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Find three consecutive odd numbers such that the sum of five times the smaller number and twicethe larger number is 33 more than six times the median number.

 Aug 27, 2020
 #1
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a=0;b=0;c=0;d=0;e=0;p=0; cycle:n= a*10000+b*1000+c*100+d*10+e;if(n%2==1 and n*5+2*(n+4)==6*(n+2)+33, goto loop, goto next); loop:printn,", ",;p=p+1; next:e++;if(e<10, goto cycle, 0);e=0;d++;if(d<10, goto cycle, 0);e=0;d=0;c++;if(c<10, goto cycle,0);c=0;d=0;e=0;b++;if(b<10, goto cycle,0);e=0;d=0;c=0;b=0;a++;if(a<10, goto cycle,0);print"Total = ",p

 

OUTPUT: Your numbers are: 37, 39, 41

 Aug 27, 2020
 #2
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Let the 3 numbers be:
n,  n + 2,  n + 4
5n + 2[n + 4] =6[n + 2] + 33, solve for n
5n + 2n + 8 = 6n + 12 + 33
5n + 2n - 6n = 12 + 33 - 8
n = 37 - the smallest number
37 + 2 = 39 - the middle number
37 + 4 = 41 - the larger number.

 Aug 27, 2020

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