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26 Jun 2023, 12:36
Kudos
Bookmarks
B
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:
15% (03:06) correct
85%
(02:44)
wrong
based on 48
sessions
History
Date
Time
Result
Not Attempted Yet
If \(x = 40^3 - 17^3 - 23^3\), how many positive factors does x have?
A. 32
B. 42
C. 64
D. 72
E. 80
Out of Scope Question. Just for fun!
A. 32
B. 42
C. 64
D. 72
E. 80
Out of Scope Question. Just for fun!
26 Jun 2023, 13:28
Kudos
Bookmarks
Bunuel
\(a^3 - b^3 = (a-b)(a^2 + b^2 + ab)\)
\(40^3 - 17^3 = (40 - 17)(40^2 + 17^2 + (40*17))\)
\(40^3 - 17^3 = (23)(1600 + 289 +680)\)
\(40^3 - 17^3 = (23)(2569)\)
\(x = 40^3 - 17^3 - 23^3\)
\(x = (23)(2569) - 23^3\)
\(x = (23)(2569 - 23^2)\)
\(x = (23)(2569 - 529)\)
\(x = (23)(2040)\)
Let's prime factorize 2040
\(x = 23*2^3*5^1*3^1*17^1\)
Number of factors = 2 * 4 * 8 = 64
Option C
27 Jun 2023, 00:22
Kudos
Bookmarks
Bunuel
Using the cubic formula \(a^3 + b^3 = (a+b)^3 - 3ab(a+b)\)
\(40^3 - 17^3 - 23^3\)
\(40^3 - (17^3 + 23^3)\)
\(40^3 - [ (17+23)^3\) - 3 x (17 x 23) (17+23) ]
\(40^3 - [ (40)^3\) - 3 x 17 x 23 x 40 ]
\(40^3 - 40^3\) + 3 x 17 x 23 x 40
3 x 17 x 23 x \(2^3\) x 5
This is now in the prime factor form and we can find the number of factors using the formula:
Total Number of Factors for N = (a+1) (b+1) (c+1), where N= \(p^a\) * \(q^b\) * \(r^c\) and p, q, r are prime factors of N
Total factors for \(2^3\) x 3 x 5 x 17 x 23
= (3+1) (1+1) (1+1) (1+1) (1+1)
= 4 x 2 x 2 x 2 x 2
= 64
