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A positive integer n when divided by 3, 4, 5, and 6 leaves a remainder

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Bunuel
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A positive integer n when divided by 3, 4, 5, and 6 leaves a remainder [#permalink] New post  24 Feb 2020, 06:25
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A positive integer n when divided by 3, 4, 5, and 6 leaves a remainder of 2, but leaves no remainder when divided by 7. What is the lowest possible value of n ?


A. 168
B. 182
C. 189
D. 186
E. 203
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chetan2u
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Re: A positive integer n when divided by 3, 4, 5, and 6 leaves a remainder [#permalink] New post  24 Feb 2020, 07:14
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Bunuel wrote:
A positive integer n when divided by 3, 4, 5, and 6 leaves a remainder of 2, but leaves no remainder when divided by 7. What is the lowest possible value of n ?


A. 168
B. 182
C. 189
D. 186
E. 203



Easiest way...CHOICES!!!
Start from the smallest...
168....Divisible by 3, by 4...Eliminate
182...Not divisible by 3, 4, 5, and 6, but divisible by 7=> 182=7*26, Also 182=180+2, and 180 is multiple of 3, 4, 5,6 ..YES

LOGICAL!!!
The number is of the form 4a+2, so surely even...Eliminate all odd
The number is also of the form 5b+2, so units digit has to be 2 or 7..
But since it is even, units digit will be 2..
Only possibility 182

EQUATIONS!!!
number = LCM(3,4,5,6)*a+2=60a+2, Also number = 7b
So 60a+2 should be a multiple of 7...
\(a=1, 60a+2=62=7*8+6\)...N0
\(a=2, 60a+2=122=7*117+3\)...N0
\(a=3, 60a+2=182=7*26\)...YES

B
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M1ke
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Re: A positive integer n when divided by 3, 4, 5, and 6 leaves a remainder [#permalink] New post  24 Feb 2020, 09:49
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The only number in the list that leaves a remainder of 2 when divided by 5 is 182.

Check the answer:
182=7*26
182=7*2*13 => does not contain a 3, 4, 5 or 6.

B is the correct answer.
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Re: A positive integer n when divided by 3, 4, 5, and 6 leaves a remainder [#permalink] New post  24 Feb 2020, 07:50
A positive integer n when divided by 3, 4, 5, and 6 leaves a remainder of 2, but leaves no remainder when divided by 7. What is the lowest possible value of n ?


A. 168
B. 182 --> correct: n= LCM(3,4,5,6)*d+2=60d+2, so unit digit of n has to be always 2 (only choice B, also 182%7=0 to verify the answer)
C. 189
D. 186
E. 203
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Re: A positive integer n when divided by 3, 4, 5, and 6 leaves a remainder [#permalink] New post  24 Feb 2020, 08:05
Bunuel wrote:
A positive integer n when divided by 3, 4, 5, and 6 leaves a remainder of 2, but leaves no remainder when divided by 7. What is the lowest possible value of n ?


A. 168
B. 182
C. 189
D. 186
E. 203


LCM of 3,4,5 and 6 is 60

Therefore valid numbers for the remainder of 2 are in the form of 60m + 2

Only B fits in hence is our answer

Answer - B
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Re: A positive integer n when divided by 3, 4, 5, and 6 leaves a remainder [#permalink] New post  20 Sep 2020, 09:30
Given: A positive integer n when divided by 3, 4, 5, and 6 leaves a remainder of 2, but leaves no remainder when divided by 7.
Asked: What is the lowest possible value of n ?

n = LCM(3,4,5,6)k + 2 = 60k + 2 = 7m
n_min = 182

IMO B
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Re: A positive integer n when divided by 3, 4, 5, and 6 leaves a remainder [#permalink]
20 Sep 2020, 09:30
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