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There are $4$ positive integers that add up to $216$.

If you add 5 to the first one, subtract 5 to the second one, multiply the third one by 5, and divide the fourth one by 5, they will be equal numbers.

What I know:

The first number is $A$, the second one $B$, third one $C$, and the last one $D$.

The first/second/third/fourth number after being processed is $a,b,c,d$

$$A<B,10$$ $$2d=A+B$$ $$A+B=a+b$$

The final equal number MUST be lower than $40$.

Please help me find the answer and explain how did you get to that answer, thank you.

Puffy
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2 Answers2

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$a+b+c+5d=216$

$a+5=b-5=5c=d$

$\implies b=5+5c,a=5c-5,d=5c$

$\implies216=5c-5+5c+5+c+5(5c)\iff c=6$

lab bhattacharjee
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Let the numbers, as you have assumed, be $a,b,c,d$. We have: $$a+b+c+d=216\tag{1}$$ $$a+5 = b-5 = 5c =\frac{d}{5}\tag{2}$$

Thus, we get from $(2)$, $a = 5c-5, b=5c+5, d=25c$. Substituting in $(1)$, gives us, $$(5c-5)+(5c+5)+c+25c =216 \implies 36c=216$$ $$\boxed{c=6, a= 25, b= 35, d=150}$$