Inspired by Ajala Singh

In Can Someone Explain?, Ajala Singh asked if there was a reason why the following equation is true.

123,456,789×9=1,111,111,111. 123, 456, 789 \times 9 = 1, 111, 111, 111 .

Given that the RHS has 10 1's, the Rule of divisibility of 9 tells us that the equation cannot be correct. I then tried to figure out what the actual equation should have been, and stumbled upon 2 interesting possibilities


1) 123,456,789×9=1,111,111,101. 123, 456, 789 \times 9 = 1, 111, 111, 101.

Furthermore, this generalizes to other bases. We have 123×23=10131234×34=1101412345×45=111015 \begin{array} {l r r } 12_3 & \times 2_3 = &101_3 \\ 123_4 & \times 3_4 = & 1101_4 \\ 1234_5 & \times 4_5 = & 11101_5 \\ \end{array}

Can someone explain why?


2) 12,345,679×9=111,111,111. 12,3 45,6 79 \times 9 = 111, 111, 111.

Furthermore, this generalizes to other bases. We have 134×34=11141245×45=1111512356×56=111116 \begin{array} {l r r } 13_4 & \times 3_4 = & 111_4 \\ 124_5 & \times 4_5 = & 1111_5 \\ 1235_6 & \times 5_6 = & 11111_6 \\ \end{array}

Can someone explain why?

Note by Calvin Lin
4 years, 10 months ago

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1) Let the base be rr. Write the multiplicand as an expansion in powers of rr' s (i.e. (1234)5=1.53+2.52+3.5+4(1234)_5=1.5^3+2.5^2+3.5+4) and the multiplier as r1r-1. Collect terms together.

2) Proceed by similar method.

Abhishek Sinha - 4 years, 10 months ago

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1) 123...(n1)n×n1n\overline{123...(n-1)}_n \times \overline{n-1}_n

=123...(n1)n×n123...(n1)n×1= \overline{123...(n-1)}_n \times n - \overline{123...(n-1)_n} \times 1

=123...(n1)0n123...(n1)n= \overline{123...(n-1)0}_n - \overline{123...(n-1)}_n

Note that the first number has one more digit than the second number.

21=1,32=1,...(n2)(n3)=12-1=1, 3-2=1, ... (n-2)-(n-3)=1

However, 0<(n-1). Thus, we move 1 from (n-1) to add n to 0. n-(n-1)=1 but (n-1-1)-(n-1)=0. In other words,

[b]r((n1)0)((n2)(n1))(01) \frac{ \begin{array}{c}[b]{r} \left( (n-1) 0 \right)\\ - \left( (n-2) (n-1) \right) \end{array} }{ \left( 01 \right) }

Thus we have 11...101.

2) Writing this again is too tiring, so i'll try to connect this to the 'lemma' above.

We divide 10 on both sides. We get 123...(n2).(n1)n×n1n\overline{123...(n-2).(n-1)}_n \times \overline{n-1}_n

=1...10.11...10.1 (there are (n-2) 1's in 1...1.)

Now, we add 0.1 to 123...(n2).(n1)n\overline{123...(n-2).(n-1)}_n.

123...(n3)(n2).(n1)n+0.1n=123...(n3)(n1).n\overline{123...(n-3)(n-2).(n-1)}_n + 0.1_n = \overline{123...(n-3)(n-1).}_n

We get 123...(n3)(n1).n×n1n\overline{123...(n-3)(n-1).}_n \times \overline{n-1}_n

=1...10.1n+[0.1n×(n1)]= 1...10.1_n + [0.1_n \times (n-1)]

(and ooh I see light!)

=1...10.1n+0.(n1)n= 1...10.1_n +0.(n-1)_n

=1...11n= 1...11_n

Yes! Q.E.D.

Aloysius Ng - 4 years, 10 months ago

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I believe you are referring to

123(n1)n\overline{123\dots(n-1)}_n

(Toggle LaTeX!) And I think your solution is right. Would it work the same for n > 10?

Justin Wong - 4 years, 10 months ago

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Yes I was. Thanks!

It would still work for n>10 as the base will change and more numbers are allowed. You can read the wiki for info.

Anyway, thanks!

Aloysius Ng - 4 years, 10 months ago

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(n-1) in base n is (10-1)base n

A Former Brilliant Member - 4 years, 10 months ago

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((1b^(b-1))+ (2b^(b-2)) +…+ (b-1) (b^0)) (b-1)

= (1b^b) + (2b^(b-1))+ (3b^(b-2)) +…+ (b-1) (b^1) + 0b^0) - (1b^(b-1))+ (2b^(b-2)) +…+ (b-2) (b^1) + (b-1) (b^0)

= (1b^b+1b^(b-1)+1b^(b-2)+…+ 0(b^1) + (1b^0)

Rodrigo Robledo Castillo - 4 years, 10 months ago

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The given equation is incorrect because 123456789*9=1111111101 not 1111111111.

Soumendra Barik - 4 years, 10 months ago

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Yes, the first given equation is incorrect, and I explained why.

I mentioned that even though this equation is incorrect, it led me on to discover 2 other interesting patterns, as described later in the note. Both Abhishek and Aloysius have provided great explanations above, and I encourage you to read them.

Calvin Lin Staff - 4 years, 10 months ago

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