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Bunuel
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Positive integer z leaves the remainder of 1 when divided by 3, the re 08 Oct 2019, 00:51
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B
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35% (medium)

Question Stats:

73% (01:45) correct 27% (02:10) wrong based on 429 sessions

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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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B
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Positive integer z leaves the remainder of 1 when divided by 3, the re 08 Oct 2019, 01:04
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Easiest way is to test the answer choices.

101 does not leave a remainder of 1 when divided by 3 - Eliminated

103 leaves a remainder of 1 when divided by 3, the remainder of 3 when divided by 5 and the remainder of 5 when divided by 7 - Correct

Answer is (B)
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Positive integer z leaves the remainder of 1 when divided by 3, the re 08 Oct 2019, 04:49
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Z leaves the remainder 3 when divided by 5.
Therefore , Z= {3,8,13,18,23,28.....}
Check the options that have 3 and 8 as unit digit.
Only two option i.e. B and E
check their divisibility by 3 and 7

Therefore answer is B
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Positive integer z leaves the remainder of 1 when divided by 3, the re 08 Oct 2019, 07:27
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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
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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Positive integer z leaves the remainder of 1 when divided by 3, the re 16 Oct 2019, 17:03
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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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Positive integer z leaves the remainder of 1 when divided by 3, the re 28 Oct 2019, 06:36
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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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Positive integer z leaves the remainder of 1 when divided by 3, the re 20 Jun 2023, 01:26
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dcummins
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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Positive integer z leaves the remainder of 1 when divided by 3, the re 18 Sep 2024, 02:05
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I agree I think there is a mistake:

208/3 = 69 and the r=1

208/5 = 41 and the r= 3

208/7 = 29 and the r= 5

anzzz
dcummins
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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Positive integer z leaves the remainder of 1 when divided by 3, the re 14 Nov 2024, 11:00
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que. asks for minimum.

rito54
I agree I think there is a mistake:

208/3 = 69 and the r=1

208/5 = 41 and the r= 3

208/7 = 29 and the r= 5

anzzz
dcummins
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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