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X and y are integers, are they even? [#permalink]
05 Oct 2006, 17:07

1

This post was BOOKMARKED

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A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct
0% (00:00) wrong based on 0 sessions

PS: not sure of the answers

1)X and y are integers, are they even?

1). 2y-x=x^2-y^2

2). X is even.

IMO: C

2)M=O+P+Q, where O, P, and Q are consecutive positive integer; M=R*S*T, where R, S and T are positive consecutive integers. What is the remainder when M is divided by 5? 1). When O is divided by 5, the remainder is 1 2). When R is divided by 5, the remainder is 1

O and R can have either 1 or 6 in unit's digit.
1)
if O is of the form X1 (i.e unit's digit 6, rest can be any numbers)
M= O+P+Q = X1+X2+X3 = Y6.
The unit's digit of M will be 6, so when M is divided by 5 remainder is 1.
if O is of the form X6:
M=O+P+Q = X6+X7+X8= Y1
remainder will be 1.
so, (1) is enough to answer

2) When R is of form X1
M=RxSxT = X1.X2.X3 = Z6 // so M%5 = 1
When R is of form X6:
M= X6.X7.X8 = Z6 // so M%5 = 1

hence D is the answer. Either of them is enough to answer

and thus x^2 - y^2 = even - ??? = even thus y^2 is even and hus x,y even

my answer is C

2)M=O+P+Q, where O, P, and Q are consecutive positive integer; M=R*S*T, where R, S and T are positive consecutive integers. What is the remainder when M is divided by 5?
1). When O is divided by 5, the remainder is 1
2). When R is divided by 5, the remainder is 1

M=O+P+Q

FROM ONE IF O = 5X+1 THUS
P AND Q ARE 5X+2 , 5X+3

ADD THE REMAINDERS = 6 DEVIDE BY FIVE REMAINDER IS 1....SUFF

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