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805+ (Hard)|   Remainders|      
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findingmyself
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How many integers between 200 and 800 have a reminder of 3 when Updated on: 29 Aug 2025, 01:40
Originally posted by findingmyself on 27 Aug 2025, 07:48.
Last edited by Bunuel on 29 Aug 2025, 01:40, edited 2 times in total.
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B
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Difficulty:

95% (hard)

Question Stats:

53% (03:03) correct 47% (02:59) wrong based on 133 sessions

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How many integers between 200 and 800 have a reminder of 3 when divided by 5 and remainder of 4 when divided by 6?

A. 18
B. 20
C. 24
D. 30
E. 150
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B
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gmatophobia
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How many integers between 200 and 800 have a reminder of 3 when 27 Aug 2025, 10:08
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findingmyself
How many integers between 200 and 800 have a reminder of 3 when divided by 5 and remainder of 4 when dividied by 5?

A)18
B)20
C)24
D)30
E)150
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findingmyself - I don't suppose the question as written is correct.
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findingmyself
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How many integers between 200 and 800 have a reminder of 3 when 27 Aug 2025, 11:40
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Thanks for pointing out, I have made the corrections.

gmatophobia


findingmyself - I don't suppose the question as written is correct.
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How many integers between 200 and 800 have a reminder of 3 when 27 Aug 2025, 11:48
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findingmyself
Thanks for pointing out, I have made the corrections.


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What is the source of the question? Thank you!
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How many integers between 200 and 800 have a reminder of 3 when 27 Aug 2025, 12:27
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This is from e-gmat mocks
Bunuel


What is the source of the question? Thank you!
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How many integers between 200 and 800 have a reminder of 3 when 28 Aug 2025, 08:53
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Bunuel If possible, can you please share the fastest way to solve this kind of question.
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How many integers between 200 and 800 have a reminder of 3 when 29 Aug 2025, 02:40
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How many integers between 200 and 800 have a reminder of 3 when divided by 5 and remainder of 4 when divided by 6?

A. 18
B. 20
C. 24
D. 30
E. 150
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A reminder of 3 when divided by 5 implies x = 5q + 3
A remainder of 4 when divided by 6 implies x = 6p + 4

We can derive a general formula for the total (of a type \(total=dk+r\), where \(d\) is the divisor and \(r\) is the remainder) based on the two formulas given: x = 5q + 3 (3, 8, 13, 18, 23, 28, ... ) and x = 6p + 4 (4, 10, 16, 22, 28, ... ).

The divisor \(d\) would be the least common multiple of the two divisors 5 and 6, hence d = 30 and the remainder r would be the first common integer in the two patterns, hence r = 28.

So, the general formula based on both pieces of information is x = 30k + 28. We get:

200 < 30k + 28 < 800

172 < 30k < 772

5.something < k < 25.somehting

Thus, k can take 20 value from 6 to 25, inclusive, which means that x can also take 20 values from 208 to 778.

Answer: B.
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