Tag: digits

Categories
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.

Other useful links

Related Program

Subscribe to Cheenta at Youtube

Categories
AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Integers | AIME I, 1993 Problem | Question 4

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1993 based on Integers.

Integer – AIME I, 1993

Find the number of four topics of integers (a,b,c,d) with 0<a<b<c<d<500 satisfy a+d=b+c and bc-ad=93.

  • is 107
  • is 870
  • is 840
  • cannot be determined from the given information

Key Concepts

Integers

Digits

Algebra

Check the Answer

Answer: is 870.

AIME I, 1993, Question 4

Elementary Algebra by Hall and Knight

Try with Hints

Let k=a+d=b+c

or, d=k-a, b=k-c,

or, (k-c)c-a(k-a)=k(c-a)-(c-a)(c+a)

=(a-c)(a+c-k)

=(c-a)(d-c)=93

(c-a)(d-c)=(1,93),(3,31),(31,3),(93,1)

solving for c

(a,b,c,d)=(c-93,c-92,c,c+1),(c-31,c-28,c,c+3),(c-1,c+92,c,c+93),(c-3,c+28,c,c+31)

taking first two solutions a<b<c<d<500

or,\(1 \leq c-93, c+1 \leq 499\)

or, \(94 \leq c \leq 498 \) gives 405 solutions

and \(1 \leq c-31, c+3 \leq 499\)

or, \(32 \leq c \leq 496\) gives 465 solutions

or, 405+465=870 solutions.

Other useful links

Related Program

Subscribe to Cheenta at Youtube

Categories
India Math Olympiad Math Olympiad Number Theory PRMO USA Math Olympiad

Sum of digits Problem | PRMO 2016 | Question 6

Try this beautiful problem from Number system based on sum of digits.

Sum of digits | PRMO | Problem 6

Find the sum of digits in decimal form of the number \((9999….9)^3\) (There are 12 nines)

  • $200$
  • $216$
  • $230$

Key Concepts

Number system

Digits

counting

Check the Answer

Answer:$216$

PRMO-2016, Problem 6

Pre College Mathematics

Try with Hints

we don’t know what will be the expression of \((9999….9)^3\). so we observe….

\(9^3\)=\(729\)

\((99)^3\)=\(970299\)

\((999)^3\)=\(997002999\)

……………..

……………

we observe that,There is a pattern such that…

In \((99)^3\)=\(970299\) there are 1-nine,1-seven,1-zero,1-two,2-nines & \((999)^3\)=\(997002999\) there are 2- nines,1-seven,2-zeros,1-two,3-nines….so in this way…..\((999….9)^3\) will be 11-nines,1-seven,11-zeros,1-two,12-nines……….

Therefore \((999….9)^3\)=\((99999999999) 7 (00000000000) 2(999999999999)\)

can you finish the problem?

Therefore \((999….9)^3\)=\((99999999999) 7 (00000000000) 2(999999999999)\)…….

total numbers of Nines are (11+12) and (7+2)=9(another one) …..so total (11+12+1)=24 nines and the sum be \((24\times 9)\)=\(216\)

Other useful links

Related Program

Subscribe to Cheenta at Youtube

Categories
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.

Other useful links

Related Program

Subscribe to Cheenta at Youtube

Categories
AIME I Algebra Arithmetic Calculus Math Olympiad USA Math Olympiad

Problem on Fibonacci sequence | AIME I, 1988 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1988 based on Fibonacci sequence.

Fibonacci sequence Problem – AIME I, 1988

Find a if a and b are integers such that \(x^{2}-x-1\) is a factor of \(ax^{17}+bx^{16}+1\).

  • is 107
  • is 987
  • is 840
  • cannot be determined from the given information

Key Concepts

Integers

Digits

Sets

Check the Answer

Answer: is 987.

AIME I, 1988, Question 13

Elementary Number Theory by David Burton

Try with Hints

Let F(x)=\(ax^{17}+bx^{16}+1\)

Let P(x) be polynomial such that

\(P(x)(x^{2}-x-1)=F(x)\)

constant term of P(x) =(-1)

now \((x^{2}-x-1)(c_1x^{15}+c_2x^{14}+….+c_{15}x-1)\) where \(c_{i}\)=coefficient

comparing the coefficients of x we get the terms

since F(x) has no x term, then \(c_{15}\)=1

getting \(c_{14}\)

\((x^{2}-x-1)(c_1x^{15}+c_2x^{14}+….+c_{15}x-1)\)

=terms +\(0x^{2}\) +terms

or, \(c_{14}=-2\)

proceeding in the same way \(c_{13}=3\), \(c_{12}=-5\), \(c_{11}=8\) gives a pattern of Fibonacci sequence

or, coefficients of P(x) are Fibonacci sequence with alternating signs

or, a=\(c_1=F_{16}\) where \(F_{16}\) is 16th Fibonacci number

or, a=987.

Other useful links

Related Program

Subscribe to Cheenta at Youtube

Categories
AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Digits and Integers | AIME I, 1990 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on Digits and Integers.

Digits and Integers – AIME I, 1990

Let T={\(9^{k}\): k is an integer, \(0 \leq k \leq 4000\)} given that \(9^{4000}\) has 3817 digits and that its first (leftmost) digit is 9, how many elements of T have 9 as their leftmost digit?

  • is 107
  • is 184
  • is 840
  • cannot be determined from the given information

Key Concepts

Integers

Digits

Sets

Check the Answer

Answer: is 184.

AIME I, 1990, Question 13

Elementary Number Theory by David Burton

Try with Hints

here \(9^{4000}\) has 3816 digits more than 9,

or, 4000-3816=184

or, 184 numbers have 9 as their leftmost digits.

Other useful links

Related Program

Subscribe to Cheenta at Youtube

Categories
AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

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.

Other useful links

Related Program

Subscribe to Cheenta at Youtube

Categories
AIME I Algebra Arithmetic Combinatorics Math Olympiad USA Math Olympiad

Row of Pascal Triangle | AIME I, 1992 | Question 4

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on Row of Pascal Triangle.

Row of Pascal Triangle – AIME I, 1992

In Pascal’s Triangle, each entry is the sum of the two entries above it. Find the row of Pascal’s triangle do three consecutive entries occur that are in the ratio 3:4:5.

  • is 107
  • is 62
  • is 840
  • cannot be determined from the given information

Key Concepts

Integers

Digits

Combinatorics

Check the Answer

Answer: is 62.

AIME I, 1992, Question 4

Elementary Number Theory by David Burton

Try with Hints

For consecutive entries

\(\frac{{n \choose (x-1)}}{3}=\frac{{n \choose x}}{4}=\frac{{n \choose {x+1}}}{5}\)

from first two terms \(\frac{n!}{3(x-1)!(n-x+1)!}=\frac{n!}{4x!(n-x)!}\)

\(\Rightarrow \frac{1}{3(n-x+1)}=\frac{1}{4x}\)

\(\Rightarrow \frac{3(n+1)}{7}=x\) is first equation

for the next two terms

\(\frac{n!}{4(n-x)!x!}=\frac{n!}{5(n-x-1)!(x+1)!}\)

\(\Rightarrow \frac{4(n-x)}{5}=x+1\)

\(\Rightarrow \frac{4n}{5}=\frac{9x}{5}+1\)

from first equation putting value of x here gives

\(\Rightarrow \frac{4n}{5}=\frac{9 \times 3(n+1)}{5 \times 7}+1\)

\(\Rightarrow n=62, x=\frac{3(62+1)}{7}=27\)

\(\Rightarrow\) n=62.

Other useful links

Related Program

Subscribe to Cheenta at Youtube

Categories
AIME I Algebra Arithmetic Functions Math Olympiad USA Math Olympiad

Digits and Order | AIME I, 1992 | Question 2

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

Digits and order – AIME I, 1992

A positive integer is called ascending if, in its decimal representation, there are at least two digits and each digit is less than any digit to its right. Find number of ascending positive integers are there.

  • is 107
  • is 502
  • is 840
  • cannot be determined from the given information

Key Concepts

Integers

Digits

Order

Check the Answer

Answer: is 502.

AIME I, 1992, Question 2

Elementary Number Theory by David Burton

Try with Hints

There are nine digits that we use 1,2,3,4,5,6,7,8,9.

Here each digit may or may not be present.

\(\Rightarrow 2^{9}\)=512 potential ascending numbers, one for subset of {1,2,3,4,5,6,7,8,9}

Subtracting empty set and single digit set

=512-10

=502.

Other useful links

Related Program

Subscribe to Cheenta at Youtube