Find what comes next in the following sequence:


Bonus 1: What is the rule?
Bonus 2:

What numbers never appear when you do the rule to all numbers?

  • 3
    $\begingroup$ Would be easier for a sequence 1,4,9,6,0,8,8,8,8,0,2,6,0,0, $\endgroup$
    – z100
    May 15 '21 at 19:44
  • $\begingroup$ You just... changed the problem? Was your original sequence a mistake? $\endgroup$
    – bobble
    May 15 '21 at 23:59
  • $\begingroup$ Yes. I tried the rule and found I need to change the sequence. $\endgroup$
    – Boesf
    May 16 '21 at 0:03

Next one is 0.

calculated as $15^2=225$ --> $2\times2\times5=20$ --> $2\times0=0$

by the rule: multiply the digits of the square; if it is over 10, repeat the multiplying

  • $\begingroup$ Bonus 2: for sure 5, as it always comes in pair with 2, so the result is 0. 3 and 7 not so easy to prove or simply try a simple program to find the first occurence in the sequence. $\endgroup$
    – z100
    May 16 '21 at 1:04
  • 1
    $\begingroup$ Can you clarify how 5 always comes with 2? 24^2 is 576, so do you mean such a number always has 2 as a factor? $\endgroup$ May 18 '21 at 7:16
  • 1
    $\begingroup$ @Kalyan Raghu : you're right, I don't know why a assume that5 should be the last digit in a square. So first condition should be that all the digits of a square are odd. And then all further products should also contain only odd numbers. One way to find is to search between numbers of form 1111...115 (similar for 1111..131 or 11111..171) if one of them is a square. $\endgroup$
    – z100
    May 18 '21 at 14:54

Answer to Bonus 2 as Bonus 1 is answered by @z100

{3, 5, 7} cannot come in this sequence.

For single digit numbers, we can easily check the above to be true

One observation is that whenever a number contains

an even digit

, the following multiples of all its digits will always be


and the end result of the sequence will be one of

{0, 2, 4, 6, 8}

Proof sketch is as follows:

From the properties of square numbers, if a multi digit number ends with a:

  • 0 its square ends with zero resulting in the multiplication becoming zero

  • 1 its square ends in 1 and the digit in the tenths place will be an even number as the preceding digit gets doubled.

  • 2 or 8 the square ends with 4 which is an even digit

  • 3 the square ends with 9 and the tenth's digit is even as 3^2 doesn't overflow to tenth's place and the product of tenth digit with 3 doubles contributing an even digit to the tenth digit

  • 4 or 6 the square ends with 6 which is an even digit

  • 5 the square ends with 25 with one even digit in tenth position

  • 7 the square of 7 contributes 4 from 49, an even number to the tenth position and by similar logic to 3, the tenth position will be an even digit

  • 9 similar logic as for 7 follows for 9 which contributes 8 to the tenth position resulting in even digit in the tenth position

As all the squares of multi-digit numbers contain

at least one even digit,

the result of the sequence of multiples will be the set

{0, 2, 4, 6, 8}

Hence, the set

{3, 5, 7}

never occurs in this sequence


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