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When n is divided by 2, 3, 5, 7 and 9, the remainder is 1. W

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by Max@Math Revolution » Tue Jun 12, 2018 1:53 am
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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
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by GMATGuruNY » Tue Jun 12, 2018 4:12 am
Max@Math Revolution wrote:[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
Since n must yield a remainder when divided by 2, n cannot be even.
Eliminate A.

To yield a remainder of 1 when divided by 9, n must be equal to ONE MORE THAN A MULTIPLE OF 9.
Thus, the correct answer choice must be equal to (MULTIPLE OF 9) + 1:
B) 211 = 210 + 1 --> Nope.
C) 421 = 420 + 1 --> Nope.
D) 631 = 630 + 1 --> Success!
E) 841 = 840 + 1 --> Nope.

The correct answer is D.
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by Max@Math Revolution » Thu Jun 14, 2018 1:35 am
=>

Since 2, 3, 5 and 7 are prime numbers, n = 2 ∙3 ∙5 ∙7 ∙k + 1 = 210 ∙k + 1 for some positive integer k.
If k = 1, then n = 210 ∙1 + 1 = 211 has remainder 4 when it is divided by 9 since 211 = 9 ∙23 + 4.
If k = 2, then n = 210 ∙2 + 1 = 421 has remainder 7 when it is divided by 9 since 421 = 9 ∙46 + 7.
If k = 3, then n = 210 ∙3 + 1 = 631 has remainder 1 when it is divided by 9 since 631 = 9 ∙70 + 1.

Therefore, the answer is D.
Answer: D
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by Scott@TargetTestPrep » Thu Jun 14, 2018 9:46 am
Max@Math Revolution wrote:[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
Since the division of n by 2, 3, 5, 7 and 9 produces a remainder of 1, n - 1 is divisible by 2, 3, 5, 7 and 9. The LCM of these numbers will give us the smallest value of n - 1.

The LCM of 2, 3, 5, 7, and 9 is:

2 x 3^2 x 5 x 7 = 630, so we see that 631 is the smallest value that will have a reminder of 1 when divided by 2, 3, 5, 7 and 9.

Answer: D

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by swerve » Thu Jun 14, 2018 10:44 am
We can eliminate A because it exacts gets divided by 2 and 5.

B = 211 = 210 + 1.
If we divide 211 by 9 we won't get remainder 1. So option B is wrong.

C = 421 = 420 + 1.
421 gets divided by 9 exactly. So C is wrong.

E = 841 = 840+1.
If we divide 840 by 9 it does not give the remainder 1. SO this option is Wrong.

D = 631 = 630 + 1.
This satisfie the question. If we divide 631 by 2, 3, 5, 7, 9 we get the reminder 1. So D is correct.

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by Jake@ThePrincetonReview » Thu Jun 14, 2018 1:38 pm
To add a bit to this smart approach by GMATGuruNY, why choose 9 as the divisor to check? Because it's very easy to tell when a number is a multiple of 9. If the sum of the digits of the number is 9 or a multiple of 9, then the number itself is a multiple of 9.

So after eliminating A because it's even and therefore a multiple of 2 (no remainder), eliminating B, C, and E becomes quite simple.

B) 211-1 = 210. Since 2+1+0 = 3, and 3 is not a multiple of 9, you know that 210 is not a multiple of 9. Eliminate.
C) 421-1 = 420. Since 4+2+0 = 6, and 6 is not a multiple of 9, you know that 420 is not a multiple of 9. Eliminate.
D) 631 -1 = 630. Since 6+3+0=9, and 9 is a multiple of 9, you know that 630 IS a multiple of 9. Keep.
E) 841-1=840. Since 8+4+0=12, and 12 is not a multiple of 9, you know that 840 is not a multiple of 9. Eliminate.




GMATGuruNY wrote:
Max@Math Revolution wrote:[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
Since n must yield a remainder when divided by 2, n cannot be even.
Eliminate A.

To yield a remainder of 1 when divided by 9, n must be equal to ONE MORE THAN A MULTIPLE OF 9.
Thus, the correct answer choice must be equal to (MULTIPLE OF 9) + 1:
B) 211 = 210 + 1 --> Nope.
C) 421 = 420 + 1 --> Nope.
D) 631 = 630 + 1 --> Success!
E) 841 = 840 + 1 --> Nope.

The correct answer is D.
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