Does It Always Hold?

Algebra Level 3

$\begin{array} {r c r c l } 1 &\times & 1 & = &1 \\ 11 & \times& 11 & = &121 \\ 111 & \times& 111 & =& 12321 \\ 1111 & \times& 1111 & =& 1234321 \\ \end{array}$

Observe the equations above.

Hence, what is the value of

$\underbrace{1111111111}_{10 \text{ 1's } } \times \underbrace{1111111111}_{10 \text{ 1's } } \; ?$

Inspiration .

12345678987654321 1234567900987654321 12345678910987654321 1234567891010987654321

2 solutions

Chew-Seong Cheong
Mar 28, 2016

Consider:

$\begin{array} {} a_1 = 1 & \Rightarrow a_1^2 = 1 \\ a_2 = 11 & \Rightarrow a_2^2 = 121 \\ a_3 = 111 & \Rightarrow a_3^2 = 12321 \\ a_4 = 1111 & \Rightarrow a_4^2 = 1234321 \\ ... & ... \\ a_9 = 111111111 & \Rightarrow a_9^2 = 12345678987654321 \end{array} $

$\begin{aligned} a_{10} & = 10a_9 + 1 \\ \Rightarrow a_{10}^2 & = (10a_9 + 1)^2 \\ & = 100a_9^2 + 20a_9 + 1 \end{aligned}$

$\begin{array} {llr} \quad \quad & = & & 1234567898765432100 & \\ & & + & 2222222220 \\ & & + & 1 \\ & = & & \boxed{1234567900987654321} \end{array}$

Well, actually the answer still fits the increasing decreasing pattern that we observed. We just have to remember to do the calculations in base 10.

IE $\overline{123456789(10)987654321} = \overline{12345678(10)0987654321} = \overline{1234567900987654321}$ .

In the first number, we can't write 10 as 2 seperate digits, but must consider them together. This essentailly means that we're carry over of 1. Added to 9, this give 10 (again not 2 digits), but a carry over of 1. Added to 8, this gives 9.

This easily extends to multiplying more number of 1's together.

Calvin Lin Staff - 5 years, 2 months ago

Great use of that recursion! +1

Nihar Mahajan - 5 years, 2 months ago

Hobart Pao
Mar 30, 2016

A fast way to solve the problem by observation is to consider that for $1 \times 1$ , the number of digits in the answer is $2-1$ , for $11 \times 11$ , the number of digits in the answer is $4-1$ , for $111 \times 111$ , the number of digits in the answer is $6 - 1$ . So then, the answer to this problem by this pattern should have 19 digits, and only 1234567900987654321 had 19 digits, so that was most likely the correct answer (of course, my pattern needs a proof)

Does It Always Hold?

2 solutions

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