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23 Sep 2024, 19:42
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
74% (01:28) correct
26%
(01:28)
wrong
based on 88
sessions
History
Date
Time
Result
Not Attempted Yet
What is the least number when divided by 3, 4 and 5 leaves a remainder of 2 in each case but when divided by 7 leaves nothing?
A. 60
B. 62
C. 122
D. 182
E. None
A. 60
B. 62
C. 122
D. 182
E. None
24 Sep 2024, 00:00
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adritatasnm
We have:
x = 3a + 2, so x could be 2, 5, 8, and so on.
x = 4b + 2, so x could be 2, 6, 10, and so on.
x = 5c + 2, so x could be 2, 7, 12, and so on.
We can derive a general formula for the total (of a type \(x=dq+r\), where \(d\) is the divisor and \(r\) is the remainder) based on the formulas given.
The divisor \(d\) would be the least common multiple of the three divisors 3, 4, and 5, hence \(d=60\) and the remainder \(r\) would be the first common integer in the three patterns, hence \(r=2\):
\(x = \text{(the LCM of 3, 4, and 5 )} + \text{(the first common value for x)}\)
So, the general formula based on both pieces of information is \(x= 60q+2\). Thus, x could be 2, 62, 122, 182, and so on.
The least number from these which is divisible by 7 is 182.
Answer: D.
Hope this helps.
27 Sep 2024, 12:08
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A much simpler way to do this would be to just check the options! the first few are definitely not divisible by 7. 122/7 isnt a whole number. The only option that works is 182
