Find 5 digit number that any 3 digit number it contains is divisible by 17 or 23

[testtest88]

Here is the question for my 12 year old.
Any idea as how to start this type of questions?
Thanks a million.

Find the number of 5-digit numbers less than 40000 in which the 3 3-digit numbers formed by consecutive digits are all divisible by either 17 or 23. Remember, zero cannot be the first digit in a 3-digit number.Remember, zero cannot be the first digit in a 3-digit number.

 

[Ishuda]

Not that I have a solution but just to see if I understand:
Let
N5 = a *10⁴ + b 10³ + c 10² + d 10 + e; a, b, and c not zero.
N3₁ = a *10² + b 10 + c
N3₂ = b *10² + c 10 + d
N3₃ = c *10² + d 10 + e
Find all N5 less than 40000 such that N3₁ , N3₂ , and N3₃ are divisible by either 17 or 23.

Seems to me that you would have eight cases you need to investigate. That is
(1) N3₁ divisible by 17, N3₂ divisible by 17, and N3₃ divisible by 17
(2) N3₁ divisible by 17, N3₂ divisible by 17, and N3₃ divisible by 23
(3) N3₁ divisible by 17, N3₂ divisible by 23, and N3₃ divisible by 17
(4) N3₁ divisible by 17, N3₂ divisible by 23, and N3₃ divisible by 23
...

Not seeing any other way right off hand, I would probably brute force it. You could set up some general equations which needed to be satisfied in a spread sheet and pick out corresponding groups. For example, for (1) you would need
a1 = n*17 - int(n*17/100)*100 [= b 10 + c above] > 10
b1 = int(10*a1/17)
c1 = (b1+1)*17
d1 = int(c1 / 10)
If d1 = a1 then you have constructed a number which satisfies N3₁ and N3₂ above. Now work on N3₃ the same way. Oh, and the eight cases could be set up in such a way as to only need to change at most 2 numbers to get each individual case.

Or you could just write a computer program to get the numbers which might be simpler.

 

[testtest88]

thanks a lot for your replies

It looks like no simple tricky solutions. I will just tell my kid to use bruce force approach.Thanks.