300 followers problem!

Number Theory Level 4

Let A = 123456789 ‾ ⏟ set: 1 A = \underbrace{\overline{123456789}}_{\text{set: 1}} and B = 987654321 ‾ ⏟ set: 2 B = \underbrace{\overline{987654321}}_{\text{set: 2}}

Let, x = 123456789987654321 … 123456789987654321 ‾ ⏟ 300 alternate sets or 150 A B ‾ ’s x = \underbrace{\overline{123456789987654321\dots 123456789987654321}}_{\text{300 alternate sets or 150}\overline{AB} \text{'s}}

Find the sum of digits of 9 x 9x .


The answer is 12159.

Sravanth C. by Sravanth C. , 21, India

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2 solutions

Sravanth C.
Jul 9, 2015

We have 9 A = 9 × 123456789 ‾ = 1111111101 ‾ and 9 B = 9 × 987654321 ‾ = 8888888889 ‾ 9A=9\times \overline{123456789}=\overline{1111111101}\\ \text{and}\quad 9B=9\times \overline{987654321}=\overline{8888888889}

Now, let's find 9 A B ‾ = 1111111101000000000 + 8888888889 = 1111111109 ⏟ constant: 1 888888889 ⏟ constant :2 ‾ 9\overline{AB}= 1111111101000000000+8888888889=\overline{\underbrace{1111111109}_{\text{constant: 1}}\underbrace{888888889}_{\text{constant :2}}}

Similarly, 9 A B A B ‾ = 1111111109 ⏟ constant: 1 888888890111111109 ⏟ repetitive number y 888888889 ⏟ constant: 2 ‾ 9\overline{ABAB}= \overline{\underbrace{1111111109}_{\text{constant: 1}}\underbrace{888888890111111109}_{\text{repetitive number y}}\underbrace{888888889}_{\text{constant: 2}}}

  • We can observe that the both the constants i.e. 1111111109 1111111109 and 888888889 888888889 are always at the ends no matter how many A B ’s AB\text{'s} we add.

  • And the number of y ’s = no. of AB’s − 1 y\text{'s}=\text{no. of AB's}-1 .

∴ 9 x = sum of digits of constant: 1 + sum of digits of constant: 2 + sum of digits of y ×149 = 17 + 73 + ( 81 × 149 ) = 90 + 12069 = 12159 \therefore 9x=\text{sum of digits of constant: 1}\\ +\text{sum of digits of constant: 2}+\text{sum of digits of y ×149}\\ =17+73+(81\times 149) \\=90+12069=\boxed{12159}

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Moderator note:

Why did you choose to present 9ABAB as "constant 1 , repetitive number y, constant 2"?

Isn't it true that 9ABAB is "repetitive number z, repetitive number z"? That would make it much easier to explain why this is true (which you have glossed over in your approach).

Som Ghosh
Jul 11, 2015

print(sum(int(i) for i in str(9 * int(('123456789' + '987654321') * 150))))

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