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If x, y, and z are positive integers such that \(6xyz+30xy+21xz+2yz+105x+10y+7z=812\), find x+y+z.

 Dec 13, 2019
 #1
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+1

a=0;b=0;c=0;p=0; cycle:d=a*100+b*10+c;if(6*a*b*c+30*a*b+21*a*c+2*b*c+105*a+10*b+7*c==812, goto loop, goto next); loop:printd," ",;p=p+1; next:c++;if(c<10, goto cycle, 0);c=0;b++;if(b<10, goto cycle, 0);b=0;c=0;a++;if(a<10, goto cycle,0);print"Total = ",p

 

x       y      z

2      2      6

P.S. I haven't got a clue on how to solve it algebraically !!

 Dec 13, 2019
 #2
avatar+109064 
+1

6xyz + 30xy + 21xz + 2yz  + 105x + 10y + 7z  = 812

 

3x ( 2yz + 10y + 7z)  +  (2yz + 10y + 7z)  + 105x  =  812

 

3x (2yz + 10y + 7z)  + (2yz + 10y + 7z)  + 3x(35)  = 812

 

3x [ 2yz +10y + 7z + 35 ] + [2yz + 10y + 7z]  =  812        add  35 to both sides

 

3x ( 2y +7)(z + 5) +  2yz + 10y + 7z  + 35   =  812  + 35

 

3x  (2y + 7) (z + 5)  +  1* (2y + 7)(z +5)  =   847      factor the left side

 

(3x + 1) (2y + 7) (z + 5)  =   847

 

Note that

 

7 * 11^2  = 847    ....so....

 

  7 * 11 * 11    =  847

 

So we can let    x  =2, y = 2   and z  = 6

 

So

 

x + y +  z  =   2  +  2  +  6   =    10

 

 

 

cool cool cool

 Dec 13, 2019
 #3
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+1

Thank you!

Guest Dec 14, 2019

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