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If N = 100x + 10y + z is a three-digit positive integer, is N divisibl

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If N = 100x + 10y + z is a three-digit positive integer, is N divisibl  [#permalink]

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New post 10 Apr 2018, 05:35
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

70% (01:28) correct 30% (01:41) wrong based on 45 sessions

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If N = 100x + 10y + z is a three-digit positive integer, is N divisibl  [#permalink]

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New post Updated on: 10 Apr 2018, 09:00
1
Bunuel wrote:
If N = 100x + 10y + z is a three-digit positive integer, is N divisible by 9?

(1) x + y + z =18
(2) N + 432 is divisible by 9.


From 1:

three digit number: xyz=100x + 10y + z

a number is divisible by 9, if sum of all of it digits is divisible by 9

x + y + z =18

it is divisible by 9

sufficient

From 2:

432 is divisible by 9 and N +432 is divisble by 9 so

N is divisible by 9

so sufficient.

hence D

Originally posted by kunalcvrce on 10 Apr 2018, 08:06.
Last edited by kunalcvrce on 10 Apr 2018, 09:00, edited 1 time in total.
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Re: If N = 100x + 10y + z is a three-digit positive integer, is N divisibl  [#permalink]

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New post 10 Apr 2018, 08:56
(1) x+y+z = 18
Since 100x + 10y + z is a three-digit number, each of the digit x,y and z is less than 9, hence N would be of the form xyz, which is divisible by 9.

Sufficient.

(2) N+432 is divisible by 9, since 432 is divisible by 9, hence N must also be divisible by 9.

Sufficient.

(D)
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Re: If N = 100x + 10y + z is a three-digit positive integer, is N divisibl  [#permalink]

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New post 19 Apr 2018, 00:06
For first statement
Take value of z from (I) and put it in equation
ie 100x + 10y +(18-x-y)=99x + 9y +18
All r multiple of 9
Sufficient

II)N +432 is divisible by 9
432 is divisible by 9 so N has to be divisible by 9
Sufficient

Ans-D

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Re: If N = 100x + 10y + z is a three-digit positive integer, is N divisibl &nbs [#permalink] 19 Apr 2018, 00:06
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If N = 100x + 10y + z is a three-digit positive integer, is N divisibl

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