Write any number. Generate a sequence of numbers by repeatedly adding 11. What would be the digital roots of this sequence of numbers? Share your observations.

Question

Homi · Accepted Answer

The digital root is the single digit obtained by repeatedly summing digits.

Step 1 — Generate the number sequence

Let us choose a starting number. We will pick 5. We will generate a sequence by repeatedly adding 11. Let $N_0$ be our starting number. Let $N_k$ be the $k$ -th number in the sequence.

$N_0 = 5$

$N_1 = N_0 + 11 = 5 + 11$

$= 16$

$N_2 = N_1 + 11 = 16 + 11$

$= 27$

$N_3 = N_2 + 11 = 27 + 11$

$= 38$

$N_4 = N_3 + 11 = 38 + 11$

$= 49$

$N_5 = N_4 + 11 = 49 + 11$

$= 60$

$N_6 = N_5 + 11 = 60 + 11$

$= 71$

$N_7 = N_6 + 11 = 71 + 11$

$= 82$

$N_8 = N_7 + 11 = 82 + 11$

$= 93$

$N_9 = N_8 + 11 = 93 + 11$

$= 104$

$N_{10} = N_9 + 11 = 104 + 11$

$= 115$

$N_{11} = N_{10} + 11 = 115 + 11$

$= 126$

$\boxed{\text{The sequence starts with: 5, 16, 27, 38, 49, 60, 71, 82, 93, 104, 115, 126, ...}}$

Step 2 — Calculate digital roots

We will find the digital root for each number. Let $DR(N)$ denote the digital root of number $N$ .

$DR(5) = 5$

$DR(16) = DR(1+6)$

$= 7$

$DR(27) = DR(2+7)$

$= 9$

$DR(38) = DR(3+8) = DR(11)$

$= DR(1+1) = 2$

$DR(49) = DR(4+9) = DR(13)$

$= DR(1+3) = 4$

$DR(60) = DR(6+0)$

$= 6$

$DR(71) = DR(7+1)$

$= 8$

$DR(82) = DR(8+2) = DR(10)$

$= DR(1+0) = 1$

$DR(93) = DR(9+3) = DR(12)$

$= DR(1+2) = 3$

$DR(104) = DR(1+0+4)$

$= 5$

$DR(115) = DR(1+1+5)$

$= 7$

$DR(126) = DR(1+2+6)$

$= 9$

$\boxed{\text{The digital roots sequence is: 5, 7, 9, 2, 4, 6, 8, 1, 3, 5, 7, 9, ...}}$

Step 3 — Observe the pattern

We can see a clear pattern in the digital roots. The sequence of digital roots is 5, 7, 9, 2, 4, 6, 8, 1, 3. After these 9 numbers, the pattern repeats. The next digital root is 5, then 7, then 9, and so on. All possible single digits from 1 to 9 appear in this cycle.

Step 4 — Explain the pattern

The digital root of a number is its remainder when divided by 9. If the remainder is 0, the digital root is 9. Let $DR(N)$ be the digital root of a number $N$ . A useful property is that $DR(A+B) = DR(DR(A) + DR(B))$ . We are repeatedly adding 11 to our numbers. Let us find the digital root of 11.

$DR(11) = DR(1+1)$

$= 2$

So, when we add 11 to a number, its digital root changes. The new digital root is found by adding 2 to the previous digital root. Then we find the digital root of this sum. Let $d_k$ be the digital root of the $k$ -th number $N_k$ . Then $d_{k+1} = DR(d_k + 2)$ . We saw this in our calculations. Starting with $d_0 = 5$ , we get: 5, $DR(5+2)=7$ , $DR(7+2)=9$ , $DR(9+2)=2$ , $DR(2+2)=4$ , $DR(4+2)=6$ , $DR(6+2)=8$ , $DR(8+2)=1$ , $DR(1+2)=3$ , $DR(3+2)=5$ . The sequence of digital roots repeats every 9 terms. This happens because we are adding 2 repeatedly. There are 9 possible digital roots (1 to 9). Adding 2 repeatedly will cycle through all 9 digits before repeating. This cycle will always include all digits from 1 to 9. The starting number only determines where the cycle begins.

Answer

(i) The sequence of numbers, starting with 5, is: 5, 16, 27, 38, 49, 60, 71, 82, 93, 104, 115, 126, ... (ii) The digital roots of this sequence are: 5, 7, 9, 2, 4, 6, 8, 1, 3, 5, 7, 9, ... (iii) We observe a repeating cycle of 9 terms. The cycle is: (5, 7, 9, 2, 4, 6, 8, 1, 3). This cycle includes every digit from 1 to 9. Adding 11 to a number adds 2 to its digital root. This causes the digital roots to cycle through all 9 digits.

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