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.

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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.

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edited Sep 11 "16 at 16:31
answer Sep 11 "16 in ~ 15:22

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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)\$.

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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)\$.