Alternating numbers

Question

Alternating numbers are numbers in which all digits alternate between even and odd. For example: 2703 and 7230 are alternating numbers, but 2730 isn't.

Numbers are very alternating when double the number is an alternating number as well, for example 3816 is very alternating, because 7632 is an alternating number as well.

The question to you is: how many 4 digits very alternating numbers are there? (the number can't start with one or multiple zeros)

Of course this problem can be solved by programming, but it can be solved mathematically as well! Good luck!

NB: I didn't come up with this puzzle myself, it's part of the Dutch Math Olympiad, see this PDF

Answer 1

There number of such numbers is

70

When you double a number, a digit of the result is even if and only if the digit to the right didn't carry.

Therefore, $2x$ is alternating if and only if the digits of $x$ are $LHLH$, where $L \in \{0,1,2,3,4\}$ and $H \in \{5,6,7,8,9\}$.

(As a side note, this implies that if $x$ is very alternating, then $2x$ will still be a four digit number, so that $x < 5000$ necessarily).

We know that if $x$ is alternating when it looks like OEOE or EOEO. For $x$ to be very alternating, it also must be $LHLH$, so let's count the number of ways to fulfill both these constraints.

• OEOE: Each odd number must be low, meaning they are each $1$ or $3$. Each even must be high, so is either $6$ or $8$. There are two choices for each digit for a total of $2\cdot2\cdot2\cdot2=16$ possibilities.

• EOEO: Now the evens are low ($0,2$ or $4$), and the odds are high ($5,7$ or $9$). There are three choices for each digit, except the first digit can't be zero, so there are $2\cdot3\cdot3\cdot3=54$ possibilities.

These two counts add to the desired answer.

Answer 2

The answer is

70

I did a manual count. Still trying to work out a mathematical solution.

Answer 3

Damn I started this ^{brute force} before the answers and I got

70

VBA

Better logic for this route would have been

if (one mod 2 = three mod 2) and (two mod 2 = four mod 2) then
    if one mod 2 <> two mod 2 then
     counter increase

Answer 4

Here we go, brute force again!

The answer is, as pointed out in other answers:

70

You may check JSFiddle for source code and full list of 4-digit very alternative numbers.

Answer 5

The correct answer is:

70

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#AnubhavBalodhi, puzzling.stackexchange,26036, 1/2/16 2341 hrs
E=['0','2','4','6','8']
O=['1','3','5','7','9']

ans=0
alters=[]

def Alt(num):
N=str(num)
if len(N)<5: #4 digit
    if (N[0] in E and N[1] in O and N[2] in E and N[3] in O) or (N[0] in O and N[1] in E and N[2] in O and N[3] in E):
        alters.append(num)
else:       #5 digit
    if (N[0] in E and N[1] in O and N[2] in E and N[3] in O and N[4] in E ) or (N[0] in O and N[1] in E and N[2] in O and N[3] in E and N[4] in O):
        alters.append(num)

for num in range(1000,10**5):
Alt(num)
#print(num)

print(len(alters))

for numb in alters:
if numb<9999 and numb*2 in alters: #if the number in alternating list is of 4 digits, and its double is also in the list.
    print(numb)
    ans+=1
print("ans is %d" %(ans))