Is there a six-digit number where every two consecutive digits make up a certain number two-digit number that is the square of an integer? Justify your answer.
Problem
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Tags: number theory, Digits, Perfect Square
math_maniac
08.04.2021 20:52
No there is not Consider all our two digit squares there are. These are 16,25,36,49,64,81 Now 25,36,81 have non-square first digit so these numbers must lie at the start of our number Now trial-error there is no number starting with 5 and a perfect square so 25 is not there 3649 u cant again take it forward 81649 again it stops for 16 49 64 there are cases but they again amount to same thing but my approach is not so good so if there are other approaches i would like to know them..
RedFireTruck
08.04.2021 21:00
Clearly, we want $5$ of these 2-digit squares. Look at the picture below to see that the max path length is $4$ So it is impossible.
somlogan
08.04.2021 21:03
nice work everyone
RedFireTruck
08.04.2021 21:05
somlogan wrote: nice work everyone thank you
sttsmet
28.05.2021 19:16
Accually... the maximum path is 5 as @math_maniac said : 81649 but there is not a 6th number! Also, after we prove that 25, 36, and 81 cant be used, we easily say that 49 must nessecary be at the end of the 6 digit number which is impossible. After that, we have only 16 and 64 left and it is really easy to prove that they cant form the desired!