you can simply find the digital root of the number i.e. the sum of all digits till you get a single digit number. If the digital root is not 1,4,7 or 9 then its not a perfect square. Here, the digital root is 3 in each of the statements, so the answer is D
Take any no. 0's 1's and 2's are always repeated 4 times hence every time the sum of digits will be 12. By the divisibility rule of 3 if sum of digits is divisible by 3 then number is divisible by 3. Plus if it is a sq number, all the primes would be in even powers on prime factorizing it. Hence n must be divisible by 3^2 i.e 9. By divisibility rule of 9 stating that the sum of digits must be divisible by 9 we see 12 is not divisible hence the number is not a perfect sq no. I hope this makes things more clear for you 😊
Just 1 doubt - Why did you leave 2 and consider only the prime number 3..?? What if I consider the number to be 120120120012 and Prime number 2 ..? or maybe I consider the number to be 120120121200 and prime number 5 .. The numbers are both divisible by (2 and 4) and (5 and 25) respetively..!! But its still not a perfect square...!! :( Where am I going wrong..??
Hello Neermalya In the first instance, 2 and 4 divided it. 2 is a prime factor of the number and its square also divided it. But is 2 the only prime factor of the number. 3 is also a prime factor of the number. 3 divides the number, but the square of 3 does not divide it. Only if all the prime factors that divide a number and the square of those prime factors divide the number will it be a perfect square. The same explanation holds good for the second example you have provided. 5 and 25 divide. Is 5 the only prime factor of the number. No, 3 is also a prime factor of the number. I chose 3 because, any permutation of the digits will have 3 as a prime factor - thanks to the way test of divisibility for 3 works. Please comment if the explanation above is not clear and you require more clarifications. Cheers
@@Wizako thank you for the Lucid explanation..!! I understood now.. Actually in the video, it wasnt mentioned that the number must be divisible by "all" of its primes and the corresponding squares of "all" its primes... hence i mistook it to check for just any 1 prime..!! Now its clear..!! Thank you
It is the other way around. The number that you have to consider should be a perfect square. For eg. Consider 400(perfect square), 400 is divisible by a prime number 5 and is divisble by the square of the prime (i.e. 25). The number that you have to consider should be a perfect square. This statement would be helpful in solving for large numbers like the one in this question.
But in an yes no data sufficiency questions we have to try alternatives to get both yes and no. But you simply took a number and did the data sufficiency. But there might be another number which might give an yes for the statements
Take any no. 0's 1's and 2's are always repeated 4 times hence every time the sum of digits will be 12. By the divisibility rule of 3 if sum of digits is divisible by 3 then number is divisible by 3. Plus if it is a sq number, all the primes would be in even powers on prime factorizing it. Hence n must be divisible by 3^2 i.e 9. By divisibility rule of 9 stating that the sum of digits must be divisible by 9 we see 12 is not divisible hence the number is not a perfect sq no. I hope this makes things more clear for you 😊
Any 12 digit number that is written using 4 1s, 4 0s and 4 2s will be divisible by 3. That is because test of divisibility for 3 is that the sum of the digits of the number is divisible by 3. So, we will not find a single 12 digit number that satisfies this condition that will not be divisible by 3. Again, none of those numbers - not even one 12 digit number that is written using 4 1s, 4 2s, and 4 0s will be divisible by 9. Test of divisibility for 9 is that the sum of the digits should be divisible by 9. You are right about not concluding based on an example. In this case, I did not take an example - I took the test of divisibility to prove a general case. Please ping back if the explanation is not clear. @Mudrit Sood has also posted a reply on the same lines.
you can simply find the digital root of the number i.e. the sum of all digits till you get a single digit number. If the digital root is not 1,4,7 or 9 then its not a perfect square. Here, the digital root is 3 in each of the statements, so the answer is D
Take any no. 0's 1's and 2's are always repeated 4 times hence every time the sum of digits will be 12. By the divisibility rule of 3 if sum of digits is divisible by 3 then number is divisible by 3. Plus if it is a sq number, all the primes would be in even powers on prime factorizing it. Hence n must be divisible by 3^2 i.e 9. By divisibility rule of 9 stating that the sum of digits must be divisible by 9 we see 12 is not divisible hence the number is not a perfect sq no.
I hope this makes things more clear for you 😊
Thanks
Just 1 doubt - Why did you leave 2 and consider only the prime number 3..??
What if I consider the number to be 120120120012 and Prime number 2 ..?
or maybe I consider the number to be 120120121200 and prime number 5 ..
The numbers are both divisible by (2 and 4) and (5 and 25) respetively..!! But its still not a perfect square...!! :(
Where am I going wrong..??
Hello Neermalya
In the first instance, 2 and 4 divided it. 2 is a prime factor of the number and its square also divided it.
But is 2 the only prime factor of the number. 3 is also a prime factor of the number. 3 divides the number, but the square of 3 does not divide it.
Only if all the prime factors that divide a number and the square of those prime factors divide the number will it be a perfect square.
The same explanation holds good for the second example you have provided.
5 and 25 divide. Is 5 the only prime factor of the number. No, 3 is also a prime factor of the number.
I chose 3 because, any permutation of the digits will have 3 as a prime factor - thanks to the way test of divisibility for 3 works.
Please comment if the explanation above is not clear and you require more clarifications.
Cheers
@@Wizako thank you for the Lucid explanation..!! I understood now..
Actually in the video, it wasnt mentioned that the number must be divisible by "all" of its primes and the corresponding squares of "all" its primes... hence i mistook it to check for just any 1 prime..!!
Now its clear..!! Thank you
27 is divisible by 3, and 9 , but 27 is perfect square ?
It is the other way around. The number that you have to consider should be a perfect square. For eg. Consider 400(perfect square), 400 is divisible by a prime number 5 and is divisble by the square of the prime (i.e. 25). The number that you have to consider should be a perfect square. This statement would be helpful in solving for large numbers like the one in this question.
@@Wizako appreciate you explanation sir.
But in an yes no data sufficiency questions we have to try alternatives to get both yes and no. But you simply took a number and did the data sufficiency. But there might be another number which might give an yes for the statements
Take any no. 0's 1's and 2's are always repeated 4 times hence every time the sum of digits will be 12. By the divisibility rule of 3 if sum of digits is divisible by 3 then number is divisible by 3. Plus if it is a sq number, all the primes would be in even powers on prime factorizing it. Hence n must be divisible by 3^2 i.e 9. By divisibility rule of 9 stating that the sum of digits must be divisible by 9 we see 12 is not divisible hence the number is not a perfect sq no.
I hope this makes things more clear for you 😊
Any 12 digit number that is written using 4 1s, 4 0s and 4 2s will be divisible by 3. That is because test of divisibility for 3 is that the sum of the digits of the number is divisible by 3.
So, we will not find a single 12 digit number that satisfies this condition that will not be divisible by 3.
Again, none of those numbers - not even one 12 digit number that is written using 4 1s, 4 2s, and 4 0s will be divisible by 9. Test of divisibility for 9 is that the sum of the digits should be divisible by 9.
You are right about not concluding based on an example. In this case, I did not take an example - I took the test of divisibility to prove a general case.
Please ping back if the explanation is not clear. @Mudrit Sood has also posted a reply on the same lines.