Difference between revisions of "2023 IOQM/Problem 9"
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Now, <math>abc\leq30</math> ⇒ <math>ac\leq30</math> also, abc is not divisible by the square of any [[prime]] so a should not divide c so all possible pairs of <math>(a,b,c)</math> here are <math>(2,1,15); (3,1,10); (5,1,6)</math> Total no. of [[ordered pairs]] = 3 here | Now, <math>abc\leq30</math> ⇒ <math>ac\leq30</math> also, abc is not divisible by the square of any [[prime]] so a should not divide c so all possible pairs of <math>(a,b,c)</math> here are <math>(2,1,15); (3,1,10); (5,1,6)</math> Total no. of [[ordered pairs]] = 3 here | ||
| − | Hence, total no. of | + | Hence, total no. of (a,b,c) are 17 here. So the answer is <math>\boxed{13}</math> |
Revision as of 08:59, 20 October 2023
Problem
Find the number of triples 
of positive integers such that
(a) 
is a prime;
(b) 
is a product of two primes;
(c) 
is not divisible by square of any prime and
(d) 
Solution1(Casework)
Since, ab is a prime, this means that one of a and b is 1 and the other is prime. So, there are 2 cases from here:
Case 1(a=1)
If a is one and b is a prime, this means that c is also a prime but different from b ( as bc is a product of 2 primes but abc is not divisible by the square of any prime)
Now, ⇒ 
, so all possible pairs of 
here are 
Total no. of ordered pairs = 14 here
Case 2(b=1)
If b is one and 
is a prime, this means that c is the product of 2 different primes ( as bc is a product of 2 primes but abc is not divisible by the square of any prime)
Now, ⇒ 
also, abc is not divisible by the square of any prime so a should not divide c so all possible pairs of here are 
Total no. of ordered pairs = 3 here
Hence, total no. of (a,b,c) are 17 here. So the answer is 
