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12 Jun 2018, 18:13
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D
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Difficulty:
15%
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Question Stats:
84% (01:30) correct
16%
(01:42)
wrong
based on 90
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[GMAT math practice question]
When n is divided by 2, 3, 5, 7 and 9, the remainder is 1. What is the smallest value of the positive integer n?
A. 100
B. 211
C. 421
D. 631
E. 841
When n is divided by 2, 3, 5, 7 and 9, the remainder is 1. What is the smallest value of the positive integer n?
A. 100
B. 211
C. 421
D. 631
E. 841
DentTheUniverse
GMAT 1: 720 Q48 V41
Posts: 7
12 Jun 2018, 19:28
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I employed logic to start:
A --> eliminate b/c it's even and 2 will not have a remainder. It's also a multiple of 5, so it won't have a remainder either.
Don't need to test 5 because each B, C, D, E will have a remainder of 1. Employing general number sense, we know that 21, 42, 63, 84 are divisible by 3 and so their multiple (each *10) are divisible by 3, so B, C, D, and E will each have a remainder 1 when divided by 3.
I'm not as strong with the number properties of 7, so I'm going to test 9.
Numbers are divisible by 9 when the digits add up to 9. So to have a remainder of 1, the digits need to add up to 8 or 10.
B--> eliminate. Sum of digits is 4 (2+1+1), which is not 8 or 10 so won't have a remainder of 1 when divided by 9 (remainder will be 4 - don't need to calculate while doing the problem!)
C--> eliminate. Sum of digits is 7 (4+2+1), which is not 8 or 10 so won't have remainder 1 when divided by 9 (remainder will be 7).
E--> eliminate. Sum of digits is 13 (8+4+1), not 8 or 10 so won't have remainder of 1 when divided by 9 (remainder will be 4).
That leaves D. To make sure 7 has a remainder 1, we can implore # sense. 63 is a multiple of 7, therefore 630 is (63*10) and 631 will have a remainder 1.
A --> eliminate b/c it's even and 2 will not have a remainder. It's also a multiple of 5, so it won't have a remainder either.
Don't need to test 5 because each B, C, D, E will have a remainder of 1. Employing general number sense, we know that 21, 42, 63, 84 are divisible by 3 and so their multiple (each *10) are divisible by 3, so B, C, D, and E will each have a remainder 1 when divided by 3.
I'm not as strong with the number properties of 7, so I'm going to test 9.
Numbers are divisible by 9 when the digits add up to 9. So to have a remainder of 1, the digits need to add up to 8 or 10.
B--> eliminate. Sum of digits is 4 (2+1+1), which is not 8 or 10 so won't have a remainder of 1 when divided by 9 (remainder will be 4 - don't need to calculate while doing the problem!)
C--> eliminate. Sum of digits is 7 (4+2+1), which is not 8 or 10 so won't have remainder 1 when divided by 9 (remainder will be 7).
E--> eliminate. Sum of digits is 13 (8+4+1), not 8 or 10 so won't have remainder of 1 when divided by 9 (remainder will be 4).
That leaves D. To make sure 7 has a remainder 1, we can implore # sense. 63 is a multiple of 7, therefore 630 is (63*10) and 631 will have a remainder 1.
12 Jun 2018, 19:28
Kudos
Bookmarks
we can eliminate A) because it exacts gets divided by 2 and 5
B) 211=210+1
If we divide 211 by 9 we wont get remainder 1.So option B is WRONG
C)421=420+1
421 gets divided by 9 exactly. SO C is WRONG
D)631=630+1
This satisfies the above question. If we divide 631 by 2,3,5,7,9 we get the remainder 1. So D is CORRECT
E)840=840+1
If we divide 840 by 9 it does not give the remainder 1. SO this option is Wrong
B) 211=210+1
If we divide 211 by 9 we wont get remainder 1.So option B is WRONG
C)421=420+1
421 gets divided by 9 exactly. SO C is WRONG
D)631=630+1
This satisfies the above question. If we divide 631 by 2,3,5,7,9 we get the remainder 1. So D is CORRECT
E)840=840+1
If we divide 840 by 9 it does not give the remainder 1. SO this option is Wrong
