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Re: Positive integer z leaves the remainder of 1 when divided by 3, the re [#permalink]
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z= -2 mod 3 or 3a-2
z= -2 mod 5 or 5b-2
z= -2 mod 7 or 7c-2

LCM(3, 5, 7)= 105
hence, z= -2 mod 105= 103 mod 105

z= 105k + 103, k is non negative integer

Smallest value z can take is 105*0+103=103

Bunuel wrote:
Positive integer z leaves the remainder of 1 when divided by 3, the remainder of 3 when divided by 5 and the remainder of 5 when divided by 7. What is the smallest possible value of z?

A. 101
B. 103
C. 105
D. 107
E. 208
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Re: Positive integer z leaves the remainder of 1 when divided by 3, the re [#permalink]
Z= 3Q + 1
Z= 5A + 3
Z= 7B + 5

where Q,A, B are positive quotients for Z divided by 3, 5 and 7 respectively.

Plug-in the answer choice to determine whether dividing that given answer choice produces a remainder of 1, 3 and 5 respectively.

101 doesn't work with the first equation, so A is gone.
103 works as you will see.
B is correct.
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Re: Positive integer z leaves the remainder of 1 when divided by 3, the re [#permalink]
Quick fire approach:

Division by 5 is easy, so just divide all options by 5 and see if condition for remainder of 3 if divided by 5 is met by one (or more) of the options.

Even before you get to dividing you should be able to spot 103 (i.e 100+3).

Divide 103 by 3 and 7, and the remainders match.

Apart from 101, all other options are greater than 103.

Answer: B
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Re: Positive integer z leaves the remainder of 1 when divided by 3, the re [#permalink]
dcummins wrote:
Z= 3Q + 1
Z= 5A + 3
Z= 7B + 5

where Q,A, B are positive quotients for Z divided by 3, 5 and 7 respectively.

Plug-in the answer choice to determine whether dividing that given answer choice produces a remainder of 1, 3 and 5 respectively.

101 doesn't work with the first equation, so A is gone.
103 works as you will see.
B is correct.


using this method then E is correct too. who can explain this?
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Re: Positive integer z leaves the remainder of 1 when divided by 3, the re [#permalink]
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