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how do I find the closest number divisible by 10

Posted on 2013-01-27
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Last Modified: 2013-01-31
I need a function where it will give me the closest number divisible by 10.  The input is a decimal so 85.92 is valid and the output would be 90.

so if the input was
81.9 = 80
84.34 = 80
89.11 = 90
91.4 = 90
85.0 = 90 (this wouldn't happen much because the input is a decimal)
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Question by:jackjohnson44
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Expert Comment

by:Robert Schutt
ID: 38825116
Try:
y = 10 * Math.Round(x / 10);

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Robert Schutt earned 250 total points
ID: 38825119
A bit more elaborate (and taking types in account):
private void Form1_Load(object sender, EventArgs e) {
    MessageBox.Show(findClosestDivisible((decimal)84.9, 10).ToString());
}

private int findClosestDivisible(decimal x, decimal divisor) {
    return (int)(divisor * Math.Round(x / divisor, MidpointRounding.AwayFromZero));
}

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Expert Comment

by:Monica P
ID: 38825696
Try this logic in your code

Divide by 10, then truncate, then multiply by 10.

229 / 10 = 22,90000...
Math.Truncate(22,9) = 22
22 * 10 = 220.
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Expert Comment

by:Robert Schutt
ID: 38825853
@AkilaPalanimuthu: shouldn't 229 be rounded up to 230?
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Expert Comment

by:Monica P
ID: 38825877
229 is just input number..

Incase of 229.5 it can be rounded to 230 and then can apply the logic
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Expert Comment

by:Robert Schutt
ID: 38825883
That's not what I meant. Maybe I read the question wrong but 229 is closer to 230 than to 220.
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Assisted Solution

by:John Claes
John Claes earned 250 total points
ID: 38825951
robert_schutt ,  AkilaPalanimuthu:

Combine your comments ;-)

229 / 10 = 22,90000...
Math.Round(22,9) = 23
23 * 10 = 230.
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Expert Comment

by:Robert Schutt
ID: 38825976
That was actually my first comment but then I found that code-wise it needed some enhancements, hence my second comment. Also just for fun adding the possibility to use another number than 10. The Math.Round documentation explains that by default 4.5 is rounded down to 4, so even though the OP specified it would be rare, I included the 'MidpointRounding.AwayFromZero' argument for what I shamelessly assume to be mathematically 'normal' behaviour.
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