How do I generate a number divisable by 5, and check it?
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Thank you very much. Sorry I missed this post, and to be honest I am a little p***** off with the negative comments I am receiving, that is why I stopped looking at the responses as it upset me a little that people on here can be so damn rude and insulting, oh, and judgmental, for no apparent reason? Anyway, thank you very much. Kind Regards, Stephen
You're welcome, Stephen. Please keep in mind that mad geniuses as many programmers may be, few people tend to lump us programmers into the "high social skills" cliche. It's been a while since I worked with scientists but I seem to recall from my university days they can be somewhat socially lacking too. I remember this one physicist... Ahem... :omg: Uh, anyway, its probably best to just take the "water off a ducks back" approach - no one really means to insult - especially if you want any of us social deviants to actually help you. :laugh:
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I was not commenting on your coding skills and I did read all the posts, however I do apologize. I was just so astounded to think someone could actually not know that (Any number ending in 5 or 0 is divisible by 5), that I reacted in a knee-jerk fashion. I also didn't solve your problem because it had already been well answered. BTW, you really need to develop thicker skin. There will always be some people who know so much more than you that they sometimes get irritated answering the same, in their opinion, dumb questions over and over again. It still happens to me occasionally.
That’s OK, I was already upset by someone else's comment, which is why I reacted the way I did.
RDSchaefer wrote:
BTW, you really need to develop thicker skin.
Actually, I have a thick skin. Growing up with the surname "Darling" ensures that you develop one quickly :) I am a little tired today, and when I received a comment from someone on here telling me that I was "NOT a scientist" when they don’t know me, and have never spoke to me, was what made me flip a little. I can understand that it happens, and normally it would be water off a ducks back, however, on this occasion it caused me to respond. Kind Regards, Stephen
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Everyone's a Legend in Their Own Lunchtime. You've learned a lesson - whether you wished to or not - about being precise in framing your question, so I hope that at least that, and the sensible answers you did get, helped. As to the others who felt it was easier to ridicule than to stop for a second and realise there was more to it, I apologise.
cheers, Chris Maunder The Code Project | Co-founder Microsoft C++ MVP
Chris Maunder wrote:
Everyone's a Legend in Their Own Lunchtime.
Agreed!
Chris Maunder wrote:
You've learned a lesson - whether you wished to or not
Actually I learned two. Yes; one was indeed to think allot more before I pose my question!
Chris Maunder wrote:
As to the others who felt it was easier to ridicule than to stop for a second and realise there was more to it, I apologise.
You have nothing to apologise for, and it was really just the one statement that I was not a scientist that caused me to react. Never mind, I have moved on :-D Kind Regards, Stephen
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You're welcome, Stephen. Please keep in mind that mad geniuses as many programmers may be, few people tend to lump us programmers into the "high social skills" cliche. It's been a while since I worked with scientists but I seem to recall from my university days they can be somewhat socially lacking too. I remember this one physicist... Ahem... :omg: Uh, anyway, its probably best to just take the "water off a ducks back" approach - no one really means to insult - especially if you want any of us social deviants to actually help you. :laugh:
Michael A. Cochran wrote:
It's been a while since I worked with scientists but I seem to recall from my university days they can be somewhat socially lacking too.
Could not agree more! At the end of the day we are all human.
Michael A. Cochran wrote:
it’s probably best to just take the "water off a ducks back" approach
I normally do, however, being told by someone that I had never met or spoke to that I was not a scientist was what got to me. It was not just an insult and ignorant statement, but it was not even to do with programming and in my own opinion was a personal attack that was not called for. Anyway, I have taken what I need from this post, got my code to work and moved on :) Kind Regards, Stephen
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Hi. First, how would I create a random number, and then add the last digit, so that it is divisable by 5? the number should always be 5 digits long. Second, how do I check it, I think I need to do something like...
if (int x MOD 5 ==0)
Or something like that. The first step is the most important though. Thank you, Steve
x = generate random number between 10000 and 99995 remainder to be added = 5 - x MOD 5 Therefore, the resultant random number should be x + 5 - x MOD 5 Here's a sample code: http://ideone.com/LxgrC[^]
a.k.a. Vite Phoenix and Vite Zeus... Proud member and co-founder of OlympianZ
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Look at what I said - add 5 or 10. If a number is divisible by 5 it must end in 5 or 0 - that's primary school arithmetic. If you don't believe me write out the 5 times table for the numbers 1 to 20.
Regards David R --------------------------------------------------------------- "Every program eventually becomes rococo, and then rubble." - Alan Perlis The only valid measurement of code quality: WTFs/minute.
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Hi. First, how would I create a random number, and then add the last digit, so that it is divisable by 5? the number should always be 5 digits long. Second, how do I check it, I think I need to do something like...
if (int x MOD 5 ==0)
Or something like that. The first step is the most important though. Thank you, Steve
You are getting all sorts of advice. Multiplying by 10 can result in overflow. The easiest is: num -= num%5; IE subtract the remainder if you divided by five from the result. Another way: num = (num/5) * 5; That will NEVER cause an overflow in most languages. If you are using vb.net, besides the syntax replace "/" with "\".
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You can't just add 5. Say your result is 135978653, adding 5, the last digit is 8, not divisable by 5
I did say multiply by 10 then add 5. I also said you could add 5 or 10 - but adding 10 is silly. The multiplication automatically makes it divisible by 5. :laugh:
Regards David R --------------------------------------------------------------- "Every program eventually becomes rococo, and then rubble." - Alan Perlis The only valid measurement of code quality: WTFs/minute.
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x = generate random number between 10000 and 99995 remainder to be added = 5 - x MOD 5 Therefore, the resultant random number should be x + 5 - x MOD 5 Here's a sample code: http://ideone.com/LxgrC[^]
a.k.a. Vite Phoenix and Vite Zeus... Proud member and co-founder of OlympianZ
Why didn't I think of that? :rolleyes: Too much business programming, I guess. We don't use modulus in business programming much - at least I haven't had the need. Correct, of course. Much more elegant. If Stephen is still listening, in C# it becomes;
// To generate as random numbers as possible, this variable must only
// be initialized once and then reused as much as possible.
Random rand = new Random();/// <summary>
/// Generates a random number between 0 and 99995 that is always divisible by 5.
/// </summary>
/// <returns>Returns a random integer up to 5 digits long that is evenly divisible by 5.</returns>
private int GetDivBy5()
{
// Get a random number between 0 and 99999.
int divBy5 = rand.Next(0, 99999);
// Subtract the remainder of (n/5) to make n divisible by 5.
return divBy5 - (divBy5 % 5);
}private void button1_Click(object sender, EventArgs e)
{
this.textBox1.Text = this.GetDivBy5().ToString("00000");}
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I did say multiply by 10 then add 5. I also said you could add 5 or 10 - but adding 10 is silly. The multiplication automatically makes it divisible by 5. :laugh:
Regards David R --------------------------------------------------------------- "Every program eventually becomes rococo, and then rubble." - Alan Perlis The only valid measurement of code quality: WTFs/minute.
Yes, I did read that, I was tweaking. Seriously, multiplying exposes an overflow possiblity. And why 10? num = (num/5)*5; does the trick num -= num%5; also assures overflow won't occur even if you are within 5 of overflow in either positive or negative direction.
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Generate a four digit random number, multiply it by 10, add 5 (or 10). The result will be divisible by 5.
Regards David R --------------------------------------------------------------- "Every program eventually becomes rococo, and then rubble." - Alan Perlis The only valid measurement of code quality: WTFs/minute.
This solution is not correct - it exhibits a possible overflow error into 6 digits. Consider: 4 Random digits all come out as 9 (multiply by 10 = 99990), then add 10 you get 100000 which does not meet the requirements... The correct solution is: Generate a four digit random number Multiply it by 10 (to shift into correct position) Then randomly choose to make the last digit a 5 or a 0 (adding 10 is the incorrect step) The result is divisible by 5. Regards David.
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I KNEW there was a reason why I liked num -= num%5; better! :) How often have you seen a sizable random number generator produce 0 as an answer? (OK, you can easily set your random number limit so *10 will never overflow.) The original request is divisable by 5 and only affecting the last digit, not the whole number. Lets see: convert the int to a string, take a substring 1 less than the length, add "5" to the string and then convert back to int. There's quite a few ways to get this to work, multiplying by 10 and adding 5 will ONLY work correctly when the random number IS zero. Exactly where did I say to divide by zero? If num is zero, num=(num/5)*5 will produce zero. The highest numbers in int16, int, and int64 all end in decimal 7 and this process will work in all languages, the lowest number ends in 8. That will work in reasonable languages, VB.NET will blow up with an overflow. (In that language use num=(num\5)*5) (Actually VB is a reasonable language, you just have to know the gotchas like 5/3 is 2, not 1. I spent a little time swearing at its math until I found "\". I haven't found a use for round-up but if I did, VB would actually work quite a bit better than most languages for that.) Also good luck executing 5/0! (Or ANY other number by 0)
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x = generate random number between 10000 and 99995 remainder to be added = 5 - x MOD 5 Therefore, the resultant random number should be x + 5 - x MOD 5 Here's a sample code: http://ideone.com/LxgrC[^]
a.k.a. Vite Phoenix and Vite Zeus... Proud member and co-founder of OlympianZ
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Hi. First, how would I create a random number, and then add the last digit, so that it is divisable by 5? the number should always be 5 digits long. Second, how do I check it, I think I need to do something like...
if (int x MOD 5 ==0)
Or something like that. The first step is the most important though. Thank you, Steve
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Hi. First, how would I create a random number, and then add the last digit, so that it is divisable by 5? the number should always be 5 digits long. Second, how do I check it, I think I need to do something like...
if (int x MOD 5 ==0)
Or something like that. The first step is the most important though. Thank you, Steve
The solution I found most elegant had already been posted by David1984 but is deleted. So I'm going to post it now: You want a number X that can be divided by 5. Any number Y can be multiplied by 5. The result X is divisible by definition (and by 5). You want X to have 5 digits. 5-digit-numbers range from 10000 to 99999. Divide both by 5 and throw away the result's decimal places to get the limits for Y (2000 to 19999). Supposed you already have an instance of
Randomobeying the tips on its use, you get X by using this code (X changed to a meaningful name)int randomDivisibleByFive = random.Next(2000, 19999) * 5;
Ciao, luker
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// To generate as random numbers as possible, this variable must only
// be initialized once and then reused as much as possible.
Random rand = new Random();
/// <summary>
/// Generates a random 5 digit number that is always divisible by 5.
/// </summary>
/// <returns>Returns a random integer 5 digits long that is evenly divisible by 5.</returns>
private int GetDivBy5()
{// rnd.Next(minValue, maxValue) returns a random between minValue // and maxValue, inclusive. Use min/max values of 1000 and 9999 to // yield a random 4 digit number between 1000 and 9999 then multiply // by 10 to get a 5 digit number. This number will always be // divisible by 5 but will always end in 0. int divBy5 = rand.Next(1000, 9999) \* 10; // To get more random and double the possible result set, randomly // add on 5. This will yield the total possible set of integers // between 10000 and 99995 that are divisible by 5. if (rand.NextDouble() < 0.5) return divBy5; else return divBy5 + 5;}
Stephen, you mention you are a coding noobie. So, a short word about random number generators - they're not really very random.
All they do is take a seed (beginning) value and run it through a mathmatical algorithm that generates a new number. If you use the same seed value every time, you will get the same "random" number every time. Most random number generators can either self-seed - generally using the current clock value - or you can pass them a seed value when you initialize them. This is why you only want to initialize your random number generator once and reuse it over and over. Once you get past that, the rest is pretty easy. Just use the math several others have already provided. See the comments in the code... <edit> Um, just looked at the thread in the other forum where you defined your requirements a little better. It seems you would like to include 0-9999 in your result set too. You can do this easily enough by changing the min value passed to rand.Next(minValue, maxValue) to 0 instead of 1000. This last bit, is a bit tricky in that in programming terms "00005" is actually not a number but is really a string. To get the leading zeros, you need to convert your final value to a string and use a formatting mask to append the leading zeros. This also assumes "00000" is valid. If it's not, test for it with an "if" and call GetDivBy5 again. So, to get a string with leading ze