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natita [175]
3 years ago
12

When a number is a multiple of 6, what are the possible values for the ones digits?

Mathematics
1 answer:
Liula [17]3 years ago
7 0
I would do this by first listing the multiples of 6 until I start to see a pattern with the one's digit.
6x0=0
6x1=6
6x2=12
6x3=18
6x4=24
6x5=30
6x6=36
6x7=42
6x8=48
...
The digits in bold are the one's digits so those are the only ones we really care about.  If you list just them it looks like: 0,6,2,8,4,0,6,2,8
Notice how the first set of 5 numbers seems as though it repeats in the 6th, 7th, and 8th numbers.  This probably means the pattern continues infinitely so the first 5 numbers are all the one's digits that can come from multiples of 6.  Thus your answer is: 0,6,2,8,or 4 
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Step-by-step explanation:

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Reduced form of 2ab^2-a^2 b^2/5
Ket [755]

Answer:

2ab^2-\frac{a^2b^2}{5}=\frac{10ab^2-a^2b^2}{5}

Step-by-step explanation:

Given the expression

2ab^2-\frac{a^2b^2}{5}

\mathrm{Convert\:element\:to\:fraction}:\quad \:2ab^2=\frac{2ab^25}{5}

2ab^2-\frac{a^2b^2}{5}=\frac{2ab^2\cdot \:5}{5}-\frac{a^2b^2}{5}

\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}

                 =\frac{2ab^2\cdot \:5-a^2b^2}{5}

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Thus,

2ab^2-\frac{a^2b^2}{5}=\frac{10ab^2-a^2b^2}{5}

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