## FoolForMath 3 years ago Just another cute problem: Suppose $$xyz$$ is a three digit number such that $$xzy + yxz+yzx+zxy+zyx = 3024$$, then can you find $$x \times y \times z$$?

1. ParthKohli

XYZ = 100x + 10y + z Is it something related to this?

2. FoolForMath

Could be, my approach is somewhat different.

3. ParthKohli

100x + 10z + y + 100y + 10x + z ...... = 3024

4. ParthKohli

let me think

5. ParthKohli

122x + 212y + 221x = 3024 I can't go any further :?

6. cwtan

need Permutation?

7. ParthKohli

Nope

8. FoolForMath

@cwtan: I used permutation.

9. karatechopper

what is permutation

10. ParthKohli

Hmm...I'm not sure which permutation to use here.

11. FoolForMath

Number of arrangement.

12. ParthKohli

Number of arrangements if order matters.

13. ParthKohli

As far as I know, you used permutations only to arrange xyz in different ways

14. karatechopper

i think im gettting to it!!

15. karatechopper

wait..whats the permutation formula?

16. karatechopper

ffm... http://www.mathwords.com/p/permutation_formula.htm in this liinik what do the dots mean in the formula

17. ParthKohli

Permutations formula: $$\Large \color{Black}{\Rightarrow _nP_r = {n! \over (n - r)!} }$$

18. FoolForMath

... this is not a straight forward permutation problem.

19. karatechopper

can u give me a hint..

20. cwtan

I like this question when i am unable to solve it......

21. joemath314159

is the answer:$x\times y\times z = 126$?

22. FoolForMath

Yes, approach? :)

23. joemath314159

very weird, im pretty sure its nowhere near the best way to look at this.

24. siddhantsharan

222 * 18 - 3024 = 972 --- Whose sum of digits = 18 too.

25. joemath314159

Note that 3024 is divisible by 9. Also note that if the three digit number xyz leaves a remainder of r when divided by 9, that any permutation of the digits must leave the same remainder. So adding up those 5 permutations will leave 5r as a remainder. Hence we have this equation:$5r \equiv 0 \mod 9$the only solution to this is r = 0 , so the number xyz is divisible by 9, which means the sum of its digits is divisible by 9. Now use what siddhantsharan posted above, using the fact that x+y+z can only be 0, 9, 18, or 27. Its guess and check from there.

26. siddhantsharan

Actually the above post by joemath shortens it down to only 18. As x + y + z must be 15 at least For it to be > 3024. And 222*27 - 3024 will obviously not leave a 3 digit no. Hence 18 has to be correct. Without any guess and check.

27. FoolForMath

Interesting approach can we generalize it?

28. FoolForMath

Why do you choose divisibility by 9 in the first place?

29. joemath314159

because of how divisibility by 9 is sorta tied to the digits of the number. You can find a numbers remainder when you divide by 9 by adding the digits together. it seemed like a lucky break that 3024 was divisible by 9.

30. joemath314159

hmm...actually i take that back, the problem will still be doable even if that sum wasnt divisible by 9. Since we are adding 5 numbers, we will always be able to solve the equation:$5r\equiv k \mod 9$Since 5 is invertible (mod 9).

31. joemath314159

where k is the remainder of the sum after division by 9.

32. cwtan

So the answer is ? 126?

33. karatechopper

i googled and got 126 but at first when i was solving problem i thought i had to find out what x y and z were..