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20 Feb 2024, 02:30
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E
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Difficulty:
75%
(hard)
Question Stats:
59% (02:38) correct
41%
(02:28)
wrong
based on 146
sessions
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If w, x, y and z are consecutive positive integers, in the same order, which of the following expressions must be odd?
I. \(wx+z*(y+2)^2\)
II \(w^x+z^y\)
III. \(2*(w^3)+3*(x^2)+5*y+6*(z^2)\)
A. I only
B. II only
C. III only
D. I and II only
E. II and III only
I. \(wx+z*(y+2)^2\)
II \(w^x+z^y\)
III. \(2*(w^3)+3*(x^2)+5*y+6*(z^2)\)
A. I only
B. II only
C. III only
D. I and II only
E. II and III only
20 Feb 2024, 02:58
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IMO (E)
Since w,x,y,z consecutive integer. now two possible case:
Case 1: E,O,E,O
Case 2: O,E,O,E
I. using case 1 we get, 1st term E and 2nd term E. Hence E+E is even so eliminate this one.
II using case 1 we get, E+O which will be odd. now check using case also, we get O+E which is also odd. Hence this must be true
III. using case 1 we get, E+O+E+E which is odd. with case 2 we will get E+E+O+E which is also odd. Hence this statement also must be true
Since w,x,y,z consecutive integer. now two possible case:
Case 1: E,O,E,O
Case 2: O,E,O,E
I. using case 1 we get, 1st term E and 2nd term E. Hence E+E is even so eliminate this one.
II using case 1 we get, E+O which will be odd. now check using case also, we get O+E which is also odd. Hence this must be true
III. using case 1 we get, E+O+E+E which is odd. with case 2 we will get E+E+O+E which is also odd. Hence this statement also must be true
20 Feb 2024, 20:47
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Substitute
Case 1: 1,2,3,4
Case 2: 2,3,4,5
Answer is E
Case 1: 1,2,3,4
Case 2: 2,3,4,5
Answer is E
