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14 May 2019, 00:16
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
34% (02:03) correct
66%
(02:28)
wrong
based on 149
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Date
Time
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How many positive integer divisors of 201^9 are perfect squares or perfect cubes (or both)?
A. 32
B. 36
C. 37
D. 39
E. 41
A. 32
B. 36
C. 37
D. 39
E. 41
14 May 2019, 00:50
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Bunuel
201^9 = 67^9 x 3^9
For perfect square positive divisors, adjust the above as (67^2)^4 x (3^2)^4 x 67 x 3
No. of Positive divisors = (4+1)*(4+1) = 25
For perfect cube positive divisors, adjust the above as (67^3)^3 x (3^3)^3
No. of Positive divisors = (3+1)*(3+1) = 16
The above 2 cases counts perfect square and cube (which is power 6 twice), so the divisors which are power 6 needs to be subtracted once,
For perfect power 6 positive divisors, adjust the above as (67^6)^1 x (3^6)^1 x 67^3 x 3^3
No. of Positive divisors = (1+1)*(1+1) = 4
So, total = 25 + 16 - 4 = 37.
C.
General Discussion
07 Sep 2020, 18:59
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2 Methods to Find how many Factors of a Number are Perfect Square or a Perfect Cube:
Method 1 - Direct Method
1st)Take the Square Root of the Number's Prime Factorization (for finding Perfect Squares) and take the Cube Root of the Number's Prime Factorization (for finding Perfect Cubes)
2nd)then Find the Total No. of Factors like you would normally
*Note* The Square Root Fractional Exponent = ^(1/2)
and
The Cube Root Fractional Exponent = ^(1/3)
Similar to looking for the how many Prime Bases divide into a N!-Factorial!, you only take the Quotient after Multiplying the Powers
Method 2 - More Conceptual Way
this is probably the better way because we will have cross-over/overlap of Factors that are BOTH Perfect Squares and Perfect Cubes
(201)^9 = 3^9 * 67^9
1st: How many Factors of the Number are PERFECT SQUARES
Rule: Perfect Squares must always have as its Prime Bases EVEN Exponents
The Goal is to find out how many options of each Prime Base we can use, then determine the Count of Different Combinations possible that can "Create" a Factor of the Number that is also a Perfect Square.
Prime Base 2: 2^0 -or- 2^2 -or- 2^4 -or- 2^6 -or- 2^8 ---- 5 Options
and
Prime Base 67: 67^0 -or- 67^2 -or- 67^4 -or- 67^6 -or- 67^8 ---- 5 Options
5 * 5 = 25 Different Combinations of Prime Bases that will Create a Perfect Square
2nd: How many of the Factors of the Number are also Perfect CUBES
Rule: Every Perfect Cube must only have Multiple of 3 as the Exponents of its Prime Bases
Using similar logic:
Prime Base 2: 2^0 -or- 2^3 -or- 2^6 -or- 2^9 ----- 4 Options
and
Prime Base 67: 67^0 -or- 67^3 -or- 67^6 -or- 67^9 --- 4 Options
4 * 4 = 16 Factors that are also Perfect Cubes
3rd: Remove the OVERLAP
When we counted the 25 Perfect Squares, we already counted 4 of the Combinations that were included in the Count of Perfect Cubes
2'0 * 67'0
2'0 * 67'6
2'6 * 67'0
2'6 * 67'6
We must SUBTRACT these 4 From the 16 Perfect Cubes that we found.
Notice that these 4 above are BOTH Perfect Squares and Perfect Cubes
Answer = 25 + 16 - 4 = 37
37 Factors of the Number that are: Perfect Squares -OR- Perfect Cubes -OR- BOTH
Answer -C-
EDIT: Can someone please advise whether this is a proper GMAT type question?
Method 1 - Direct Method
1st)Take the Square Root of the Number's Prime Factorization (for finding Perfect Squares) and take the Cube Root of the Number's Prime Factorization (for finding Perfect Cubes)
2nd)then Find the Total No. of Factors like you would normally
*Note* The Square Root Fractional Exponent = ^(1/2)
and
The Cube Root Fractional Exponent = ^(1/3)
Similar to looking for the how many Prime Bases divide into a N!-Factorial!, you only take the Quotient after Multiplying the Powers
Method 2 - More Conceptual Way
this is probably the better way because we will have cross-over/overlap of Factors that are BOTH Perfect Squares and Perfect Cubes
(201)^9 = 3^9 * 67^9
1st: How many Factors of the Number are PERFECT SQUARES
Rule: Perfect Squares must always have as its Prime Bases EVEN Exponents
The Goal is to find out how many options of each Prime Base we can use, then determine the Count of Different Combinations possible that can "Create" a Factor of the Number that is also a Perfect Square.
Prime Base 2: 2^0 -or- 2^2 -or- 2^4 -or- 2^6 -or- 2^8 ---- 5 Options
and
Prime Base 67: 67^0 -or- 67^2 -or- 67^4 -or- 67^6 -or- 67^8 ---- 5 Options
5 * 5 = 25 Different Combinations of Prime Bases that will Create a Perfect Square
2nd: How many of the Factors of the Number are also Perfect CUBES
Rule: Every Perfect Cube must only have Multiple of 3 as the Exponents of its Prime Bases
Using similar logic:
Prime Base 2: 2^0 -or- 2^3 -or- 2^6 -or- 2^9 ----- 4 Options
and
Prime Base 67: 67^0 -or- 67^3 -or- 67^6 -or- 67^9 --- 4 Options
4 * 4 = 16 Factors that are also Perfect Cubes
3rd: Remove the OVERLAP
When we counted the 25 Perfect Squares, we already counted 4 of the Combinations that were included in the Count of Perfect Cubes
2'0 * 67'0
2'0 * 67'6
2'6 * 67'0
2'6 * 67'6
We must SUBTRACT these 4 From the 16 Perfect Cubes that we found.
Notice that these 4 above are BOTH Perfect Squares and Perfect Cubes
Answer = 25 + 16 - 4 = 37
37 Factors of the Number that are: Perfect Squares -OR- Perfect Cubes -OR- BOTH
Answer -C-
EDIT: Can someone please advise whether this is a proper GMAT type question?
