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Difference between revisions of "2023 IOQM/Problem 9"

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'''Case 2(<math>b=1</math>)'''
 
'''Case 2(<math>b=1</math>)'''
  
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If b is [[one]] and <math>a</math> is a [[prime]], this means that c is the product of 2 different [[primes]] ( as <math>bc</math> is a product of 2 primes but <math>abc</math> is not divisible by the square of any [[prime]])
+
If <math>b</math> is [[one]] and <math>a</math> is a [[prime]], this means that <math>c</math> is the product of 2 different [[primes]] ( as <math>bc</math> is a product of 2 primes but <math>abc</math> is not divisible by the square of any [[prime]])
  
 
Now, <math>abc\leq30</math> ⇒ <math>ac\leq30</math> also, <math>abc</math> is not divisible by the square of any [[prime]] so a should not divide <math>c</math> 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, <math>abc</math> is not divisible by the square of any [[prime]] so a should not divide <math>c</math> 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

Latest revision as of 10:52, 25 October 2023

Problem

Find the number of triples $(a, b, c)$ of positive integers such that (a) $ab$ is a prime;

(b) $bc$ is a product of two primes;

(c) $abc$ is not divisible by square of any prime and

(d) $abc\leq30$

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, $abc\leq30$ ⇒ $bc\leq30$, so all possible pairs of $(a,b,c)$ here are $(1,2,3); (1,2,5); (1,2,7); (1,2,11); (1,2,13); (1,3,2); (1,3,5); (1,3,7); (1,5,2); (1,5,3); (1,7,2); (1,7,3); (1,11,2); (1,13,2)$ Total no. of ordered pairs = 14 here

Case 2($b=1$)

If $b$ is one and $a$ 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, $abc\leq30$ ⇒ $ac\leq30$ also, $abc$ is not divisible by the square of any prime so a should not divide $c$ so all possible pairs of $(a,b,c)$ here are $(2,1,15); (3,1,10); (5,1,6)$ Total no. of ordered pairs = 3 here

Hence, total no. of triplets $(a,b,c)$ = 14+3= $\boxed{17}$ ~SANSGANKRSNGUPTA

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