3 variables 1 equation?

Number Theory Level 2

Let r , s r,s and k k be positive integers such that k + s + r + r s + s k + r k + r s k = 30148 k+s+r+rs+sk+rk+rsk=30148

then find the value of r + s + k r+s+k .


The answer is 136.

Ravneet Singh by Ravneet Singh , 30, India

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1 solution

Jason Martin
Aug 26, 2015

Adding 1 to both sides and factoring yields ( r + 1 ) ( s + 1 ) ( k + 1 ) = 30149 (r+1)(s+1)(k+1)=30149 . Since 30149 = 7 ⋅ 59 ⋅ 73 30149=7 \cdot 59 \cdot 73 and r , s , k > 0 r, s, k>0 , we know ( r + 1 ) , ( s + 1 ) , ( k + 1 ) (r+1), (s+1), (k+1) are each nontrivial factors of 30149 30149 , so WLOG, r = 6 , s = 58 , k = 72 r=6, s=58, k=72 . Therefore, r + s + k = 136 r+s+k= \boxed{136} .

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