Question #283474

An engineer designs at least one robot a day for 30 days. If a total of 45

robots have been designed, then show that there must have been a series of consecutive

days when exactly 14 robots were designed.


1
Expert's answer
2021-12-31T02:25:06-0500

Let aia_i be the number of robots designed on the iith day. Then we have 30 numbers a1,a2,,a30a_1,a_2,…,a_{30} and ai1a_i\geq 1 for all 1i301\leq i\leq 30 .

Let sk=i=1kais_k=\sum_{i=1}^ka_i , it is the number of robots designed on or before the kkth day.

We have a sequence of 30 numbers s1,s2,,s30, where s_1,s_2,…,s_{30},\ \text{where } 1s1<s2<<s30=451\leq s_1<s_2<…<s_{30}=45 .

Let us consider new sequence: s1+14, s2+14, , s30+14s_1+14, \ s_2+14,\ …,\ s_{30}+14 . There are 30 numbers and s1+14<s2+14<<s30+14=45+14=59s_1+14<s_2+14<…<s_{30}+14=45+14=59 .

There are totally 60 numbers: s1,s2,,s30,s1+14,s2+14,,s30+14s_1,s_2,…,s_{30},s_1+14,s_2+14,…,s_{30}+14 and all of them are less or equal than 5959 (si45s_i\leq 45 and si+1459s_i+14\leq 59 ).

By the Pigeonhole Principle, at least two of these numbers are equal.

Since sisjs_i\neq s_j and si+14sj+14s_i+14\neq s_j+14 for all iji\neq j, it follows that si=sj+14s_i=s_j+14 for some ii and jj .

Now we can conclude that exactly 14 robots were designed from day j+1j+1 to day ii.


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