Hi sir .... Thank you so much for the amezing explanation ... BTW ... I took an alternate approach to solve this one. Sharing it here. We know that prime no.s should be in the form 6N(+/-)1 for any Prime no. greater than 6 ... Hence if we draw a no. line placing p, p+2 and p-2 then by analogy p+2 must be in the form 6N-1 and p-2 must be in the form 6N+1 as only 3 integers can exist between them ... So in any way P must be of the form 6N (+/-) 3 ... And must be divisible by 3 ... Same approach for 2 nd condition ...
The idea is right. But, between 6N - 1 and 6N + 1 you will have only one integer 6N, if N is the same in both cases. However, between p - 2 and p + 2, you will have 3 integers. You may want to look at the numbers as 6N + 1 and 6N + 5.
Dear professor, could you please explain the rationale behind choosing the number 3 in 4:13 "property: one out of 3 consecutive odd integers is divisible by 3"?. Thank you in advance for your explanation. Your answer will be much appreciated.
Hi, Take any set of 3 consecutive odd integers. One of them will be definitely divisible by 3. For instance, take 23, 25, 27 - 27 is divisible by 3 91, 93, 95 - 93 is divisible by 3
@@Wizako Thank you professor for your explanation. However, why not choose a different number such as 5 or 7? Thank you in advance for your explanation. Your answer will be much appreciated.
hey, i have another proposed solution for this question adding the statements (P+2) + (P-2)= prime +prime, 2P = 2*prime number, P=prime number, I just came up with this, although i know this is not mathematically sound.
Hello Pratik, An interesting workaround you've come up with here. There is no two-digit number that satisfies that equation, and so you can immediately conclude that P is not prime. The only triplet of P-2, P, P+2, all of which are prime, is 3,5,7, and they're all single-digit numbers. Well done.
no 2 digit prime numbers can be divisible by 3 @ 4:21. Is this a fact or there is some logic which can be inferred from there for other numbers as well ?
If a number is divisible by 3, it cannot be prime unless the number is 3. 3 is a single digit number. So, any two digit number that is divisible by 3 will have 1, the number, and 3 as factors. So, it will not be prime.
Dear Professor, could you please explain the rationale behind using p+1 and p-1?. alternatively, is it possible to arrive at the same result by using the consecutive numbers p-4,p,p+4 in an AP with a d=4?. thank you in advance for your answer, your feedback is much appreciated.
(p - 4), p, and (p + 4) will be in an AP with d = 4. We can deduce that (p - 4) and (p + 4) are not divisible by 3. Now if we are to make an assertion that p will, therefore, be divisible by 3 - the dots remain unconnected. When we bring in (p - 1) and (p + 1) in the picture, the reasoning is complete.
Hello Baskar Sir, Isn't it a right way to put the values of P-2 and P+2 equal to some prime numbers and then deduce the values of P? For ex, if I put the value of P-2 = 17, I will get P=19 which is a prime number and if I put P-2 = 13, I get P=15, which is not a prime number.... So, can't I say that statement one is not sufficient enough??? Thanks in advance...
Hello Abbas Two points to note 1. In the example you took: P - 2 = 17 is prime. P = 19 is prime. But P + 2 = 21 is not prime. Any example in which P - 2 and P + 2 are not simultaneously prime will not be valid ones. 2. If you pick the right example, you will realize for quite a few the answer that you get is that P is not prime. But that leaves one with a nagging feeling as to whether a value exists in which P is prime. The approach taken in the video is a generic approach using properties of numbers. So, it will work in all situations.
I did it the dumb way and wrote out like first 20 prime numbers and saw theres no 3 prime numbers separated by 2 each or 4 each, and thought its save to conclude that both conditions are impossible and answered D :P
If we take 10 seconds to write down the first 10 or so positive 2 digit prime numbers, it takes another 10 seconds to notice 2 clear pattern: (1) Either (p-2) will be prime or (p+2) (2) Either (p-4) will be prime or (p+4) Both observations (1) and (2) combined, we can go with option C.
Absolutely spot on!! But our focus was on developing the process of solving the question. GMAT may or may not test with integers under 10. This method might help you in any question that is of this type. Brute force method like yours is very helpful in certain questons but not so in a few. so it is better to always learn to solve methodically as well. Cheers!!!
A gem of a question, offering learnings on multiple dimensions!!! Thanks .
Amazing sir, loved the explanation.. 💯 thanks for posting a 700 level question.. eager to have more..
Thanks Mudrit. I am shooting another 700 level question for coming Thursday. Cheers.
You are too good!!! a teacher Sir!!! 👌👌👌 par excellence!!! Sir!!! 🙏🙏 I will be joining up soon Sir.👍👍
Thanks Prasenjit for your feedback.
incredibly helpful, you're an awesome teacher thank you!
Hi sir .... Thank you so much for the amezing explanation ... BTW ... I took an alternate approach to solve this one. Sharing it here. We know that prime no.s should be in the form 6N(+/-)1 for any Prime no. greater than 6 ... Hence if we draw a no. line placing p, p+2 and p-2 then by analogy p+2 must be in the form 6N-1 and p-2 must be in the form 6N+1 as only 3 integers can exist between them ... So in any way P must be of the form 6N (+/-) 3 ... And must be divisible by 3 ... Same approach for 2 nd condition ...
The idea is right.
But, between 6N - 1 and 6N + 1 you will have only one integer 6N, if N is the same in both cases.
However, between p - 2 and p + 2, you will have 3 integers.
You may want to look at the numbers as 6N + 1 and 6N + 5.
Yes sir ... BTW ... Thank you so much for replying ...
Dear professor, could you please explain the rationale behind choosing the number 3 in 4:13 "property: one out of 3 consecutive odd integers is divisible by 3"?. Thank you in advance for your explanation. Your answer will be much appreciated.
Hi,
Take any set of 3 consecutive odd integers. One of them will be definitely divisible by 3.
For instance, take 23, 25, 27 - 27 is divisible by 3
91, 93, 95 - 93 is divisible by 3
@@Wizako Thank you professor for your explanation. However, why not choose a different number such as 5 or 7? Thank you in advance for your explanation. Your answer will be much appreciated.
Hi Baskar Sir, I am planning to take my gmat before October. Could I get in touch with you regarding the prep if possible?
Sure. Please send an email to learn at wizako dot com with your queries.
hey, i have another proposed solution for this question
adding the statements (P+2) + (P-2)= prime +prime, 2P = 2*prime number, P=prime number, I just came up with this, although i know this is not mathematically sound.
Hello Pratik,
An interesting workaround you've come up with here. There is no two-digit number that satisfies that equation, and so you can immediately conclude that P is not prime.
The only triplet of P-2, P, P+2, all of which are prime, is 3,5,7, and they're all single-digit numbers.
Well done.
no 2 digit prime numbers can be divisible by 3 @ 4:21. Is this a fact or there is some logic which can be inferred from there for other numbers as well ?
If a number is divisible by 3, it cannot be prime unless the number is 3. 3 is a single digit number. So, any two digit number that is divisible by 3 will have 1, the number, and 3 as factors. So, it will not be prime.
Definition of prime number is your answer
Dear Professor, could you please explain the rationale behind using p+1 and p-1?. alternatively, is it possible to arrive at the same result by using the consecutive numbers p-4,p,p+4 in an AP with a d=4?. thank you in advance for your answer, your feedback is much appreciated.
(p - 4), p, and (p + 4) will be in an AP with d = 4. We can deduce that (p - 4) and (p + 4) are not divisible by 3. Now if we are to make an assertion that p will, therefore, be divisible by 3 - the dots remain unconnected. When we bring in (p - 1) and (p + 1) in the picture, the reasoning is complete.
Hello Baskar Sir,
Isn't it a right way to put the values of P-2 and P+2 equal to some prime numbers and then deduce the values of P? For ex, if I put the value of P-2 = 17, I will get P=19 which is a prime number and if I put P-2 = 13, I get P=15, which is not a prime number.... So, can't I say that statement one is not sufficient enough???
Thanks in advance...
Hello Abbas
Two points to note
1. In the example you took: P - 2 = 17 is prime. P = 19 is prime. But P + 2 = 21 is not prime. Any example in which P - 2 and P + 2 are not simultaneously prime will not be valid ones.
2. If you pick the right example, you will realize for quite a few the answer that you get is that P is not prime. But that leaves one with a nagging feeling as to whether a value exists in which P is prime. The approach taken in the video is a generic approach using properties of numbers. So, it will work in all situations.
@@Wizako Understood... thanks for the explanation
I did it the dumb way and wrote out like first 20 prime numbers and saw theres no 3 prime numbers separated by 2 each or 4 each, and thought its save to conclude that both conditions are impossible and answered D :P
Elon 😮
Hi Sir Why to introduce (P-1) and (P+1) for statement 2.Why not use the same logic of statement 1.Out of Ap with d=4 one Wil be a multiple of 3
Hello Premkumar
Certainly possible. And it will work for all values of 'd' as well.
If we take 10 seconds to write down the first 10 or so positive 2 digit prime numbers, it takes another 10 seconds to notice 2 clear pattern:
(1) Either (p-2) will be prime or (p+2)
(2) Either (p-4) will be prime or (p+4)
Both observations (1) and (2) combined, we can go with option C.
Absolutely spot on!! But our focus was on developing the process of solving the question. GMAT may or may not test with integers under 10. This method might help you in any question that is of this type. Brute force method like yours is very helpful in certain questons but not so in a few. so it is better to always learn to solve methodically as well. Cheers!!!
Wizako GMAT Prep Sure. Thanks