# Given the digit (0, 1,3,5, 6), how many diff. 4 digit numbers can be made that aredivisible by 4 if repetitions are alloweed? If repetitions are not allowed?

**Talent34**- 7 Answers

Hi,

There are totally 100 such integers (4 digit numbers, where 1st digit is nonzero and is one of 1,3,5,6 and the 2nd through 4th digit are one of 0,1,3,5,6 and which are exacrly divisible by 4).

starting with 1000 and ending 6660.

The values are as under:

1000

1016

1036

1056

1060

1100

1116

1136

1156

1160

1300

1316

1336

1356

1360

1500

1516

1536

1556

1560

1600

1616

1636

1656

1660

3000

3016

3036

3056

3060

3100

3116

3136

3156

3160

3300

3316

3336

3356

3360

3500

3516

3536

3556

3560

3600

3616

3636

3656

3660

5000

5016

5036

5056

5060

5100

5116

5136

5156

5160

5300

5316

5336

5356

5360

5500

5516

5536

5556

5560

5600

5616

5636

5656

5660

6000

6016

6036

6056

6060

6100

6116

6136

6156

6160

6300

6316

6336

6356

6360

6500

6516

6536

6556

6560

6600

6616

6636

6656

6660

The macro that got these is as under, embed it into xl and run it.

Sub Macro1()

'

' Macro1 Macro

'

' Keyboard Shortcut: Ctrl+j

'

For i = 1000 To 9999

DoEvents

a$ = Trim(Str(i))

one = Left(a$, 1)

two = Mid(a$, 2, 1)

three = Mid(a$, 3, 1)

four = Right(a$, 1)

myflag = 0

If Val(one) = 1 Or Val(one) = 3 Or Val(one) = 5 Or Val(one) = 6 Then

myflag = myflag + 1

Else

myflag = myflag - 1

End If

If Val(two) = 0 Or Val(two) = 1 Or Val(two) = 3 Or Val(two) = 5 Or Val(two) = 6 Then

myflag = myflag + 1

Else

myflag = myflag - 1

End If

If Val(three) = 0 Or Val(three) = 1 Or Val(three) = 3 Or Val(three) = 5 Or Val(three) = 6 Then

myflag = myflag + 1

Else

myflag = myflag - 1

End If

If Val(four) = 0 Or Val(four) = 1 Or Val(four) = 3 Or Val(four) = 5 Or Val(four) = 6 Then

myflag = myflag + 1

Else

myflag = myflag - 1

End If

If myflag = 4 Then

r = i Mod 4

If r = 0 Then

j = j + 1

Cells(j, 1) = i

End If

End If

Next i

MsgBox "Done"

End Sub

The above 100 are with repetitions allowed.

Now if repetitions are not allowed.

There are only 18!

1036

1056

1356

1360

1536

1560

3016

3056

3156

3160

3516

3560

5016

5036

5136

5160

5316

5360

The macro used is as under in xl module.

Sub Macro1()

'

' Macro1 Macro

'

' Keyboard Shortcut: Ctrl+j

'

For i = 1000 To 9999

DoEvents

a$ = Trim(Str(i))

one = Left(a$, 1)

two = Mid(a$, 2, 1)

three = Mid(a$, 3, 1)

four = Right(a$, 1)

myflag = 0

If Val(one) = 1 Or Val(one) = 3 Or Val(one) = 5 Or Val(one) = 6 Then

myflag = myflag + 1

Else

myflag = myflag - 1

End If

If Val(two) = 0 Or Val(two) = 1 Or Val(two) = 3 Or Val(two) = 5 Or Val(two) = 6 Then

myflag = myflag + 1

Else

myflag = myflag - 1

End If

If Val(three) = 0 Or Val(three) = 1 Or Val(three) = 3 Or Val(three) = 5 Or Val(three) = 6 Then

myflag = myflag + 1

Else

myflag = myflag - 1

End If

If Val(four) = 0 Or Val(four) = 1 Or Val(four) = 3 Or Val(four) = 5 Or Val(four) = 6 Then

myflag = myflag + 1

Else

myflag = myflag - 1

End If

'the rule for checking repetition

If Val(one) = Val(two) Or _

Val(one) = Val(three) Or _

Val(one) = Val(four) Then

myflag = 0

End If

If Val(two) = Val(three) Or _

Val(two) = Val(four) Then

myflag = 0

End If

If Val(three) = Val(four) Then

myflag = 0

End If

If myflag = 4 Then

r = i Mod 4

If r = 0 Then

j = j + 1

Cells(j, 1) = i

End If

End If

Next i

MsgBox "Done"

End Sub

Comment the "rule" to get the inclusive values again (100 of them).
...

First, if repetetion is allowed:

Given digits are 0, 1, 3, 5 and 6.

If we group these numbers into groups of 2, then we can get the following numbers:

00, 01, 03, 05, 06, 10, 11, 13, 15, 16, 30, 31, 33, 35, 36, 50, 51, 53, 55, 56, 60, 61, 63, 65, 66.

Out of these numbers, the numbers that are divisible by 4 are 00, 16, 36, 56 and 60. Thus there are 5 two digit numbers divisible by 4 from the above given digits.

All the numbers that end in 00, 16, 36, 56 or 60 are divisible by 4.

When the last two digits of a number are 0 and 0, then there can be 5 * 4 = 20 such numbers.

When the last two digits of a number are 1 and 6, then there can be 5 * 4 = 20 such numbers.

When the last two digits of a number are 3 and 6, then there can be 5 * 4 = 20 such numbers.

When the last two digits of a number are 5 and 6, then there can be 5 * 4 = 20 such numbers.

When the last two digits of a number are 6 and 0, then there can be 5 * 4 = 20 such numbers.

Therefore the total number of four digit numbers that can be formed from the digits 0, 1, 3, 5 and 6 that are divisible by 4 are 20 * 5 = 100 when repetetion is allowed

Then, if repetetion is not allowed:

When the last two digits of a number are 1 and 6, then there can be 2 * 2 = 4 such numbers.

When the last two digits of a number are 3 and 6, then there can be 2 * 2 = 4 such numbers.

When the last two digits of a number are 5 and 6, then there can be 2 * 2 = 4 such numbers.

When the last two digits of a number are 6 and 0, then there can be 3 * 2 = 6 such numbers.

Therefore the total number of four digit numbers that can be formed from the digits 0, 1, 3, 5 and 6 that are divisible by 4 are 4 * 3 + 6 = 18 when repetetion is not allowed.

If repetitions are allowed, then there are 5*4*3*2 = 120 different 4-digit numbers possible. Of these 120 different numbers the only ones divisble by 4 are those where the last 2 digits are 00, 16, 36, 56, or 60 (per the hint on divisibility by 4). That means you only have the first two digits to play with now. So you have 5*5 = 25 possibilities for each ending 2-digit sequence. Because there are 5 of these (i.e., 00, 16, 36, 56, and 60) you have 5*25=125 different possibilities of a 4-digit number that is divisible by 4.

Without repetition there are Permut(5,4) = (5*4*3*2*1)/(4*3*2*1) = 5 different 4-digit numbers for each of the 5 different possibilities of the first digit. Thus there are 5*5 = 25 different 4-digit numbers you can make out of the integers in question (without any of the digits being used more than once). If you reverse this idea so that the first and second digit are the 16, 36, 56, or 60 (reversed to be 61, 63, 65, and 06 [note that 00 has been removed as a potential base due to repetition of 0]) then you have 3 remaining integers to choose the remaining 2 digits from. Thus you have (3*2)=6 possibilities for each of the 4 bases. Therefore, without repetition, you have 6*4=24 different 4-digit numbers that can be made from (0,1,3,5,6) without repeating any of the digits.

If repetitions are allowed, then there are 5*4*3*2 = 120 different 4-digit numbers possible. Of these 120 different numbers the only ones divisble by 4 are those where the last 2 digits are 00, 56, or 60 (per the hint on divisibility by 4). That means you only have the first two digits to play with now. So you have 5*4 = 20 possibilities for each ending 2-digit sequence. Because there are 3 of these (i.e., 00, 56, and 60) you have 3*20=60 different possibilities of a 4-digit number that is divisible by 4.

Without repetition there are Permut(5,4) = (5*4*3*2*1)/(4*3*2*1) = 5 different 4-digit numbers for each of the 5 different possibilities of the first digit. Thus there are 5*5 = 25 different 4-digit numbers you can make out of the integers in question (without any of the digits being used more than once). If you reverse this idea so that the first and second digit are the 00, 56, or 60 (reversed to be 00, 65, and 06) then you have 3 remaining integers to choose the remaining 2 digits from. Thus you have (3*2*1)/(2*1)=3 possibilities for each of the 3 bases. Therefore, without repetition, you have 3*3=9 different 4-digit numbers that can be made from (0,1,3,5,6) without repeating any of the digits.

In short, there are only 100 digits with repetition. thanks to all who post and clarify their answer!

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