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

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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION.  [#permalink]

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New post 02 Mar 2018, 23:56
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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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION.  [#permalink]

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New post 03 Mar 2018, 01:13
2
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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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION.  [#permalink]

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New post 07 Mar 2018, 22:27
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.
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STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION.  [#permalink]

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New post 07 Mar 2018, 22:39
1
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,
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.  [#permalink]

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New post 09 Mar 2018, 05:07
This is a gold mine of information!
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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION.  [#permalink]

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New post 11 Apr 2018, 10:51
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Re: STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION.  [#permalink]

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New post 25 Apr 2018, 07:12
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.  [#permalink]

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New post 26 May 2018, 21:57
1
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.  [#permalink]

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New post 26 May 2018, 23:59
2
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.  [#permalink]

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New post 27 May 2018, 00:16
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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STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION.  [#permalink]

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New post 27 May 2018, 19:31
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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STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION.  [#permalink]

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New post 02 Aug 2018, 19:30
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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STONECOLD'S MATH CHALLENGE - PS AND DS QUESTION COLLECTION. &nbs [#permalink] 02 Aug 2018, 19:30

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