A Unique 10-digit Number

Took a long time with a lot of small steps that I won't all write down explicitly. But basically, you want to complete this diagram:

where each letter is matched to exactly one number.

First step: since the entire number has to be divisible by 10, we must have $j=0$ . This leaves $e=5$ since $\overline{abcde}$ must be divisible by 5. Similarly, even letters have to equal even numbers and vice versa. Then we must have that 3 divides $a+b+c$ , $d+e+f$ as well as $g+h+i$ . Combining this with divisibility conditions for 4 and 8, you eventually arrive at these possibilities:

Checking $\overline{abcdefg}$ for divisibility by 7 for each of these, you will find that only $3816547290$ suffices. Hence the solution is $\boxed{381}$ .

$3816547290$

Why 381,654,729 is awesome

Guys first of all donot go with the hit-&-trial method. We can figure out the 5th and 10th digit easily. ----5----0 Now, As the no. uptill 3,6&9 digits are divisibleby 3 then sum of digits 1,2&3 ; 4,5&6 ; 7,8&9 will be multiple 3. Each alternate is even from 2nd digit from left. Now, digit 4 + digit 5 + digit 6 =multiple of 3 even +5+ even = multiple of 3 Four cases possible, (2,8),(6,4),(8,2),(4,6) However digit 4 can never be 8&4. As there is no multiple of 4 which has tens digit odd and ones digit 4&8. Now I end the first part of my solution , I will soon post the second part.

Clearly we have e=5 and j=0. Now note that every even digit of the number is also an even number. Since the first 3 digits are divisible by 3 and the first 6 by 6, the fourth digit can either be 2 or 6 as c is odd and f is 8 or 4 respectively. But if d=2,we get b=4, which makes abc not divisible by 3 for the remaining values of numbers. Hence d=6 and h=2.Now note that a+8+b is divisible by 3 which means (a, b) is either (3,1) (3,7) or (9, 1) or the pairs other way round. But 3, 7 will make the 8 digit number not divisible by 8 and 1, 9 will do so for the 7digit number. Now by some trial and error, we get abc=381!!

Clearly e,j=5,0, and the remaining even digits are even and odd digits odd. Since 3|(d+e+f), d+f=1mod3, so {d,f} = {2,8} or {4,6}. Since f is even and 8|fgh it follows that 8|gh. Since g is odd and 8|gh, gh = 16, 32, 72, or 96. So h is 2 or 6.

Suppose, for reductio, that h is 6. Then gh = 96, so g=9. Since 3|ghi, i=3. So {a,c} = {1,7}, and b=1mod3 so b=4. Since 4|cd, and 10 and 70 are 2mod4, d=2mod4, so d=2. Since 3|(d+e+f), f=2mod3, so f=8. This leaves two options for a and c, and neither one yields 7|abcdefg. So the assumption that h is 6 is incorrect, and so h=2 (and we have only identified e,h, and j so far).

Since 3|def d+f=1mod3, {d,f} = {4,6} and b=8. Since h=2, gh is 32 or 72, so g = 3 or 7. Since 3|ghi, ghi is one of 321, 327, 723, or 729. Similarly a+c=1mod3, so {a,c} is one of {1,3}, {1,9}, and {7,1} ({3,7} is ruled out because g is one of 3 and 7). So cd is one of 16, 36, 76, and 96, and so d=6 and f=4. Putting in zeros for a, c, and g, we find that 7|0806540, and so 7|(a*10^6 + c*10^4 + g). Knowing that g is 3 or 7 and that either a or c is 1, the only way to achieve divisibility by 7 is for a,c, and g to be 3, 1, and 7, because 3*10^6=3mod7 and 1*10^4=4mod7.

That gives us 3816547290.

There is actually a good way to think about it.

The 5th digit must be 5 and the tenth digit must be 0.

The sum of the first three digits, middle 3-digits an last 4-digits are all multiples of 3.

So take the middle 3:

x + 5 + y = 3N

x + y = 3N + 1

This can be 4, 7, 10, 13, or 16.

(This is just a start)

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