ns am stuck to the question, i have tried a couple of random approaches yet none of them is correct. The answer is 2. Please assist if you know exactly how to resolve this question.

You are watching: What can 38 be divided by


*

*

Using the fact that 2nd and third component should both it is in either even or odd, number of possibilities are reduced very much. For first part =8, 21, 9 room the solution.For very first part= 16, 7, 15 space the solution


*

Assuming that each part must it is in an integer, this is equivalent to the number of non an unfavorable integral services of$$8x+7y+3z=38$$

We have $x,y,zge1$. So, permit $a=x-1,b=y-1,c=z-1$ such the $a,b,cge0$.Then,$$8a-8+7b-7+3c-3=38$$$$8a+7b+3c=56$$Let $X=8a,Y=7b,Z=3c$.$$X+Y+Z=56$$Allowed values of $X=0,8,16,24,32,40,48,56$.

Allowed values of $Y=0,7,14,21,28,35,42,49,56$.

Allowed worths of $Z=0,3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54$.

Number that solutions=Coefficient of $x^56$ in the growth of$$(1+x^8+x^16+cdots+x^56)(1+x^7+x^14+cdots+x^56)(1+x^3+x^6+cdots+x^54)$$that is coefficient that $x^56$ in$$left(frac1-x^641-x^8 ight)left(frac1-x^631-x^7 ight)left(frac1-x^571-x^3 ight)$$

which can be more simplified making use of binomial expansions.

Using this, girlfriend can find all feasible solutions in general. In this case, however, trial and also error may be much more convenient.


re-superstructure
cite
monitor
edited Sep 11 "16 at 16:31
answer Sep 11 "16 in ~ 15:22
*

GoodDeedsGoodDeeds
11.1k33 gold badges1919 silver- badges4242 bronze title
$endgroup$
7
| present 2 an ext comments
0
$egingroup$
We desire $8A + 7B + 3C = 38;A,B,C ge 1$

Method 1:

Note: $8 = 2*7 - 2*3$ and $3*7 = 7*3$

So if $8A + 7B + 3C = 38$ is a solution. So is $8(Apm1) + 7(Bpm 2)+ 3(Cmp 2)=38$. And so is $8A + 7(Bpm 3) + 3(Cmp 7)=38$.

From this if we gain one solution we can tweak the to obtain all solutions, also if our an initial solution was the end of range.

$8*0 + 7*5 + 3*1 = 38$. Yet that can"t be a solution because $A= 0 $2A + 3j + k = 6; 1 le A le 3; j ge (2-A)/3; k ge A/7; k equiv A mod 3$.

......

If $A = 1;$

$3j + k = 4$

$jge 1; k ge 1; k equiv 1 mod 3$. If $k ge 4$ climate $j le 0$. Therefore $k=1; j = 1; B =3; C =3$ is only solution. I.e. $(8, 21, 9)$.

See more: Why Do Cats Lick Their Bottoms, Why Cats Lick Their Privates

If $A = 2;$

$3j + k = 2$

$jge 0; k ge 2; k equiv 2 mod 3$. Therefore $k = 2;j=0; B= 1; C = 5$ is only solution i.e. $(16,7,15)$.