Random numbers and division

[DrNoose]

I'm helping a friend with a program that uses random numbers. The user
tells the program how many questions to ask and the program will either
add, sub, multiply, or divide 2 numbers. The range of random numbers is
from 0-100. The problem, if there is one, is that many times the 2nd
number is larger when dividing the numbers causing you to have to give a
decimal answer. Is there a way to code the division case selection so
that it only returns a number without being a decimal?

Here is the code for case 4, which does the division of the random numbers.

**********************************************************************

Case 4

'Generate question.
sUserAnswer = InputBox("What is " & iOperand1 & _
" / " & iOperand2)

' Determine if user's answer was correct and add an
' appropriate item to the multi-column list box.
If Val(sUserAnswer) = iOperand1 / iOperand2 Then
lstResults.AddItem iOperand1 & " / " & _
iOperand2 & ";" & sUserAnswer & ";Correct"
Else
lstResults.AddItem iOperand1 & " / " & _
iOperand2 & ";" & sUserAnswer & ";Incorrect"
End If

Thanks!

 

[Van T. Dinh]

Well ... division in the set of natural numbers is not closed according to
the Number Theory whether the dividend is greater or smaller than the
divisor. For example:

Dividend / Divisor
31 / 5 = 6.2
5 / 8 = 0.625

(I think you referred to the second case dividend < divisor)

The results of both divisions are not in the set of natural numbers, i.e.
the operator / is not closed.

If you want to force this to a natural number, you will need to either round
or trucate the results. In Access, there is a integer division operator \
(backlash or slosh) which gives integer result. There are also functions
Int(), CInt(), CLng() and Round() to convert from a decimal value to an
integral value.

Check Access Help topics for these functions / operator ...

 

[DrNoose]

Well ... division in the set of natural numbers is not closed according to
the Number Theory whether the dividend is greater or smaller than the
divisor. For example:

Dividend / Divisor
31 / 5 = 6.2
5 / 8 = 0.625

(I think you referred to the second case dividend < divisor)

The results of both divisions are not in the set of natural numbers, i.e.
the operator / is not closed.

If you want to force this to a natural number, you will need to either round
or trucate the results. In Access, there is a integer division operator \
(backlash or slosh) which gives integer result. There are also functions
Int(), CInt(), CLng() and Round() to convert from a decimal value to an
integral value.

Check Access Help topics for these functions / operator ...

Hi!

Thanks for your help!

 

[Marshall Barton]

I'm helping a friend with a program that uses random numbers. The user
tells the program how many questions to ask and the program will either
add, sub, multiply, or divide 2 numbers. The range of random numbers is
from 0-100. The problem, if there is one, is that many times the 2nd
number is larger when dividing the numbers causing you to have to give a
decimal answer. Is there a way to code the division case selection so
that it only returns a number without being a decimal?

Here is the code for case 4, which does the division of the random numbers.

**********************************************************************

Case 4

'Generate question.
sUserAnswer = InputBox("What is " & iOperand1 & _
" / " & iOperand2)

' Determine if user's answer was correct and add an
' appropriate item to the multi-column list box.
If Val(sUserAnswer) = iOperand1 / iOperand2 Then
lstResults.AddItem iOperand1 & " / " & _
iOperand2 & ";" & sUserAnswer & ";Correct"
Else
lstResults.AddItem iOperand1 & " / " & _
iOperand2 & ";" & sUserAnswer & ";Incorrect"
End If

How about changing the problem by using one of the random
numbers as the answer?

sUserAnswer = InputBox("What is " & _
iOperand1 * iOperand2 & " / " & iOperand2)
If Val(sUserAnswer) = iOperand1 Then
. . .