### Dhaka Secondary 2011/2 (Junior 2011/4)

Posted:

**Fri Jan 28, 2011 9:44 pm**$A$ is the product of seven odd prime numbers. $A \times B$ is a perfect even square. What is the minimum number of prime factors of $B$?

The Official Online Forum of BdMO

https://matholympiad.org.bd/forum/

Page **1** of **1**

Posted: **Fri Jan 28, 2011 9:44 pm**

$A$ is the product of seven odd prime numbers. $A \times B$ is a perfect even square. What is the minimum number of prime factors of $B$?

Posted: **Fri Jan 28, 2011 10:04 pm**

are the seven odd prime numbers distinct?

Posted: **Fri Jan 28, 2011 11:33 pm**

yes, they are distinct

Posted: **Fri Jan 28, 2011 11:46 pm**

then the ans is 8.

Let, $A=p_1.p_2...p_7$ none of them are same or 2. So $B=2.2.p_1.p_2...p_7$ then the ans is 8

Let, $A=p_1.p_2...p_7$ none of them are same or 2. So $B=2.2.p_1.p_2...p_7$ then the ans is 8

Posted: **Fri Aug 05, 2011 2:55 am**

bro i did not get your solution, can u describe it even more clearly??@hasib bro

Posted: **Wed Nov 30, 2011 12:46 pm**

shifat, i am giving an example. suppose $A= 3\times 5\times 7\times 11\times 13\times 17\times 19$ and as $A\times B$ is an even perfect square, it should look like this $A\times B= 2^2\times 3^2\times 5^2\times 7^2\times 11^2\times 13^2\times 17^2\times 19^2\times $(any other perfect square) . thus we get $B=2^2\times 3\times 5\times 7\times 11\times 13\times 17\times 19\times $(any other perfect square). so the minimum prime factors of $B$ are $2$, $3$, $5$, $7$, $11$, $13$, $17$, $19$. so the answer is $8$. Hasib has just used $p_{1}$, $ p_{2}$, $ p_{3} $..... in lieu of $3$, $5$, $7$.... that's it

Posted: **Sun Dec 18, 2011 1:46 am**

oh, got it, thanks.....

Posted: **Thu Jan 07, 2016 2:58 pm**

What's a perfect even square?

Posted: **Sun Jan 10, 2016 4:34 pm**

Perfect even squares = The squares of even numbers.