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Question
Let N=10800. Then the correct option(s) among the following is/are
Let N=10800. Then the correct option(s) among the following is/are
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Solution
The correct options are
A Number of even divisors is 48
B Number of odd divisors is 12
C Number of divisors that are multiple of 3 is 45
Here N=243352.
number of divisors =(4+1)(3+1)(2+1)=60.
(1) To from an even factor, we must select atleast one ′2′
from the lot, which will ensure that whatever be the remaining selection, their multiplication will always result in an even factor.
The number of ways to select atleast one ′2′ from a lot of four indentical ′2′s will be 4 (i.e. select 1 or select 2 or select 3 or select 4.) And, we will select any number of ′3′s and ′5′s, in 4 and 3
ways respectively.
The required number of ways will be =4×4×3=48
Now, To count the odd factors, we will get rid of the ′2′s. We will make the selection from the ′3′s and the ′5′s only. The number of selections (or factors) will therefore
be (3+1)(2+1)=12.
Note that this could also have been obtained by subtracting the even factors from the total, i.e. 60−48, which will give the same answer.
(iii) This one is similar to (i). For factors to be a multiple of 3, we must select at least one ′3′ (3 ways), and any number of ′2′s and ′5′s (5 and 3 ways respectively).
The required number will be 5×3×3=45.
(iv) For factors to be a multiple of 15, we must select atleast one ′3′ (3 ways) and one ′5′ (2 ways), and any number of ′2′s (5 ways). The required number is 5×3×2=30.
A Number of even divisors is 48
B Number of odd divisors is 12
C Number of divisors that are multiple of 3 is 45
Here N=243352.
number of divisors =(4+1)(3+1)(2+1)=60.
(1) To from an even factor, we must select atleast one ′2′
from the lot, which will ensure that whatever be the remaining selection, their multiplication will always result in an even factor.
The number of ways to select atleast one ′2′ from a lot of four indentical ′2′s will be 4 (i.e. select 1 or select 2 or select 3 or select 4.) And, we will select any number of ′3′s and ′5′s, in 4 and 3
ways respectively.
The required number of ways will be =4×4×3=48
Now, To count the odd factors, we will get rid of the ′2′s. We will make the selection from the ′3′s and the ′5′s only. The number of selections (or factors) will therefore
be (3+1)(2+1)=12.
Note that this could also have been obtained by subtracting the even factors from the total, i.e. 60−48, which will give the same answer.
(iii) This one is similar to (i). For factors to be a multiple of 3, we must select at least one ′3′ (3 ways), and any number of ′2′s and ′5′s (5 and 3 ways respectively).
The required number will be 5×3×3=45.
(iv) For factors to be a multiple of 15, we must select atleast one ′3′ (3 ways) and one ′5′ (2 ways), and any number of ′2′s (5 ways). The required number is 5×3×2=30.
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