Categories

## 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

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.

Categories

## 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

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.

Categories

## 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

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.