Rounding to nearest 100

Question

First number needs to be rounded to nearest second number. There are many ways of doing this, but whats the best and shortest algorithm? Anyone up for a challenge :-)

Answer 1

Is this homework?

Generally, mod 100, then if >50 add else subtract.

Answer 2

This will do it, given you're using integer math:

n = (n + 50) / 100 * 100

Of course, you didn't specify the behavior of e.g., 1350 and 1450, so I've elected to round up. If you need round-to-even, that'll not work.

Answer 3

Ruby's round method can consume negative precisions:

n.round(-2)

In this case -2 gets you rounding to the nearest hundred.

Answer 4

100 * round(n/100.0)

Answer 5

As per Pawan Pillai's comment above, rounding to nearest 100th in Javascript:
100 * Math.floor((foo + 50) / 100);

Answer 6

I know it's late in the game, but here's something I generally set up when I'm dealing with having to round things up to the nearest nTh:

Number.prototype.roundTo = function(nTo) {
    nTo = nTo || 10;
    return Math.round(this * (1 / nTo) ) * nTo;
}
console.log("roundto ", (925.50).roundTo(100));

Number.prototype.ceilTo = function(nTo) {
    nTo = nTo || 10;
    return Math.ceil(this * (1 / nTo) ) * nTo;
}
console.log("ceilTo ", (925.50).ceilTo(100));

Number.prototype.floorTo = function(nTo) {
    nTo = nTo || 10;
    return Math.floor(this * (1 / nTo) ) * nTo;
}
console.log("floorTo ", (925.50).floorTo(100));

I find myself using Number.ceilTo(..) because I'm working with Canvas and trying to get out to determine how far out to scale.