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 #1
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Let c+1/a=z. Note that (22/7)(8)(z)=(abc+1/abc)+(a+1/b)+(b+1/c)+z=21+1/21+22/7+8, so z=169/132.

 Dec 11, 2020
 #2
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a=1;p=0; b=1;c=1;n=a+1/b;m=b+1/c;r=a*b*c;if(n==22/7 and m==8 and r==21, goto loop, goto next);loop:p=p+1;printp," =",a,b,c;next:a++;if(a<100, goto4,0);a=1;b++;if(b<100, goto4, 0);a=1;b=1;c++;if(c<100, goto4,0)

 

OUTPUT:  a = 3,  b =7,  c =1

c + 1 /a ==1 + 1/3 =4 / 3  or  1 1/3

 Dec 11, 2020

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