Try this beautiful problem from PRMO, 2019, Problem 18 based on Ordered Pairs.
Orderd Pairs | PRMO | Problem-18
How many ordered pairs \((a, b)\) of positive integers with \(a < b\) and \(100 \leq a\), \(b \leq 1000\) satisfy \(gcd (a, b) : lcm (a, b) = 1 : 495\) ?
$20$
$91$
$13$
\(23\)
Key Concepts
Number theory
Orderd Pair
LCM
Check the Answer
Answer:\(20\)
PRMO-2019, Problem 18
Pre College Mathematics
Try with Hints
At first we assume that \( a = xp\) \(b = xq\) where \(p\) & \(q\) are co-prime
Therefore ,
\(\frac{gcd(a,b)}{LCM(a ,b)} =\frac{495}{1}\)
\(\Rightarrow pq=495\) Can you now finish the problem ..........
Therefore we can say that
\(pq = 5 \times 9 \times 11\) \(p < q\)
when \( 5 < 99\) (for \(x = 20, a = 100, b = 1980 > 100\)),No solution when \(9 < 55\) \((x = 12\) to \(x = 18)\),7 solution when,\(11 < 45\) \((x = 10\) to \(x = 22)\),13 solution Can you finish the problem........
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1985 based on GCD and Sequence.
GCD and Sequence - AIME I, 1985
The numbers in the sequence 101, 104,109,116,.....are of the form \(a_n=100+n^{2}\) where n=1,2,3,-------, for each n, let \(d_n\) be the greatest common divisor of \(a_n\) and \(a_{n+1}\), find the maximum value of \(d_n\) as n ranges through the positive integers.
is 107
is 401
is 840
cannot be determined from the given information
Key Concepts
GCD
Sequence
Integers
Check the Answer
Answer: is 401.
AIME I, 1985, Question 13
Elementary Number Theory by David Burton
Try with Hints
\(a_n=100+n^{2}\) \(a_{n+1}=100+(n+1)^{2}=100 + n^{2} +2n +1\) and \(a_{n+1}-a_{n}=2n +1\)
\(d_{n}|(2n+1)\) and \(d_{n}|(100 +n^{2})\) then \(d_{n}|[(100+n^{2})-100(2n+1)]\) then \(d_{n}|(n^{2}-200n)\)
here \(n^{2} -200n=0\) then n=200 then \(d_{n}\)=2n+1=2(200)+1=401.
Try this beautiful problem from the PRMO, 2017 based on GCD and Primes.
GCD and primes - PRMO 2017
For each positive integer n, consider the highest common factor \(h_n\) of the two numbers n!+1 and (n+1)! for n<100, find the largest value of \(h_n\).
is 107
is 97
is 840
cannot be determined from the given information
Key Concepts
GCD
Primes
Inequalities
Check the Answer
Answer: is 97.
PRMO, 2017, Question 29
Elementary Number Theory by David Burton
Try with Hints
n! +1 is not divisible by 1,2,.....,n (n+1)! divisible by 1,2,....,n then \(hcf \geq (n+1)\) and (n+1)! not divisible by n+2, n+3,...... then hcf= (n+1)
let n=99, 99! +1 and (100)! hcf=100 not possible as 100 |99! and 100 is non prime
let n=97 96! + 1 and 97! both divisible by 97 then hcf=97.
Try this beautiful problem from Algebra based on LCM from AMC-8, 2016.
Problem based on LCM - AMC 8, 2016
The least common multiple of $a$ and $b$ is $12$, and the least common multiple of $b$ and $c$ is $15$. What is the least possible value of the least common multiple of $a$ and $c$?
\(26\)
\(20\)
\(28\)
Key Concepts
Algebra
Divisor
multiplication
Check the Answer
Answer:20
AMC-8, 2016 problem 20
Challenges and Thrills of Pre College Mathematics
Try with Hints
We have to find out the least common multiple of $a$ and $c$.if you know the value of \(a\) and \(c\) then you can easily find out the required LCM. Can you find out the value of \(a\) and \(c\)?
Can you now finish the problem ..........
Given that the least common multiple of $a$ and $b$ is $12$, and the least common multiple of $b$ and $c$ is $15$ .then b must divide 12 and 15. There is only one possibility that b=3 which divide 12 and 15. therefore \(a\)=\(\frac{12}{3}=4\)
can you finish the problem........
so\(b\)=3. Given that LCM of \(b\) and \(c\) is 15. Therefore c=5
Now lcm of \(a\) and \(c\) that is lcm of 4 and 5=20
Consider the equation 2019x + 2020y = 2018. Are there integers x and y that satisfy this equation?
Concepts in this lesson will help you to answer this question and more.
Concept - GCD, Bezout Theorem
GCD of two numbers a and b is their greatest common divisor. For example for 10 and 15, GCD is 5.
Bezout Theorem, in essence, describes the equation 10x + 15y = 5. It ensures that there are integer solutions to this equation. In fact for any two integers a and b, if GCD(a, b) = d, Bezout Theorem says that there are integer solutions to the equation: ax + by = d
