Difference between revisions of "2023 IOQM/Problem 9"
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| − | 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: | + | Since, <math>ab</math> is a [[prime]], this means that one of <math>a</math> and <math>b</math> is [[1]] and the other is [[prime]]. So, there are 2 cases from here: |
| − | '''Case 1(a=1)''' | + | '''Case 1(<math>a=1</math>)''' |
| − | 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]]) | + | If <math>a</math> is [[one]] and <math>b</math> is a [[prime]], this means that <math>c</math> 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, <math>abc\leq30</math> ⇒ <math>bc\leq30</math>, so all possible pairs of <math>(a,b,c)</math> here are <math>(1,2,3); (1,2,5); (1,2,7); (1,2,11); (1,2,13); (1,3,2); (1,3,5); (1,3,7); (1,5,2); (1,5,3); (1,7,2); (1,7,3); (1,11,2); (1,13,2)</math> Total no. of [[ordered pairs]] = 14 here | Now, <math>abc\leq30</math> ⇒ <math>bc\leq30</math>, so all possible pairs of <math>(a,b,c)</math> here are <math>(1,2,3); (1,2,5); (1,2,7); (1,2,11); (1,2,13); (1,3,2); (1,3,5); (1,3,7); (1,5,2); (1,5,3); (1,7,2); (1,7,3); (1,11,2); (1,13,2)</math> Total no. of [[ordered pairs]] = 14 here | ||
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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 (a,b,c) | + | Hence, total no. of triplets (a,b,c) = 14+3= <math>\boxed{17}</math> |
| + | ~SANSGANKRSNGUPTA | ||
Revision as of 09:07, 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, is a prime, this means that one of 
and 
is 1 and the other is prime. So, there are 2 cases from here:
Case 1(
)
If is one and is a prime, this means that 
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 triplets (a,b,c) = 14+3= 
~SANSGANKRSNGUPTA
