Find the smallest number that is divisible by 2, 3, 4, 5, 6 and add 1.
We need the least common multiple of 2, 3, 4, 5, 6.
2 = 2
3 = 3
4 = 2^2
5 = 5
6 = 2 * 3
LCM = product of common and not common prime factors with larger exponent.
LCM = 2^2 * 3 * 5 = 4 * 3 * 5 = 60
To always have a remainder of 1, you need of add 1 to 60.
The number is 61.
Check:
61/2 = 30 remainder 1
61/3 = 20 remainder 1
61/4 = 15 remainder 1
61/5 = 12 remainder 1
61/6 = 10 remainder 1
Answer:
Valid
Step-by-step explanation:
Answer: Choice D
b greater-than 3 and StartFraction 2 over 15 EndFraction
In other words,
b > 3 & 2/15
or

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Explanation:
Let's convert the mixed number 2 & 3/5 into an improper fraction.
We'll use the rule
a & b/c = (a*c + b)/c
In this case, a = 2, b = 3, c = 5
So,
a & b/c = (a*c + b)/c
2 & 3/5 = (2*5 + 3)/5
2 & 3/5 = (10 + 3)/5
2 & 3/5 = 13/5
The inequality
is the same as 
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Let's multiply both sides by 15 to clear out the fractions

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Now isolate the variable b

Side note: Another way to go from 47/15 to 3 & 2/15 is to notice how
47/15 = 3 remainder 2
The 3 is the whole part while 2 helps form the fractional part. The denominator stays at 15 the whole time.
Answer:
b
Step-by-step explanation:
parabola's aren't functions