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19 Feb 2021, 05:45
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
54% (02:44) correct
46%
(02:24)
wrong
based on 129
sessions
History
Date
Time
Result
Not Attempted Yet
What is the number of positive integers that are divisors of at least one of the number 10^10; 15^7 and 18^11 ?
A. 17
B. 85
C. 125
D. 250
E. 435
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
A. 17
B. 85
C. 125
D. 250
E. 435
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
Updated on: 19 Feb 2021, 07:53
Originally posted by CrackverbalGMAT on 19 Feb 2021, 07:05.
Last edited by CrackverbalGMAT on 19 Feb 2021, 07:53, edited 1 time in total.
Last edited by CrackverbalGMAT on 19 Feb 2021, 07:53, edited 1 time in total.
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Bunuel
Hi Bunuel, can you please check the answer options? IMO it should be 435
For any number, N broken into prime factors and their powers in the form \(N = a^p * b^q*c^r\) and so on, then the total number of factors n = (p + 1) * (q + 1) * (r + 1)...
\(10^{10} = 2^{10} * 5^{10}\), \(15^7 = 3^7 * 5^7\) and \(18^{11} = 2^{11} * (3^2)^{11} = 2^{11} * 3^{22}\)
The total number of divisors of \(2^{10} * 5^{10}\) = (10 + 1) * (10 + 1) = 121
The total number of divisors of \(3^7 * 5^7\) = (7 + 1) * (7 + 1) = 64
The total number of divisors of \(2^{11} * 3^{22}\) = (11 + 1) * (22 + 1) = 276
Totals of individual divisors = 121 + 64 + 276 = 461. We must remember that 1 is a divisor for each and therefore it has been counted thrice.
From this we need to remove the number of divisors common to the pairs.
Divisors common to \(2^{10} * 5^{10}\) and \(3^7 * 5^7\) = \(5^7\). The number of divisors = 7 + 1 = 8
Divisors common to \(2^{10} * 5^{10}\) and \(2^{11} * 3^{22}\) = \(2^{10}\). The number of divisors = 10 + 1 = 11
Divisors common to \(3^{7} * 5^{7}\) and \(2^{11} * 3^{22}\) = \(3^{7}\). The number of divisors = 7 + 1 = 8
Total such divisors = 11 + 8 + 8 = 27. Each of these divisors contain 1. So again 1 has been counted thrice.
Divisors of at least 1 number = Total divisors - divisors common to the pairs = 461 - 27 = 434
Since 1 gets completely cancelled, we need to add back 1, as it is a common divisor to all 3.
Required number = 434 + 1 = 435
Arun Kumar
19 Feb 2021, 07:48
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18^11 = (2 * 3^2)^11 = 2^11 * 3^22 which has 12*23 = 276 divisors. So the answer to the question must be 276 or greater, just looking at 18^11 on its own, and only answer E is possible.
When I do complete the question, though, I'm getting 435 as the answer, not 425. We get 121 divisors from 10^10, and 64 divisors from 15^7, for a total of 276+121+64=461 divisors, but we're double-counting 2^0, 2^1, ... 2^10, along with 3^0, 3^1, ... 3^7, and along with 5^1, 5^2, ... 5^7, so we need to discard those 26 divisors, leaving 435.
That arithmetic is slightly annoying so it's possible I made a mistake somewhere.
When I do complete the question, though, I'm getting 435 as the answer, not 425. We get 121 divisors from 10^10, and 64 divisors from 15^7, for a total of 276+121+64=461 divisors, but we're double-counting 2^0, 2^1, ... 2^10, along with 3^0, 3^1, ... 3^7, and along with 5^1, 5^2, ... 5^7, so we need to discard those 26 divisors, leaving 435.
That arithmetic is slightly annoying so it's possible I made a mistake somewhere.
