Enter a;

enter b;

enter c;

if a>=b and a>c

print "a is the greatest number"

then if b>=a and b>c

print "b is the greatest number"

then if a=b and a=c

print "a, b, c are equal"

if not print "c is the greatest number"

this is pseudocode.. or my drunken attempt to remember pseudocode... but I think the logic could work.

Posted on Nov 27, 2008

To display an even number when an odd number is entered you can add 1.

n= val(text1.text) 'Take the value of n from the textbox.

If n mod 2 =1 then 'If n is an odd number n=n+1 'You make it an even number end if

text2.text = n 'Display it in another textbox.

Please give a good rating to this answer so that I will be encouraged to support people like you having problems with VB6.

n= val(text1.text) 'Take the value of n from the textbox.

If n mod 2 =1 then 'If n is an odd number n=n+1 'You make it an even number end if

text2.text = n 'Display it in another textbox.

Please give a good rating to this answer so that I will be encouraged to support people like you having problems with VB6.

Sep 22, 2011 | Operating Systems

Download link Visual Basic 6.0 Version.

1. Microsoft Visual Basic 6.0 Service Pack 6 Cumulative Update

2. Service Pack 6 for Visual Basic 6.0, Visual C++ 6.0 with Visual Source Safe 6.0d

3. Service Pack 6 for Visual Basic 6.0: Run-Time Redistribution Pack (vbrun60sp6.exe)

4. Service Pack 6 for Visual Basic 6.0

Hope it helps.

1. Microsoft Visual Basic 6.0 Service Pack 6 Cumulative Update

2. Service Pack 6 for Visual Basic 6.0, Visual C++ 6.0 with Visual Source Safe 6.0d

3. Service Pack 6 for Visual Basic 6.0: Run-Time Redistribution Pack (vbrun60sp6.exe)

4. Service Pack 6 for Visual Basic 6.0

Hope it helps.

May 21, 2011 | Operating Systems

<a href="http://www.brothersoft.com/downloads/visual-basic-6.0.html">http://www.brothersoft.com/downloads/visual-basic-6.0.html</a><br /><br /><br />go to the above given link and download the file,<br />if its compressed extract it, and run the setup.<br /><br />hope this one helps you out.....<br /><br />

Aug 29, 2010 | Microsoft Operating Systems

this would be explained on the basis of two FACTS below.......

sum of two odd numbers are always even and

sum of to even numbers are also always even

then how 6 odd numbers (consider 3 pairs, 9 5 3 1 being odd numbers) can make an odd number 21..!!

SO SIMPLE

Make your basics strong...

sum of two odd numbers are always even and

sum of to even numbers are also always even

then how 6 odd numbers (consider 3 pairs, 9 5 3 1 being odd numbers) can make an odd number 21..!!

SO SIMPLE

Make your basics strong...

Aug 14, 2010 | Operating Systems

Check the product pack, copy all the serial numbers you see. Then call microsoft support via the phone numbers given. Once you quote these numbers, they give you instructions.

Mar 18, 2010 | Microsoft Windows Vista Home Basic for PC

okay its easy do this click start then go to all programs check all programs untill u come to microsoft visual studio 6.0 follow the right arrow and screw down to microsoft visual basic 6.0 thats it u are done.click ok ok and u are home

Nov 11, 2009 | Operating Systems

Dear, the following page contains the tutorial for beginner to visual basic 6, I hope that would be very helpful to you.

http://www.hitmill.com/programming/vb/beginner.html

Best regards.

http://www.hitmill.com/programming/vb/beginner.html

Best regards.

May 26, 2009 | Microsoft Windows XP Professional

This code generates some random number to test.

#include <limits.h>

#include <stdio.h>

#include <stdlib.h>

int main()

{ findLargest();

return 0;

}

int findLargest()

{ int someNumbers[50];

int i;

// to generate some numbers for us to test

for(i=0; i<50; i++)

// generate a random number between 0-100

someNumbers[i] = rand() % 100;

// for keeping track of numbers, set as smallest possible

int largest = INT_MIN;

int largest2 = INT_MIN;

// go through each item in the array

for(i=0; i<50; i++)

{ // if bigger than our previous max, set as new max

if (someNumbers[i] > someNumbers[largest])

largest = i;

// if it's not been set as new max, and is bigger than current 2nd largest

else if(someNumbers[i] > someNumbers[largest2])

largest2 = i;

// for printing all numbers in the array

printf("%d | %d\n",i, someNumbers[i]);

}

// print largest numbers and their position in the array

printf("largest %d (pos %d).\n2nd largest %d (pos %d)",

someNumbers[largest],

largest,

someNumbers[largest2],

largest2

);

return 0;

}

#include <limits.h>

#include <stdio.h>

#include <stdlib.h>

int main()

{ findLargest();

return 0;

}

int findLargest()

{ int someNumbers[50];

int i;

// to generate some numbers for us to test

for(i=0; i<50; i++)

// generate a random number between 0-100

someNumbers[i] = rand() % 100;

// for keeping track of numbers, set as smallest possible

int largest = INT_MIN;

int largest2 = INT_MIN;

// go through each item in the array

for(i=0; i<50; i++)

{ // if bigger than our previous max, set as new max

if (someNumbers[i] > someNumbers[largest])

largest = i;

// if it's not been set as new max, and is bigger than current 2nd largest

else if(someNumbers[i] > someNumbers[largest2])

largest2 = i;

// for printing all numbers in the array

printf("%d | %d\n",i, someNumbers[i]);

}

// print largest numbers and their position in the array

printf("largest %d (pos %d).\n2nd largest %d (pos %d)",

someNumbers[largest],

largest,

someNumbers[largest2],

largest2

);

return 0;

}

May 07, 2009 | Operating Systems

This is a common starter challenge in college and high school programming classes. While i'm not willing to code it for you, I can show you the methods you'll need to understand in order to approach the problem. The easiest way of doing this is by using a brute force method. Just keep testing the number to see if the remainder after dividing it with a number lower than it is zero. If so, it's a factor.

For example, If the number is 35 then pick a number just one lower than it that's not the number 1. Now divide the two numbers. 35/34 = a number with many decimal places after it, which means its not one of the factors of the number. However, eventually, you'll approach the number 7, and 35/7 = 5. This number has no decimal, meaning it IS a factor of 35, and also the largest non-zero factor of 35. Your algorithm can now stop, since it found the largest non-zero number.

Here's some pseudo-code:

int largestfactor(int number) {

int temp = number;

for(temp - 1; temp>=1; temp--) {

if (number % temp == 0) return temp

}

return temp

}

Something like that, with some bug tweaks will do it. The "%" sign is called a modulus. It's a standard c++ operator. Good luck, and remember to vote Fixya if you liked it.

The wikipedia article that helps you with better and more effecient algorithms can be found here.

Steven

For example, If the number is 35 then pick a number just one lower than it that's not the number 1. Now divide the two numbers. 35/34 = a number with many decimal places after it, which means its not one of the factors of the number. However, eventually, you'll approach the number 7, and 35/7 = 5. This number has no decimal, meaning it IS a factor of 35, and also the largest non-zero factor of 35. Your algorithm can now stop, since it found the largest non-zero number.

Here's some pseudo-code:

int largestfactor(int number) {

int temp = number;

for(temp - 1; temp>=1; temp--) {

if (number % temp == 0) return temp

}

return temp

}

Something like that, with some bug tweaks will do it. The "%" sign is called a modulus. It's a standard c++ operator. Good luck, and remember to vote Fixya if you liked it.

The wikipedia article that helps you with better and more effecient algorithms can be found here.

Steven

Sep 15, 2008 | Operating Systems

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