Inspired by Ajala Singh

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

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

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? Note by Calvin Lin
4 years, 10 months ago

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1) 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.

- 4 years, 10 months ago

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,

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

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

=$1...10.1$ (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.

- 4 years, 10 months ago

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?

- 4 years, 10 months ago

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!

- 4 years, 10 months ago

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

- 4 years, 10 months ago

((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)

- 4 years, 10 months ago

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

- 4 years, 10 months ago

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.

Staff - 4 years, 10 months ago