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# STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION.

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BSchool Forum Moderator
Joined: 12 Aug 2015
Posts: 2494
GRE 1: 323 Q169 V154
Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION. [#permalink]

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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 $$T = √a+b + √a-b$$

$$T = √a+b + √a-b$$
$$T^2 = a+b + a - b + 2√a+b * √a-b = 2a + 2√a^2-b^2$$

Now, put back the values of a and b
$$T^2 = 50 + 2√625-600$$
$$T^2 = 50 + 2*√25$$
$$T^2 = 50 + 10 = 60$$
$$T = √60$$
$$T = 2√15$$

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.
$$x = √(6+x)$$
$$x^2 = 6+x$$
$$x^2-x-6 = 0$$
$$x^2-3x+2x-6=0$$
$$x(x-3)+2(x-3)=0$$
$$(x-3)(x+2)=0$$
$$x=3$$ 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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Joined: 02 Sep 2009
Posts: 44290
Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION. [#permalink]

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03 Mar 2018, 02:13
2
KUDOS
Expert's post
loserunderachiever wrote:
Stone ,

Can you give me a solution of these two ?

First question is discussed here: https://gmatclub.com/forum/new-tough-an ... l#p1029216

Second question is discussed here: https://gmatclub.com/forum/new-tough-an ... l#p1029228

PLEASE FOLLOW THE RULES WHEN POSTING QUESTIONS: https://gmatclub.com/forum/rules-for-po ... 33935.html Questions should NOT be posted in this thread they should be posted in respective forums! Thank you.
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Joined: 10 Feb 2017
Posts: 14
Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION. [#permalink]

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07 Mar 2018, 23:27
Hi stonecold

just 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.
BSchool Forum Moderator
Joined: 12 Aug 2015
Posts: 2494
GRE 1: 323 Q169 V154
STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION. [#permalink]

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07 Mar 2018, 23:39
Bobzi wrote:
Hi stonecold

just 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,

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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Intern
Joined: 09 Oct 2017
Posts: 2
Location: India
WE: Analyst (Consulting)
Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION. [#permalink]

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09 Mar 2018, 06:07
This is a gold mine of information!
Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION.   [#permalink] 09 Mar 2018, 06:07

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