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24 Sep 2016, 13:37
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I encountered this question on the Kaplan CAT, and I wasted a lot of time trying to interpret "multiple of any of x, y, and z". After reading the answer, I understand what they're asking for, but shouldn't it be written as "multiple of x, y, or z"? I started down the path of trying to evaluate the answers with all 3 primes being factors of the final result.
stonecold
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Last visit: 09 Apr 2024
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Schools: Boston U '20 (M)
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15 Jan 2017, 23:27
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Great Question.
Here is what i did in this Question.
Method 1->
Using test cases->
Option 1=>
x=2 y=3 and z=5
Clearly x+y=5 which is a multiple of 5.
WRONG
option 2=>
x=3
y=5
z=2
clearly y-x is a multiple of z
WRONG.
Option 3=>
x=2
y=3
z=7
Clearly xy+1 is a multiple of z.
WRONG.
Option 4=>
No test cases fulfilled.
MAY be right.
Lets analyse option 5 =>
x=2
y=3
z=13
Clearly x^2+y^2 is multiple of z.
WRONG.
So option 4 must be correct.
Hence D.
Method 2->
Any two consecutive integers are always coprimes.
So xyz will be divisible by all x,y,z and hence xyz+1 will never be divisible by x or y or z.
Hence D.
Here is what i did in this Question.
Method 1->
Using test cases->
Option 1=>
x=2 y=3 and z=5
Clearly x+y=5 which is a multiple of 5.
WRONG
option 2=>
x=3
y=5
z=2
clearly y-x is a multiple of z
WRONG.
Option 3=>
x=2
y=3
z=7
Clearly xy+1 is a multiple of z.
WRONG.
Option 4=>
No test cases fulfilled.
MAY be right.
Lets analyse option 5 =>
x=2
y=3
z=13
Clearly x^2+y^2 is multiple of z.
WRONG.
So option 4 must be correct.
Hence D.
Method 2->
Any two consecutive integers are always coprimes.
So xyz will be divisible by all x,y,z and hence xyz+1 will never be divisible by x or y or z.
Hence D.
Archit3110
Major Poster
31 Mar 2021, 11:46
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karishmatandon
let three numbers be 2,3,5
out of given options only at D we observe that the resulting number is not multiple of any of 2,3,5 similarly test with other numbers 3,5,7
option D is correct.
