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31 Jan 2022, 02:30
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Difficulty:
55%
(hard)
Question Stats:
65% (02:12) correct
35%
(02:23)
wrong
based on 190
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What is the remainder when the positive three-digit number xyz is divided by 9?
(1) The three-digit number whose hundreds digit is (x+2), tens digit is (y+1) and units digit is z, is divisible by 9
(2) xyz is divisible by 7
(1) The three-digit number whose hundreds digit is (x+2), tens digit is (y+1) and units digit is z, is divisible by 9
(2) xyz is divisible by 7
31 Jan 2022, 03:29
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(1) Only
(x+2)+(y+1)+z is divisible by 9
=> x+y+z+3 is divisible by 9
=> The remainder when (x+y+z) is divided by 9 should be 6.
Sufficient.
(2) Only
700, which is divisible by 7, cannot be divided by 9.
Can we come up with a three-digit number that is divisible by both 7 and 9. How about 7*9*5=315.
Insufficient.
The answer is (A).
(x+2)+(y+1)+z is divisible by 9
=> x+y+z+3 is divisible by 9
=> The remainder when (x+y+z) is divided by 9 should be 6.
Sufficient.
(2) Only
700, which is divisible by 7, cannot be divided by 9.
Can we come up with a three-digit number that is divisible by both 7 and 9. How about 7*9*5=315.
Insufficient.
The answer is (A).
kungfury42
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31 Jan 2022, 03:51
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Rem (xyz) with 9 = Rem (x+y+z) with 9
1. x+y+z+3 is divisible by 9 which means x+y+z leaves a remainder of 6 upon division with 9 which means xyz leaves a remainder of 6 upon division with 9. Clearly, (1) is sufficient.
2. xyz is divisible by 7. Let's take a few examples of 3-digit numbers that are divisible by 7 - 105, 112, 119 and so on.. notice that each of these leave a different remainder upon division with 9. 105 leaves 6, 112 leaves 4, 119 leaves 2 and so on.. clearly, (2) is insufficient.
Hence, our answer is A.
Posted from my mobile device
1. x+y+z+3 is divisible by 9 which means x+y+z leaves a remainder of 6 upon division with 9 which means xyz leaves a remainder of 6 upon division with 9. Clearly, (1) is sufficient.
2. xyz is divisible by 7. Let's take a few examples of 3-digit numbers that are divisible by 7 - 105, 112, 119 and so on.. notice that each of these leave a different remainder upon division with 9. 105 leaves 6, 112 leaves 4, 119 leaves 2 and so on.. clearly, (2) is insufficient.
Hence, our answer is A.
Posted from my mobile device
