Situation 7911

How many years will elapse from 1 January 2013 before the situation occurs that the product of the digits of a given year is greater than the sum of the digits of that year? A) 87 b) 98 c) 101 d) 102 e) 103

Correct answer:

n = C

Step-by-step explanation:

$$\begin{gathered} r_1=2013+87=2100 \\ {r}_2=2013+101=2114 \\ {r}_3=2013+102=2115 \\ {r}_4=2013+103=2116 \\ {x}_1=2\cdot 1\cdot 0\cdot 0=0 \\ {x}_2=2\cdot 1\cdot 1\cdot 4=8 \\ {x}_3=2\cdot 1\cdot 1\cdot 5=10 \\ {x}_4=2\cdot 1\cdot 1\cdot 6=12 \\ {s}_2=2+1+1+4=8 \\ {s}_3=2+1+1+5=9 \\ {s}_4=2+1+1+6=10 \\ {x}_3>s_3 \\ n=C \end{gathered}$$

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