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Re: STONECOLD'S MATH CHALLENGE  PS AND DS QUESTION COLLECTION.
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03 Mar 2018, 00:56
loserunderachiever wrote: Stone ,
Can you give me a solution of these two ? Sure! Let me try > Question 1) Here, just for the ease of calculation > let us assume that \(25 = a\) and \(10√6= b\) such that \(T = √a+b + √ab\)
\(T = √a+b + √ab\) \(T^2 = a+b + a  b + 2√a+b * √ab = 2a + 2√a^2b^2\)
Now, put back the values of a and b \(T^2 = 50 + 2√625600\) \(T^2 = 50 + 2*√25\) \(T^2 = 50 + 10 = 60\) \(T = √60\) \(T = 2√15\)
SMASH that C!
Question 2) Now, This one is tricky. But notice that this is an infinite loop. So there must be a major trick to solve this. We can actually replace the value in the root symbol after the first 6 with x. \(x = √(6+x)\) \(x^2 = 6+x\) \(x^2x6 = 0\) \(x^23x+2x6=0\) \(x(x3)+2(x3)=0\) \((x3)(x+2)=0\) \(x=3\) or \(x=2\)
But x cannot be negative since its a root value and a root is always passive. So x≠2
Hence x = 3 SMASH that B!
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Re: STONECOLD'S MATH CHALLENGE  PS AND DS QUESTION COLLECTION.
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03 Mar 2018, 02:13
loserunderachiever wrote: Stone ,
Can you give me a solution of these two ? First question is discussed here: https://gmatclub.com/forum/newtoughan ... l#p1029216Second question is discussed here: https://gmatclub.com/forum/newtoughan ... l#p1029228 PLEASE FOLLOW THE RULES WHEN POSTING QUESTIONS: https://gmatclub.com/forum/rulesforpo ... 33935.html Questions should NOT be posted in this thread they should be posted in respective forums! Thank you.
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Re: STONECOLD'S MATH CHALLENGE  PS AND DS QUESTION COLLECTION.
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07 Mar 2018, 23:27
Hi stonecoldjust a quick ask... this question is from Mock 1 and I am just wondering if the answer is correct. Please let me know as I am getting 270 total factors. 2^8*3^4*5^2*7^1 so basically = (8+1)(4+1)(2+1)(1+1) 48)If n is the product of all the integers from 1 to 10 exclusive,how many factors does n have? [Obscure] Spoiler: 160.



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STONECOLD'S MATH CHALLENGE  PS AND DS QUESTION COLLECTION.
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07 Mar 2018, 23:39
Bobzi wrote: Hi stonecoldjust a quick ask... this question is from Mock 1 and I am just wondering if the answer is correct. Please let me know as I am getting 270 total factors. 2^8*3^4*5^2*7^1 so basically = (8+1)(4+1)(2+1)(1+1) 48)If n is the product of all the integers from 1 to 10 exclusive,how many factors does n have? [Obscure] Spoiler: 160.
Hi, Your solution is incorrect.
See >
\(2*3*4*5*6*7*8*9= 2^7*3^4*5*7\)
=> # of factors = \(8*5*2*2 = 160\)
Your mistake > You did not read the question properly. Notice the word EXCLUSIVE. You must EXCLUDE 1 and 10.
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Re: STONECOLD'S MATH CHALLENGE  PS AND DS QUESTION COLLECTION.
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09 Mar 2018, 06:07
This is a gold mine of information!



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Re: STONECOLD'S MATH CHALLENGE  PS AND DS QUESTION COLLECTION.
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11 Apr 2018, 11:51
DebUSA wrote: This is a gold mine of information! I am glad you found it useful. I will be updating the tests soon.
All the best
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Re: STONECOLD'S MATH CHALLENGE  PS AND DS QUESTION COLLECTION.
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25 Apr 2018, 08:12
14)If p is a positive integer and 200 multiplies by p is square of an integer,what is the value of p? A)2 B)3 C)5 D)10 E)Cannot be determined. HI stonecold, \(x^2 = p*200\) In this question why can't p =2 ?
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Re: STONECOLD'S MATH CHALLENGE  PS AND DS QUESTION COLLECTION.
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26 May 2018, 22:57
Firstly, thanks for having these questions  they are a great help!
Wanted to ask, though, about the question below:
7)What is the units digit of 23^99∗14^352+9002^1003∗918^437986
I keep getting 7 * 1 + 8 * 4 = 9. I'm using cyclicity here. Thanks in advanced.



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STONECOLD'S MATH CHALLENGE  PS AND DS QUESTION COLLECTION.
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27 May 2018, 00:59
wchin24 wrote: Firstly, thanks for having these questions  they are a great help!
Wanted to ask, though, about the question below:
7)What is the units digit of 23^99∗14^352+9002^1003∗918^437986
I keep getting 7 * 1 + 8 * 4 = 9. I'm using cyclicity here. Thanks in advanced. Hey wchin24 , The mistake you did is highlighted above. 14^352 The pattern of 4 is 4,6 This means cyclicity is 2. When you divide the power with 2, you will get 0 remainder. That means last digit is last place of the pattern, which is 6 here. Hence, the last digit of 14^352 is 6. Does that make sense?
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Re: STONECOLD'S MATH CHALLENGE  PS AND DS QUESTION COLLECTION.
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27 May 2018, 01:16
abhimahna wrote: wchin24 wrote: Firstly, thanks for having these questions  they are a great help!
Wanted to ask, though, about the question below:
7)What is the units digit of 23^99∗14^352+9002^1003∗918^437986
I keep getting 7 * 1 + 8 * 4 = 9. I'm using cyclicity here. Thanks in advanced. Hey wchin24 , The mistake you did is highlighted above. 14^352 The pattern of 4 is 4,6 This means cyclicity is 2. When you divide the power with 2, you will get 0 remainder. That means last digit is last place of the pattern, which is 6 here. Hence, the last digit of 14^352 is 6. Does that make sense? OMG YES! Thank you! I was tripping up since I was seeing it as 4^0, but in reality this just means that it'll end up at the last place of the pattern here.



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27 May 2018, 20:31
NandishSS wrote: 14)If p is a positive integer and 200 multiplies by p is square of an integer,what is the value of p? A)2 B)3 C)5 D)10 E)Cannot be determined. HI stonecold, \(x^2 = p*200\) In this question why can't p =2 ? Hi, You are right in saying that p CAN be 2. But you missing the big picture here. Ask yourself, is that the only possible value of p? What if p is \(2*5^2\) or \(2*11^2\) or \(2*101^2\) ?
You see, there are various value of the variable p that are possible in thus question. Hence the OA is E.
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02 Aug 2018, 20:30
121)Data Sufficiency>How many divisors does the positive integer N have?
(1)27N^3 has 16 factors. (2)90<N^3<200
as explained, how can we say that N is prime from Statement 1. N could be 3^4 as well. If N= 3^4 then 27N^3 = 3^3*((3^4)^3)= 3^3*3^12= 3^15 Hence Factors will be 15+1=16. And Nos of Divisor of N = 3^4 is Five. Not 2 as in the case if N is prime.
Please explain if anything wrong in my understanding



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Re: STONECOLD'S MATH CHALLENGE  PS AND DS QUESTION COLLECTION.
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28 Jan 2019, 02:38
Hi stonecoldFor question 18 It looks like answer should be C and not E as only possible option after combining both the statement is P = 3, Could you please advise if I am missing something. chetan2u, Bunuel, VeritasKarishma, gmatbusters, amanvermagmatstonecold wrote: 18)Data Sufficiency>What is the value of positive integer p? A)300 multiplied by p is square of an integer. B)p is a factor of 75
Spoiler: :: E. Combing the two statements > p can be 3 or 3*5^2
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Re: STONECOLD'S MATH CHALLENGE  PS AND DS QUESTION COLLECTION.
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28 Jan 2019, 02:42
Gmatprep550 wrote: Hi stonecoldFor question 18 It looks like answer should be C and not E as only possible option after combining both the statement is P = 3, Could you please advise if I am missing something. chetan2u, Bunuel, VeritasKarishma, gmatbusters, amanvermagmatstonecold wrote: 18)Data Sufficiency>What is the value of positive integer p? A)300 multiplied by p is square of an integer. B)p is a factor of 75
Spoiler: :: E. Combing the two statements > p can be 3 or 3*5^2 No, both 3 and 75 will fit in.. A)300 multiplied by p is square of an integer...300*3=900=30^2 and 300*75=22500=150^2 B)p is a factor of 75... Factors of 75 are 1,3,5,15,25,75, so both 3 and 75 fit in
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Re: STONECOLD'S MATH CHALLENGE  PS AND DS QUESTION COLLECTION.
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28 Jan 2019, 04:27
Hi chetan2u, Bunuel, VeritasKarishma, Gladiator59, generisFor question 23 statement 78 I am only able to find that "All the prime numbers greater than 3 can be written as either 6n+1 or 6n1." Not able to find anything for 6+1 and 4n+1 or 4n1. Could you please review and advise if I am missing something. stonecold wrote: 23)Which of the following statements must be true> 1)A prime number must be positive. 2)For any prime number p,there is no x such that 1<x<p and x is a divisor of p. 3)The product of first ten primes is even. 4)All prime numbers greater than 71 are odd. 5)2 and 3 are the only consecutive integers that are also prime numbers. 6)p is a prime number and x and y are positive integers.If p=x*y then one out of x or y must be one. 7)All the prime numbers greater than 3 can be written as either 4n+1 or 4n1. 8)All the prime numbers greater than 3 can be written as either 6+1 or 6n1.
Spoiler: :: All statements are true.
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28 Jan 2019, 05:09
Gmatprep550 wrote: Hi chetan2u, Bunuel, VeritasKarishma, Gladiator59, generisFor question 23 statement 78 I am only able to find that "All the prime numbers greater than 3 can be written as either 6n+1 or 6n1." Not able to find anything for 6+1 and 4n+1 or 4n1. Could you please review and advise if I am missing something. stonecold wrote: 23)Which of the following statements must be true> 1)A prime number must be positive. 2)For any prime number p,there is no x such that 1<x<p and x is a divisor of p. 3)The product of first ten primes is even. 4)All prime numbers greater than 71 are odd. 5)2 and 3 are the only consecutive integers that are also prime numbers. 6)p is a prime number and x and y are positive integers.If p=x*y then one out of x or y must be one. 7)All the prime numbers greater than 3 can be written as either 4n+1 or 4n1. 8)All the prime numbers greater than 3 can be written as either 6+1 or 6n1.
Spoiler: :: All statements are true. Hi.. All prime numbers have to be of type of 6n+1 or 6n1 Now 4n+1 and 4n1 is nothing but set of all odd numbers.. When n=1, 4n+1 and 4n1 becomes 3 and 5 and when n=2, 4n+1 and 4n1 becomes 7 and 9..so 3,5,7,9,11,... And all primes above 2 are odd numbers..so primes are also of type 4n+1 or 4n1, basically they will be odd
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Re: STONECOLD'S MATH CHALLENGE  PS AND DS QUESTION COLLECTION.
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28 Jan 2019, 05:43
Thanks chetan2u for valuable response. It helped me
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STONECOLD'S MATH CHALLENGE  PS AND DS QUESTION COLLECTION.
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28 Jan 2019, 09:24
As perfectly explained by chetan2u Sir, All prime numbers greater than 3 can be written as 6n+/1 or 4n+/1. But vice versa is not truefor example, for n =4, 6*4+1= 25 , but it is not prime Similarly, for n = 2, 4n+1 = 2*4+1= 9, which is not prime In fact, there is no direct formula, which can find whether a number is prime or not. All prime numbers greater than 3 can be written as 6n+/1 or 4n+/1, but all numbers in form of 6n+/1 or 4n+/1 are not prime. Gmatprep550 wrote: Hi chetan2u, Bunuel, VeritasKarishma, Gladiator59, generisFor question 23 statement 78 I am only able to find that "All the prime numbers greater than 3 can be written as either 6n+1 or 6n1." Not able to find anything for 6+1 and 4n+1 or 4n1. Could you please review and advise if I am missing something. stonecold wrote: 23)Which of the following statements must be true> 1)A prime number must be positive. 2)For any prime number p,there is no x such that 1<x<p and x is a divisor of p. 3)The product of first ten primes is even. 4)All prime numbers greater than 71 are odd. 5)2 and 3 are the only consecutive integers that are also prime numbers. 6)p is a prime number and x and y are positive integers.If p=x*y then one out of x or y must be one. 7)All the prime numbers greater than 3 can be written as either 4n+1 or 4n1. 8)All the prime numbers greater than 3 can be written as either 6+1 or 6n1.
Spoiler: :: All statements are true.
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Re: STONECOLD'S MATH CHALLENGE  PS AND DS QUESTION COLLECTION.
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18 Feb 2019, 14:18
@14)If p is a positive integer and 200 multiplies by p is square of an integer,what is the value of p? A)2 B)3 C)5 D)10 E)Cannot be determined.
Hi, can you explain, how have you written p=2*2^2 expression in the first set of questions, numbered 1517.. What does this expression mean? i am confused in this and the DS questions that contain 300 and 200 too




Re: STONECOLD'S MATH CHALLENGE  PS AND DS QUESTION COLLECTION.
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