Divisibility Rules! (Brainteaser)

Otis

Otis

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Find a nine-digit number containing no zero or repeated digit such that removing the rightmost digits – one digit at a time – results in remaining numbers that are divisible in turn by 8, 7, 6, 5, 4, 3, 2 and 1.

For example, 921654387 almost works, but removing the rightmost digit does not leave a number that's divisible by 8. However, the remaining steps work:

The 7-digit number 9216543 is divisible by 7
The 6-digit number 921654 is divisible by 6
The 5-digit number 92165 is divisible by 5
The 4-digit number 9216 is divisible by 4
The 3-digit number 921 is divisible by 3
The 2-digit number 92 is divisible by 2
The 1-digit number 9 is divisible by 1

The thread title contains a clue that leads to shortcuts.

?

[imath]\;[/imath]
 
Otis said:
Find a nine-digit number containing no zero or repeated digit such that removing the rightmost digits – one digit at a time – results in remaining numbers that are divisible in turn by 8, 7, 6, 5, 4, 3, 2 and 1.

For example, 921654387 almost works, but removing the rightmost digit does not leave a number that's divisible by 8. However, the remaining steps work:

The 7-digit number 9216543 is divisible by 7
The 6-digit number 921654 is divisible by 6
The 5-digit number 92165 is divisible by 5
The 4-digit number 9216 is divisible by 4
The 3-digit number 921 is divisible by 3
The 2-digit number 92 is divisible by 2
The 1-digit number 9 is divisible by 1

The thread title contains a clue that leads to shortcuts.

?

[imath]\;[/imath]
Click to expand...
One helpful clue could be that:

A number divisible by 8 must have the last 3 digits from RHS be divisible by 8. Those numbers could be 128, 136, 152, etc.
 
Subhotosh Khan said:
One helpful clue ... A number divisible by 8 must have the last 3 digits from RHS be divisible by 8.
Click to expand...
deyesed ("yes, indeed") !

When I worked through the puzzle, I was eliminating digit choices while moving from left to right. Hence, the eighth and ninth digit possibilities fell out at the end of each go. But the divisibility rule above is very handy for other approaches. Another one for 8 is that the number contains at least three factors of 2.

Similarly, divisibility by 4 requires that the two rightmost digits form a multiple of 4. (I used that one a lot.)

Before all that begins, there's a major simplification as well. Yet, it's too early for big clues. ;)

[imath]\;[/imath]
 
Otis said:
deyesed ("yes, indeed") !

When I worked through the puzzle, I was eliminating digit choices while moving from left to right. Hence, the eighth and ninth digit possibilities fell out at the end of each go. But the divisibility rule above is very handy for other approaches. Another one for 8 is that the number contains at least three factors of 2.

Similarly, divisibility by 4 requires that the two rightmost digits form a multiple of 4. (I used that one a lot.)

Before all that begins, there's a major simplification as well. Yet, it's too early for big clues. ;)

[imath]\;[/imath]
Click to expand...
Is that pertaining to locations of 2 & 5?
 
Otis said:
Ya got it! ⭐

Realize the beginning digit-parity must be OEOE5EOEO

[imath]\;[/imath]
Click to expand...

I looked at many combinations before finding one that works.
a=odd
b=even
c=odd
d=even
e=5
f=even
g=?
10g+h = multiple of 8 (note f=even implies 8| 100f = 200*(f/2)
h=even
i had to be odd, since we used the 4 even values
 
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