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17 Jul 2020, 04:50
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D
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Difficulty:
65%
(hard)
Question Stats:
54% (02:09) correct
46%
(02:07)
wrong
based on 106
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If z is a positive three-digit integer, is z a multiple of 7?
(1) All of the factors of 133 are factors of z.
(2) If the number of hundreds in z is multiplied by 2 and then added to the remaining two-digit number, the result is divisible by 7
Are You Up For the Challenge: 700 Level Questions
(1) All of the factors of 133 are factors of z.
(2) If the number of hundreds in z is multiplied by 2 and then added to the remaining two-digit number, the result is divisible by 7
Are You Up For the Challenge: 700 Level Questions
17 Jul 2020, 05:02
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Bunuel
Question: Is z a multiple of 7?
Statement 1: All of the factors of 133 are factors of z.
Factors of 133 = {1, 7, 19, 133}
i.e. 7 is a factors of z
SUFFICIENT
Statement 2: If the number of hundreds in z is multiplied by 2 and then added to the remaining two-digit number, the result is divisible by 7
Multiples of 7 are 105, 112, 119, 126...
for 105, required number = 2*1+05 = 7 i.e. Divisible by 7
for 112, required number = 2*1+12 = 14 i.e. Divisible by 7
for 119, required number = 2*1+19 = 21 i.e. Divisible by 7
This is the test of divisibility of 7 hence the statement confirms that the number z is divisible by 7
SUFFICIENT
Answer: Option D
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18 Jul 2020, 04:01
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Let the digits of number z be abc (a-hundreds digit, b-tens digit, c-ones digit) so the number will be 100a + 10b + c.
(1) All factors of 133 are factors of z.
Factors of 133 are --> 1, 133, 7, 19. This implies 7 is a factor of z.
SUFFICIENT
(2) If the number of hundreds in z is multiplied by 2 and then added to the remaining two-digit number, the result is divisible by 7
--> 2a + 10b + c = 7q {for some integer q}
Add 98a on both sides :
--> 98 a + 2a + 10b + c = 7q + 98a
--> 100a + 10b + c = 7q + 7*14*a = 7[q + 14a]
--> z = 7c {c is some integer such that c = q + 14a}
Hence, z is multiple of 7
SUFFICIENT
(1) All factors of 133 are factors of z.
Factors of 133 are --> 1, 133, 7, 19. This implies 7 is a factor of z.
SUFFICIENT
(2) If the number of hundreds in z is multiplied by 2 and then added to the remaining two-digit number, the result is divisible by 7
--> 2a + 10b + c = 7q {for some integer q}
Add 98a on both sides :
--> 98 a + 2a + 10b + c = 7q + 98a
--> 100a + 10b + c = 7q + 7*14*a = 7[q + 14a]
--> z = 7c {c is some integer such that c = q + 14a}
Hence, z is multiple of 7
SUFFICIENT
