I am not able to apply my vtp 8.0.1.

it is showing error the exception integer division by zero.(0x0000094)occured in the application at location0x10001f6d. please suggest me some solution to overcome this problem.

Hi,

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Posted on Jan 02, 2017

Try rolling back with System Restore. It can be found in help, right clicking My Computer, or typing System Restore in Search.

You may need to try a few dates, prior to the problem starting.

You may need to try a few dates, prior to the problem starting.

Dec 24, 2012 | Computers & Internet

The cause, according to Borland:</font color=orange> (http://info.borland.com/devsupport/pascal/><font)

< font color=blue>"Applications that use the CRT unit may generate this error message when running on very fast machines (i.e. Pentium Pro 180 and above). The cause of this error is a timing loop that occurs as part of the initialization of the CRT unit. This timing loop counts how many clock ticks occur within the loop and then that number is divided by 55. The result of this division is a value that is too large to fit into an integer value. The 'Divide by 0' error message is the catch-all error that is displayed when this overflow occurs."</font color=blue>

Here is the more in depth information about the Runtime Error 200</font color=orange> (http://www.merlyn.demon.co.uk/pas-r200.htm><font) including various fixes/patches to download.

I realize that most, if not all, of this information is older but so is the problem.

Hope it helps,

< font color=blue>"Applications that use the CRT unit may generate this error message when running on very fast machines (i.e. Pentium Pro 180 and above). The cause of this error is a timing loop that occurs as part of the initialization of the CRT unit. This timing loop counts how many clock ticks occur within the loop and then that number is divided by 55. The result of this division is a value that is too large to fit into an integer value. The 'Divide by 0' error message is the catch-all error that is displayed when this overflow occurs."</font color=blue>

Here is the more in depth information about the Runtime Error 200</font color=orange> (http://www.merlyn.demon.co.uk/pas-r200.htm><font) including various fixes/patches to download.

I realize that most, if not all, of this information is older but so is the problem.

Hope it helps,

Dec 23, 2012 | Computers & Internet

Below is a listing of the more commonly experienced processor exceptions ranging from 00 to 0F.

**00 = Divide Fault**

Occurs if division by zero is attempted or if the result of the operation does not fit in the destination operand.

**02=NMI interrupt**

Interrupt 2 is reserved for the hardware Non-Maskable-Interrupt condition. No exceptions trap through interrupt 2.

**04=Overflow trap**

Occurs after an INTRO instruction has executed and the OF bit is set to 1.

**05=Bounds Check fault**

The array Index is out of range

**06=Invalid Opcode fault**

This error can be caused by one of the below conditions.

- Processor attempting to decode a bit pattern that does not correspond to any legal computer instruction.
- Processor attempts to execute an instruction that contains invalid operands.
- Processor attempts to execute a protected-mode instruction while running in virtual 8086 mode.
- Processor attempts to execute a LOCK prefix with an instruction that cannot be locked.

**07=Copressor not available fault.**

This error can occur if no math coprocessor is present. This error can also occur when the math coprocessor is used and a task switch is executed.

**08=Double Fault.**

This error occurs when processing an exception triggers a second exception.

**09(OD)=Copressor Segment Overrun.**

Floating point operand is outside the segment.

**10(0Ah/0A)=Invalid Task State Segment Fault**

Can be caused by a number of possibilities as Task State Segment contains a number of descriptors.

**11(0Bh)=Not Present Fault**

The Not Present interrupt allows the operating system to implement virtual memory through the segmentation mechanism. 0B fault occurs when this segment is not available.

**12(0Ch)=Stack Fault**

Occurs when instruction refers to memory beyond the limit of the stack segment.

**13(Odh)=General Protection Fault**

Caused
by any condition that is not covered by any of the other processor
exceptions. The exception indicates that this program has been corrupted
in memory generally resulting in the immediate termination of the
program.

**14(Oeh)=Page Fault**

Occurs when a paging
protection rule is violated (when the retrieve fails, data retrieved is
invalid or the code that issued the fault broke the protection rule for
the processor).

**16(10h)=Coprocessor error fault**

Occurs when an unmasked floating-point exception has signaled a previous instruction.

**17(11h)=Alignment Check Fault**

Only
used on 80486 computers. Caused when code executing at ring privilege 3
attempts to access a word operand that is not divisible by four, or a
long real or temp real whose address is not divisible by eight.

Find out the reason of Fatal Error if that is due to Software related then go in safe mode and uninstall the software recently installed or udate them/patch them. or if the same is related to hardware then unplug the recently added hardware,

please vote

Aug 21, 2011 | Computers & Internet

I'm not going to do your homework for you, but here's a hint. You can use the Java modulus operator % to find if a number is divisible by 5. It provides the remainder after dividing, which will be zero if the number is divisible by 5. Here's a pseudocode example:

if ( (testNumber % 5) == 0)

{

numberIsDivisibleBy5 = true

}

else

{

numberIsDivisibleBy5 = false

}

You'll need to provide the rest of the work to put this test into a loop that goes from >40 to <250, checking each number and doing whatever you are supposed to (print a message, probably) when you find one divisible by 5.

Good luck and thanks for using Fixya.

if ( (testNumber % 5) == 0)

{

numberIsDivisibleBy5 = true

}

else

{

numberIsDivisibleBy5 = false

}

You'll need to provide the rest of the work to put this test into a loop that goes from >40 to <250, checking each number and doing whatever you are supposed to (print a message, probably) when you find one divisible by 5.

Good luck and thanks for using Fixya.

Jan 26, 2011 | Computers & Internet

This is from wikipedia:

When division is explained at the elementary arithmetic level, it is often considered as a description of dividing a set of objects into equal parts. As an example, consider having ten apples, and these apples are to be distributed equally to five people at a table. Each person would receive = 2 apples. Similarly, if there are 10 apples, and only one person at the table, that person would receive = 10 apples.

So for dividing by zero - what is the number of apples that each person receives when 10 apples are evenly distributed amongst 0 people? Certain words can be pinpointed in the question to highlight the problem. The problem with this question is the "when". There is no way to distribute 10 apples amongst 0 people. In mathematical jargon, a set of 10 items cannot be partitioned into 0 subsets. So , at least in elementary arithmetic, is said to be meaningless, or undefined.

Similar problems occur if we have 0 apples and 0 people, but this time the problem is in the phrase "**the** number". A partition is possible (of a set with 0 elements into 0 parts), but since the partition has 0 parts, vacuously every set in our partition has a given number of elements, be it 0, 2, 5, or 1000. If there are, say, 5 apples and 2 people, the problem is in "evenly distribute". In any integer partition of a 5-set into 2 parts, one of the parts of the partition will have more elements than the other.

In all of the above three cases, , and , one is asked to consider an impossible situation before deciding what the answer will be, and that is why the operations are undefined in these cases.

To understand division by zero, we must check it with multiplication: multiply the quotient by the divisor to get the original number. However, no number multiplied by zero will produce a product other than zero. To satisfy division by zero, the quotient must be bigger than all other numbers, i.e., infinity. This connection of division by zero to infinity takes us beyond elementary arithmetic (see below).

A recurring theme even at this elementary stage is that for every undefined arithmetic operation, there is a corresponding question that is not well-defined. "How many apples will each person receive under a fair distribution of ten apples amongst three people?" is a question that is not well-defined because there can be no fair distribution of ten apples amongst three people.

There is another way, however, to explain the division: if we want to find out how many people, who are satisfied with half an apple, can we satisfy by dividing up one apple, we divide 1 by 0.5. The answer is 2. Similarly, if we want to know how many people, who are satisfied with nothing, can we satisfy with 1 apple, we divide 1 by 0. The answer is infinite; we can satisfy infinite people, that are satisfied with nothing, with 1 apple.

Clearly, we cannot extend the operation of division based on the elementary combinatorial considerations that first define division, but must construct new number systems.

[edit]

Oct 08, 2010 | Puzzle Massey Ferguson Tractor

Hi,

The following information on "Hierarchy of Operations" was taken from Pages 45-46 of the "Microsoft QuickBasic 4.0: Basic Language Reference" manual for Versions 4.00 and 4.00b. This information also applies to the following products:

For more information on the "Hierarchy of Operations," consult the Basic language reference manual for your version of Basic. Please post your feedback and Vote if the problem resolved as per your satisfaction.

The following information on "Hierarchy of Operations" was taken from Pages 45-46 of the "Microsoft QuickBasic 4.0: Basic Language Reference" manual for Versions 4.00 and 4.00b. This information also applies to the following products:

- Microsoft GW-Basic Versions 3.20, 3.22, and 3.23
- QuickBasic Versions 1.00, 1.01, 1.02, 2.00, 2.01, 3.00, 4.00, 4.00b, and 4.50
- Microsoft Basic Compiler Versions 5.35, 5.36, 6.00, and 6.00b
- Microsoft Basic PDS Version 7.00.

- Arithmetic operations

- Exponential (^)
- Negation (-)
- Multiplication and division (*, /)
- Integer division (\)
- Modula arithmetic (MOD)
- Addition and subtraction (+, -)

- Relational operations (=, >, <, <>, <=, >=)
- Logical operations

- NOT
- AND
- OR
- XOR
- EQV
- IMP

For more information on the "Hierarchy of Operations," consult the Basic language reference manual for your version of Basic. Please post your feedback and Vote if the problem resolved as per your satisfaction.

Jul 13, 2010 | Microsoft Windows XP Professional

his problem can occur when you give the workbook a defined name and then copy the worksheet several times without first saving and closing the workbook, as in the following sample code:
Sub CopySheetTest()
Dim iTemp As Integer
Dim oBook As Workbook
Dim iCounter As Integer
' Create a new blank workbook:
iTemp = Application.SheetsInNewWorkbook
Application.SheetsInNewWorkbook = 1
Set oBook = Application.Workbooks.Add
Application.SheetsInNewWorkbook = iTemp
' Add a defined name to the workbook
' that RefersTo a range:
oBook.Names.Add Name:="tempRange", _
RefersTo:="=Sheet1!$A$1"
' Save the workbook:
oBook.SaveAs "c:\test2.xls"
' Copy the sheet in a loop. Eventually,
' you get error 1004: Copy Method of
' Worksheet class failed.
For iCounter = 1 To 275
oBook.Worksheets(1).Copy After:=oBook.Worksheets(1)
Next
End Sub

Nov 18, 2009 | Computers & Internet

TRY,, DRIVER MAX free program i always had problems with driver detevtive

May 22, 2009 | HP Pavilion m7480n (882780389410) PC...

The picture link didn't work for me... and seeing as you asked in August of 2008, I'm certain you don't need this answer, but here you go...

According to the TI-89 Guidebook:

This notation indicates an “arbitrary integer” that represents any integer. When an arbitrary integer occurs multiple times in the same session, each occurrence is numbered consecutively. After it reaches 255, arbitrary integer consecutive numbering restarts at @n0.

This notation indicates an “arbitrary constant” that represents any integer. When an arbitrary constant occurs multiple times in the same session, each occurrence is numbered consecutively. After it reaches 255, arbitrary integer consecutive numbering restarts at @0.

So it's the same concept, just a little different. It all depends if it is shown as @n1 or @1 where 1 can be any integer to 255

Hope this helps

According to the TI-89 Guidebook:

This notation indicates an “arbitrary integer” that represents any integer. When an arbitrary integer occurs multiple times in the same session, each occurrence is numbered consecutively. After it reaches 255, arbitrary integer consecutive numbering restarts at @n0.

This notation indicates an “arbitrary constant” that represents any integer. When an arbitrary constant occurs multiple times in the same session, each occurrence is numbered consecutively. After it reaches 255, arbitrary integer consecutive numbering restarts at @0.

So it's the same concept, just a little different. It all depends if it is shown as @n1 or @1 where 1 can be any integer to 255

Hope this helps

Aug 28, 2008 | Texas Instruments TI-89 Calculator

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