All powers of 11 are related to the Pascal Triangle

Number Theory Level 1

$1 \\ 1 \ \ 1 \\ 1 \ \ 2 \ \ 1 \\ 1 \ \ 3 \ \ 3 \ \ 1 \\ 1 \ \ 4 \ \ 6 \ \ 4 \ \ 1 \\ 1 \ 5 \ 10 \ 10 \ 5 \ 1 \\$

Above shows the first 6 rows of a Pascal Triangle. By concatenating the numbers in their respective rows, we can see that the first five equations are true.

$\begin{aligned} 11^0 &=& 1 \\ 11^1 &=& 11 \\ 11^2 &=& 121 \\ 11^3 &=& 1331 \\ 11^4 &=& 14641 \end{aligned}$

What is the value of $11^5$ ?

The answer is 161051.

3 solutions

Patrick Heebels
May 30, 2015

In this problem the number 10 cannot take up 2 spaces in the decimal system and therefore the number 10 will result in a carry , which we can represent with a subscript, i.e. " $_10$ ".

By concatenating " $1, 5, 10, 10, 5, 1$ " and utilizing writing the carry as a subscript the result is:

$11^5 = 15_10_1051$

Adding up from right to left and taking the carry along to the left we get: $15_10_1051 = 161051$

Moderator note:

Right on point!

Bonus question 1 : You are given the numbers in the 10th row of the Pascal Triangle (from left to right) to be $1,10,45,120,210,252,210,120,45,10,1$ , can you work out the value of $11^{10}$ in a similar manner?

Bonus question 2 : For positive integer $n$ , is it possible to work out the value of the $11^{n+1}$ if you're given all the numbers in the $n^\text{th}$ row of the Pascal Triangle? If yes, can you work out the value of $11^{n+1}$ without knowing the value of $11^n$ ?

Thank you mr. Chung!

Nice bonus questions as well.

Bonus question 1: reply

The solution is basically equivalent for the $10^{th}$ row as I described earlier as long as you keep in mind that the carry does not have be a single-digit number.

When we concatenate the numbers in the $10^{th}$ row from right to left, we only jot down the unit-digit of each number, say $a$ and carry the remaining digits (this will be $\frac{a - (a \text{ mod } 10)}{10}$ to the number adjacent to the left of it.

So we get:

The $1^{st}$ digit from the right is just the 1 with no carry .

The $2^{nd}$ digit is $10 \text{ (mod } 10) = 0$ and the carry is $\frac{10-0}{10} = 1$

The $3^{rd}$ digit is $45 + 1 \text{ (mod } 10) = 6$ and the carry is $\frac{46-6}{10} = 4$

The $4^{th}$ digit from the right is $120 + 4 \text{ (mod } 10) = 4$ and the carry is $\frac{124-4}{10} = 12$

The $5^{th}$ digit from the right is $210 + 12 \text{ (mod } 10) = 2$ and the carry is $\frac{222-2}{10} = 22$

et cetera, et cetera, ...

If we continue is this fashion all the unit-digits from right to left are: $1, 0, 6, 4, 2, 4, 7, 3, 9, 5, 2$

Writing them from left to right gives $11^{10} = 25937424601$ .

Note: I realize is it easier done than said ;-) !

Addendum:

Realizing that the numbers in the $n^{th}$ row of Pascal's triangle are the binomial coefficients ${n \choose 0}$ , ${n \choose 1}$ , ${n \choose 2}$ , ..., ${n \choose n}$ and using the binomial theorem we can visually figure out $11^{10}$ as follows disregarding the powers of 10, but using these as place-holders:

1
1 0
  4 5
  1 2 0
    2 1 0
      2 5 2
        2 1 0
          1 2 0
              4 5
                1 0
                    1
---------------------- +
2 5 9 3 7 4 2 4 6 0 1

Bonus question 2: reply

$1^{st}$ part:

The given strategy will work on any row and will give the value for $11^{n+1}$ provided the $n^{th}$ row of Pascal's triangle is given.

$2^{nd}$ part:

If we remember the binomial theorem:

$(a + b)^n = \sum_{k=0}^n {n \choose k} \cdot a^k \cdot b^{n-k}$

and setting $a = 10$ and $b = 1$ then all we basically need to do is calculate ${n \choose k}$ because $a^k = 10^k$ is easy enough and $b^{n-k} = 1^{n-k} = 1$ .

These binomial coefficients ${n+1 \choose 0}$ , ${n+1 \choose 1}$ , ${n+1 \choose 2}$ , ..., ${n+1 \choose n+1}$ will give us the numbers in the $(n+1)^{th}$ row of Pascal's triangle and the method described above will give the result of $11^{n+1}$ .

This is by all means not a very fast method when the exponent is large, so I have doubts my method is the one you have in mind.

Patrick Heebels - 6 years ago

Smruti Charkha
May 31, 2015

This is just the addition to the other solution, I think this might be helpful in understanding why exactly this happens?

11 X 11 = 11 + 110 = 121

Here, 2 is the addition of 1 at unit's and 1 at ten's place. This is all because, the required number is simply the addition of 10 times the number and the number itself.

number X 11 = number X (10 + 1) = 10 X number + number

A = 10 X number and

B = number

In A, every digit is shifted to left by one place of significance than B. Hence the solution is in such a way. In Pascals triangle as well, we just add two numbers and add carry if it's there like in row two 1 and 1 to get 2, in row 3, 1 and 2 to get 3. Thus, the equity is attained.

Altaf Ahmed
Dec 10, 2015

1 5 10 10 5 1

=1 (5+1) (0+1) 0 5 1

=161051