Sequence with equal digit sums

For some n n and k k let a 1 , a 2 , . . . , a n a_{1}, a_{2}, ..., a_{n} be a sequence where:

  • a i a_{i} is a positive integer

  • a 1 = k a_{1} = k

  • a i + 1 = 2 × a i a_{i+1} = 2 \times a_{i}

  • Digit sums of all a i a_{i } are equal (base 10 10 )

Can we construct such sequence for any positive integer n n ( k k is of your choice)?

No Yes

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1 solution

Robert Szafarczyk
Mar 13, 2018

Hint : Prove that their digit sum must be divisible by 9. Knowing that you can show that we can lengthen a given sequence by adding a zero on the end of the first term and then dividing it by 2 (e.g. 9 , 18 , 36 , 72 , 144 > 45 , 90 , 180 , 360 , 720 , 1440 9, 18, 36, 72, 144 --> 45, 90, 180, 360, 720, 1440 )

If we continue your process a few more steps, we get 1125,2250,4500,9000,18000,36000,72000,144000. How do you extend the sequence from here? 5625 does not have the same digit sum as 11250,22500,...

John Ross - 3 years, 2 months ago

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You are right. I might have missed something in my proof. I'll try to come up with a fix.

I think this sequence can still be made arbitrarly large despite this flaw.

Robert Szafarczyk - 3 years, 2 months ago

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