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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.
Last edited by Bunuel on 29 Aug 2025, 01:40, edited 2 times in total.
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
53% (03:00) correct
47%
(02:59)
wrong
based on 139
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
A. 18
B. 20
C. 24
D. 30
E. 150
29 Aug 2025, 02:40
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findingmyself
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.
General Discussion
27 Aug 2025, 10:08
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