Adding 0 makes no change, Really ?

Number Theory Level 3

Given that in the 8-digit number $\overline{\mathrm{ABCDEFGH}}$ ,

(i) $\overline{\mathrm{ABCD}}=\overline{\mathrm{EFGH}}$
(ii) The numbers $\overline{\mathrm{ABCDEFGH}}$ and $\overline{\mathrm{ABCD0EFGH}}$ are both divisible by $11$ .
Let the sum of all possible values of $\overline{\mathrm{ABCDEFGH}}$ be $N$ .

Find the digit sum of $N$ .

Details and assumptions :-

$\overline{\mathrm{ABC}}$ means the number in decimal representation with digits $A,B,C$ i.e. $\overline{\mathrm{ABC}} = \mathrm{100A+10B+C}$
The letters $A$ to $H$ do not necessarily stand for distinct digits.
In the second number, the digit $0$ is added in the middle of the 8-digit number, which makes it a 9-digit number.
Digit sum is sum of all digits in decimal representation, digit sum of $12023$ is $1+2+0+2+3=8$
00123 is not a 5-digit number.

The answer is 36.

3 solutions

Aditya Raut
Oct 21, 2014

Because it's given that $\overline{ABCD}=\overline{EFGH}$ , we have

$\overline{ABCDEFGH} = \overline{ABCDABCD} = 10001\times \overline{ABCD}$ .

Because $11 \nmid 10001$ , we obtain that $11 \mid\overline{ABCD}$ .

The $\overline{ABCD0EFGH}$ part is just a troll, because it will always be divisible by 11

$\overline{ABCD0EFGH} = \overline{ABCD0ABCD} = \overline{ABCD} \times 100001$ and as $100001$ is divisible by 11, this is always divisible by $11$ .

From this we conclude that $\overline{ABCD}$ is a 4-digit number divisible by $11$ , that is $(1001,1012,....,9999)$

Hence sum of all the values of $\overline{ABCDEFGH}$ is

$\displaystyle 10001\times \sum_{k=91}^{909} 11k \\ = 10001\times \biggl( 819 \times \dfrac{1001+9999}{2} \biggr)\\ = 10001\times 4504500 \\ = \color{#3D99F6}{\mathrm{45049504500}}$

Digit sum of $45049504500$ is $\boxed{36}$

Nice solution, same way here: except I used a slightly different summation

$10001\displaystyle \sum_0^{818} 11x+1001$

Trevor Arashiro - 6 years, 7 months ago

@Aditya Raut as ,You said we could use python for this soo below is my python code

def sd(n):  #sum of digits
    s = 0
    while n:
        s += n % 10
        n /= 10
    return s
k=0
for a in range(1,10):
    for b in range(10):
        for c in range(10):
            for d in range(10):
                for e in range(10):
                    for f in range(10):
                        for g in range(10):
                            for h in range(10):
                                if (1000.0*a+100.0*b+10.0*c+1.0*d==1000.0*e+100.0*f+10.0*g+1.0*h) and (10**7*a+10**6*b+10**5*c+10**4*d+10**3*e+10**2*f+10**1*g+10**0*h)%(11.0) == 0:
                                    if (10**8*a+10**7*b+10**6*c+10**5*d+10**3*e+10**2*f+10**1*g+10**0*h) % (11.0) == 0:
                                        k+=(10**7*a+10**6*b+10**5*c+10**4*d+10**3*e+10**2*f+10**1*g+10**0*h)
print 'answer is',sd(k)

can u tell me yours, Aditya

Mehul Chaturvedi - 6 years, 2 months ago

You want a programming solution? Here's the one that takes less than 1 second for giving the answer.... @Mehul Chaturvedi you will surely like this.... I have first made the integers into strings, then I added strings which is way easier and faster for the computer, then I converted the strings back to integers for checking divisibility by 11, and finally made all of them append to a list, asking sum of that list...

>>> a=[]
>>> b=[]
>>> c=[]
>>> for i in range(1000,10000):
    a.append(''.join(str(i)))

>>> for i in a:
    if int(i+i)%11==0 and int(i+'0'+i)%11==0:
        b.append(i+i)

>>> for i in b:
    c.append(int(i))


>>> sum(c)
45049504500L

Now easily get the sum of digits which is 36. Please read and contact if you don't understand something, this is way better than giving every letter all values and then checking loads of calculations you're gonna love this way...

Aditya Raut - 6 years, 2 months ago

I guess your program must have taken a lottttt time to give answer, also it is not bug free, it's not printed intended things

This is one inelegant way because the computer has to make lot calculations, for each letter 10 choices, then the * and + function, then == function.... Will post mine soon...

Aditya Raut - 6 years, 2 months ago

No, @Aditya Raut It took just 1 minute

Mehul Chaturvedi - 6 years, 2 months ago

@Mehul Chaturvedi – 1 minute is way out of efficiency for a program of this kind man, this problem must get solution in maximum 5 seconds...

Aditya Raut - 6 years, 2 months ago

First of all, let me give you one more way of getting sum of digits, it is a hint towards my programming solution for the problem:-

def sd(n):
    li=[]
    x=''.join(str(n))
    for i in x:
        li.append(int(i))
    k=0
    for i in li:
        k+=i
    return k

@Mehul Chaturvedi

Aditya Raut - 6 years, 2 months ago

@Aditya Raut I just know basic commands in Python ,therefore my solution was long....would you suggest me some source from where i can learn python,or from where did u learn. And please check the message on slack

Mehul Chaturvedi - 6 years, 2 months ago

Michael Occhipinti
Apr 10, 2016

Its easy for me to visualize the concept, but not the summation, so here's python brute force

for $i$ in range( $1000$ $,10000$ ):

$x=i*10000 + i$

$if ((x/11)==x/11.):$

      z+=x

print $z$

then do digit sum in head :)

K T
Mar 4, 2019

From(ii) it follows that the difference $90000×\overline{ABCD}$ must also be divisible by 11. Because 90000 and 11 are coprime, $\overline{ABCD}$ itself must be divisible by 11.

Multiples of 11 that form 4-digit numbers are $91×11=1001, 92×11, ... \text{through } 909×11=9999$ .

The sum of all these is $11×(91+92+...+909)=11×\frac{(909+91)×(909-91+1)}{2}=4504500$ .

The sum of all $\overline{ABCDABCD}$ then is 10001 times this, so 45049504500 of which the digit sum is $\boxed{36}$ .

Note: I misread the question first, so I calculated N itself.

Im 7th grade and i fon't understend furst part when u said "comprime" can u help plz haha

Syio t - 2 years ago

'Coprime' means the same as 'relative prime'. Saying a and b are coprime is equivalent to saying gcd(a,b)=1. The only factor they share is 1.

K T - 2 years ago

Adding 0 makes no change, Really ?

The answer is 36.

3 solutions

then do digit sum in head :)

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