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

\[ 123, 456, 789 \times 9 = 1, 111, 111, 111 . \]

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

Furthermore, this generalizes to other bases. We have \[ \begin{array} {l r r } 12_3 & \times 2_3 = &101_3 \\ 123_4 & \times 3_4 = & 1101_4 \\ 1234_5 & \times 4_5 = & 11101_5 \\ \end{array} \]

Can someone explain why?

2) \( 12,3 45,6 79 \times 9 = 111, 111, 111. \)

Furthermore, this generalizes to other bases. We have \[ \begin{array} {l r r } 13_4 & \times 3_4 = & 111_4 \\ 124_5 & \times 4_5 = & 1111_5 \\ 1235_6 & \times 5_6 = & 11111_6 \\ \end{array} \]

Can someone explain why?

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## Comments

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TopNewest1) Let the base be \(r\). Write the multiplicand as an expansion in powers of \(r\)' s (i.e. \((1234)_5=1.5^3+2.5^2+3.5+4\)) and the multiplier as \(r-1\). Collect terms together.

2) Proceed by similar method.

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

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

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

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

\(2-1=1, 3-2=1, ... (n-2)-(n-3)=1\)

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,

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

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 \(\overline{123...(n-2).(n-1)}_n \times \overline{n-1}_n\)

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

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

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

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

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

(and ooh I see light!)

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

\(= 1...11_n\)

Yes! Q.E.D.

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I believe you are referring to

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

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

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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!

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

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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.

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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)

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

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