## FoolForMath Group Title Just another cute problem: Suppose $$xyz$$ is a three digit number such that $$xzy + yxz+yzx+zxy+zyx = 3024$$, then can you find $$x \times y \times z$$? 2 years ago 2 years ago

1. ParthKohli Group Title

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

2. FoolForMath Group Title

Could be, my approach is somewhat different.

3. ParthKohli Group Title

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

4. ParthKohli Group Title

let me think

5. ParthKohli Group Title

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

6. cwtan Group Title

need Permutation?

7. ParthKohli Group Title

Nope

8. FoolForMath Group Title

@cwtan: I used permutation.

9. karatechopper Group Title

what is permutation

10. ParthKohli Group Title

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

11. FoolForMath Group Title

Number of arrangement.

12. ParthKohli Group Title

Number of arrangements if order matters.

13. ParthKohli Group Title

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

14. karatechopper Group Title

i think im gettting to it!!

15. karatechopper Group Title

wait..whats the permutation formula?

16. karatechopper Group Title

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

17. ParthKohli Group Title

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

18. FoolForMath Group Title

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

19. karatechopper Group Title

can u give me a hint..

20. cwtan Group Title

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

21. joemath314159 Group Title

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

22. FoolForMath Group Title

Yes, approach? :)

23. joemath314159 Group Title

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

24. siddhantsharan Group Title

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

25. joemath314159 Group Title

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 Group Title

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 Group Title

Interesting approach can we generalize it?

28. FoolForMath Group Title

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

29. joemath314159 Group Title

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 Group Title

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 Group Title

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

32. cwtan Group Title

So the answer is ? 126?

33. karatechopper Group Title

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..