Number and Series | Number Theory | AIME I, 2015
Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2015 based on Number Theory and Series.
Number Theory and Series - AIME 2015
The expressions A = \((1 \times 2)+(3 \times 4)+....+(35 \times 36)+37\) and B = \(1+(2 \times 3)+(4 \times 5)+....+(36 \times 37)\) are obtained by writing multiplication and addition operators in an alternating pattern between successive integers. Find the positive difference between integers A and B.
- is 107
- is 648
- is 840
- cannot be determined from the given information
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Answer: is 648.
AIME, 2015, Question 1
Elementary Number Theory by David Burton
Try with Hints
B-A=\(-36+(2 \times 3)+....+(2 \times 36)\)
=\(-36+4 \times (1+2+3+....+18)\)
=\(-36+(4 \times \frac{18 \times 19}{2})\)=648.
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Probability Biased and Unbiased | AIME I, 2010 Question 4
Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2010 based on Probability Biased and Unbiased.
Probability Biased and Unbiased - AIME 2010
Ramesh and Suresh have two fair coins and a third coin that comes up heads with probability \(\frac{4}{7}\),Ramesh flips the three coins, and then Suresh flips the three coins, let \(\frac{m}{n}\) be the probability that Ramesh gets the same number of heads as Suresh, where m and n are relatively prime positive integers. Find m+n.
- is 107
- is 250
- is 840
- cannot be determined from the given information
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Answer: is 107.
AIME, 2010, Question 4
Combinatorics by Brualdi
Try with Hints
No heads TTT is \(\frac{1.1.1}{2.2.7}=\frac{3}{28}\)and \((\frac{3}{28})^{2}=\frac{9}{784}\)
One Head HTT THT TTH with \(\frac{3}{28}\) \(\frac {3}{28}\) and \(\frac{4}{28}\) then probability is \(\frac{4(3.3)+4(3.4)+1(4.4)}{28^{2}}\)=\(\frac{100}{784}\)
Two heads HHT \(\frac{4}{28}\) HTH \(\frac{4}{28}\) THH \(\frac{3}{28}\) then probability is \(\frac{1(3.3)+4(3.4)+4(4.4)}{28^{2}}\)=\(\frac{121}{784}\).
Three heads HHH is \(\frac{4}{28}\) then probability \(\frac{16}{784}\)
Then sum is \(\frac{9+100+121+16}{784}=\frac{123}{392}\) then 123+392=515.
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Number Theory of Primes | AIME I, 2015
Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 based on Number Theory of Primes.
Number Theory of Primes - AIME 2015
There is a prime number p such that 16p+1 is the cube of a positive integer. Find p.
- is 307
- is 250
- is 840
- cannot be determined from the given information
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Answer: is 307.
AIME, 2015
Elementary Number Theory by Sierpinsky
Try with Hints
Notice that 16p+1must be in the form \((a+1)^{3}=a^{3}+3a^{2}+3a\), or \(16p=a(a^{2}+3a+3)\). Since p must be prime, we either have p=a or a=16
p not equal to a then we have a=16,
p\(=16^{2}+3(16)+3=307
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Complex Numbers | AIME I, 2009 | Problem 2
Try this beautiful problem from AIME, 2009 based on complex numbers.
Complex Numbers - AIME, 2009
There is a complex number z with imaginary part 164 and a positive integer n such that $\frac{z}{z+n}=4i$, Find n.
Key Concepts
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Answer: 697.
AIME, 2009, Problem 2
Complex Numbers from A to Z by Titu Andreescue .
Try with Hints
Taking z=a+bi
then a+bi=(z+n)4i=-4b+4i(a+n),gives a=-4b b=4(a+n)=4(n-4b)
then n=$\frac{b}{4}+4b=\frac{164}{4}+4.164=697$
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Combinations | AIME I, 2009 |Problem 9
Try this problem from American Invitational Mathematics Examination, AIME, 2019 based on Combinations
Combinations- AIME, 2009
A game show offers a contestant three prizes A B and C each of which is worth a whole number of dollars from $1 to $9999 inclusive. The contestant wins the prizes by correctly guessing the price of each prize in the order A B and C. As a hint the digits of three prizes are given. On a particular day the digits given were 1,1,1,1,3,3,3. Find the total number of possible guesses for all three prizes consistent with the hint.
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Answer: 420.
AIME I, 2009, Problem 9
Combinatorics by Brualdi .
Try with Hints
Number of possible ordering of seven digits is$\frac{7!}{4!3!}$=35
these 35 orderings correspond to 35 seven-digit numbers, and the digits of each number can be subdivided to represent a unique combination of guesses for A B and C. Thus, for a given ordering, the number of guesses it represents is the number of ways to subdivide the seven-digit number into three nonempty sequences, each with no more than four digits. These subdivisions have possible lengths 1/2/4,2/2/3,1/3/3, and their permutations. The first subdivision can be ordered in 6 ways and the second and third in three ways each, for a total of 12 possible subdivisions.
then total number of guesses is 35.12=420
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