on digits and numbers.

Digits and numbers – AIME I, 2012


Let S be set of all perfect squares whose rightmost three digits in base 10 are 256. T be set of numbers of form \(\frac{x-256}{1000}\) where x is in S, find remainder when 10th smallest element of T is divided by 1000.

  • is 107
  • is 170
  • is 840
  • cannot be determined from the given information

Key Concepts


Digits

Algebra

Numbers

Check the Answer


Answer: is 170.

AIME I, 2012, Question 10

Elementary Number Theory by David Burton

Try with Hints


x belongs to S so perfect square, Let x=\(y^{2}\), here \(y^{2}\)=1000a+256 \(y^{2}\) element in S then RHS being even y=2\(y_1\) then \(y_1^{2}=250a+64\) again RHS being even \(y_1=2y_2\) then \(y_2^{2}\)=125\(\frac{a}{2}\)+16 then both sides being integer a=2\(a_1\) then \(y_2^{2}=125a_1+16\)

\(y_2^{2}-16=125a_1\) then \((y_2-4)(y_2+4)=125a_1\)

or, one of \((y_2+4)\) and \((y_2-4)\) contains a non negative multiple of 125 then listing smallest possible values of \(y_2\)

or, \(y_2+4=125\) gives \(y_2=121\) or, \(y_2-4=125\) gives \(y_2=129\) and so on

or, \(y_2=4,121,129,upto ,621\) tenth term 621

\(y=4y_2\)=2484 then \(\frac{2483^{2}-256}{1000}\)=170.

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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Percentage Problem | AIME I, 2008 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2008 based on Percentage.

Percentage Problem – AIME I, 2008


Of the students attending a party, 60% of the students are girls, and 40% of the students like to dance. After these students are joined by 20 more boy students, all of whom like to dance, the party is now 58% girls, find number of students now at the party like to dance.

  • is 107
  • is 252
  • is 840
  • cannot be determined from the given information

Key Concepts


Ratios

Percentage

Numbers

Check the Answer


Answer: is 252.

AIME I, 2008, Question 1

Elementary Number Theory by David Burton

Try with Hints


Let number of girls and boys be 3k and 2k, out of 3k girls, 2k likes to dance and 2k+20(boys) like to dance

here given that \(\frac{3k}{5k+20}\)=\(\frac{58}{100}\) then k=116

then 2k+20=252.

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