+0  
 
+1
112
3
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Let (a x b x c) / (a + b + c) = 341 be an equation where a, b and c are consecutive positive integers. What is the least possible value of a?

 Nov 3, 2019
 #1
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0

Let x be the middle number

a = x-1

c = x+1

 

(x-1)(x)(x+1) / [(x-1)+(x)+(x+1)]   = (x^3 +x)/3x =====>  (x^2-1)/3

(x^2-1) /3 = 341

x^2 = 1024

x= +- 32

 

smallest x would be  -32    smallest a would be - 33

 Nov 3, 2019
 #2
avatar+19897 
0

or   +31    since it has to be positive     31   32  and 33

ElectricPavlov  Nov 3, 2019
 #3
avatar+8810 
+2

Let (a x b x c) / (a + b + c) = 341 be an equation where a, b and c are consecutive positive integers. What is the least possible value of a?

 

Sei (a x b x c) / (a + b + c) = 341 eine Gleichung, in der a, b und c aufeinanderfolgende positive ganze Zahlen sind. Was ist der geringstmögliche Wert von a?

 

Hello Guest!

 

\((a\times b \times c)/ (a + b + c) = 341\\ a\times (a+1)\times (a+2)/(a+a+1+a+2)=341\\ \color{blue}a=31\\ \color{blue}(31\times 32\times 33)/(31+32+33)=341\)

 

laugh  !

 Nov 3, 2019
edited by asinus  Nov 3, 2019

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