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Sum of Digits / Digital Root

Introduction

The Sum of Digits and Digital Root patterns let you reduce large numbers to a single-digit representative quickly. They're useful for divisibility checks (especially 3 and 9), checksum problems, and simplifying repeated-sum tasks on exams.

Pattern: Sum of Digits / Digital Root

Pattern

The digital root of a number is the single-digit value obtained by iteratively summing its digits until one digit remains. Use sum-of-digits to test divisibility by 3 and 9, and use the modular formula for fast calculation.

  • Sum of digits (S): For n with decimal digits d_k...d_1d_0, S(n) = d_0 + d_1 + ... + d_k.
  • Digital root (DR): Repeatedly apply S until result is a single digit (0-9).
  • Fast modular formula:
    For n > 0, DR(n) = 1 + ((n - 1) mod 9). For n = 0, DR(0) = 0.
  • Divisibility tests using digit sum:
    • If S(n) is divisible by 3 → n divisible by 3.
    • If S(n) is divisible by 9 → n divisible by 9.
  • Casting out 9s: Subtracting 9 from a digit-sum does not change remainder mod 9; useful for simplifying large digit sums mentally.
  • Relation to mod 9: Digital root (except when result 9) corresponds to n mod 9; specifically DR(n) = 0 when n ≡ 0 (mod 9) and DR(n) = n mod 9 otherwise (or use the 1+((n-1) mod 9) formula).

Step-by-Step Example

Question

(a) Find the digital root of 987654.
(b) Using the sum-of-digits, decide whether 987654 is divisible by 3 and by 9.

Solution

  1. Step 1: Compute the sum of digits:

    S(987654) = 9 + 8 + 7 + 6 + 5 + 4 = 39.

  2. Step 2: Reduce to digital root (iterative method):

    S(39) = 3 + 9 = 12.
    S(12) = 1 + 2 = 3.
    Therefore, the digital root DR(987654) = 3.

  3. Step 3: Using modular formula (quick check):

    DR = 1 + ((987654 - 1) mod 9). Compute 987654 mod 9: since S = 39 and 39 mod 9 = 3, we get DR = 3 (matches iterative result).

  4. Step 4: Divisibility conclusions:

    Since DR = 3 (i.e., sum-of-digits = 39 is divisible by 3 but not by 9), 987654 is divisible by 3 but not by 9.

  5. Final Answer:

    (a) Digital root = 3.
    (b) Divisible by 3: Yes. Divisible by 9: No.

  6. Quick Check:

    Verify: 987654 ÷ 3 = 329218 (integer). 987654 ÷ 9 = 109739.333... (not integer) → conclusions confirmed. ✅

Quick Variations

1. Missing digit problems: If a number with unknown digit x has sum-of-digits S and must be divisible by 9, pick x so that (S + x) ≡ 0 (mod 9).

2. Negative numbers: Use absolute value for digit-sum and digital root.

3. Base-b generalization: Digital root ideas extend to other bases using modulus (b-1) instead of 9.

Trick to Always Use

  • Step 1 → Compute digit-sum; if large, cast out 9s (subtract 9 repeatedly) to shrink the number quickly.
  • Step 2 → Use DR = 1 + ((n - 1) mod 9) for immediate single-step result when comfortable with modulus.
  • Step 3 → For missing-digit divisibility by 3 or 9, solve (known sum + x) ≡ 0 (mod 3 or 9) to find x quickly.

Summary

Summary

  • Use sum-of-digits to quickly determine divisibility by 3 and 9.
  • Compute the digital root by repeatedly summing the digits until one digit remains.
  • Use the fast formula DR(n) = 1 + ((n - 1) mod 9) for instant single-digit results.
  • For missing-digit problems, set up congruences (mod 3 or mod 9) using the digit sum.

Example to remember:
The number 987654 has digit-sum 39 → digital root 3 → divisible by 3 but not by 9.

Practice

(1/5)
1. Find the digital root of 4729.
easy
A. 4
B. 5
C. 3
D. 6

Solution

  1. Step 1: Compute sum of digits:

    4 + 7 + 2 + 9 = 22.

  2. Step 2: Reduce to single digit (iterative):

    2 + 2 = 4 → digital root = 4.

  3. Final Answer:

    Digital root = 4 → Option A.

  4. Quick Check:

    Using formula DR = 1 + ((n - 1) mod 9): 4729 mod 9 = 22 mod 9 = 4 → DR = 4 ✅
Hint: Sum digits and reduce (or use DR = 1 + ((n-1) mod 9)).
Common Mistakes: Stopping after one sum without reducing to a single digit when necessary.
2. Which of the following numbers is divisible by 9?
easy
A. 738
B. 739
C. 740
D. 742

Solution

  1. Step 1: Rule for 9:

    A number is divisible by 9 if the sum of its digits is divisible by 9.

  2. Step 2: Check options (example for 738):

    7+3+8 = 18 → 18 is divisible by 9, so 738 is divisible by 9. (739 → 7+3+9=19 not; 740 → 7+4+0=11 not; 742 → 7+4+2=13 not.)

  3. Final Answer:

    738 → Option A.

  4. Quick Check:

    738 ÷ 9 = 82 → integer ✅
Hint: Use digit-sum test for 9 instead of long division.
Common Mistakes: Checking only the last digit or using the rule for 3 instead of 9.
3. Find the digit x (0-9) such that 4x6 is divisible by 9.
easy
A. 1
B. 8
C. 9
D. 7

Solution

  1. Step 1: Sum-of-digits condition for 9:

    4 + x + 6 must be divisible by 9.

  2. Step 2: Solve congruence:

    4 + x + 6 = 10 + x ≡ 0 (mod 9) → x ≡ -10 ≡ -1 ≡ 8 (mod 9). So x = 8 (single-digit solution).

  3. Final Answer:

    x = 8 → Option B.

  4. Quick Check:

    4 + 8 + 6 = 18 → 18 ÷ 9 = 2 → 486 is divisible by 9 ✅
Hint: Solve (known sum + x) ≡ 0 (mod 9) for missing digit x.
Common Mistakes: Forgetting to reduce modulo 9 and testing incorrect candidate digits.
4. Find the digital root of 7^5.
medium
A. 3
B. 5
C. 4
D. 9

Solution

  1. Step 1: Use mod 9 property:

    Digital root corresponds to value mod 9 (with 9 mapped to 9 or 0 case). Compute 7^5 (mod 9).

  2. Step 2: Reduce powers modulo 9:

    7 ≡ 7 (mod 9). 7^2 ≡ 49 ≡ 4 (mod 9). 7^3 ≡ 7^2 × 7 ≡ 4 × 7 = 28 ≡ 1 (mod 9). So cycle length 3 here; 7^5 = 7^3 × 7^2 ≡ 1 × 4 = 4 (mod 9).

  3. Final Answer:

    Digital root = 4 → Option C.

  4. Quick Check:

    Since 7^5 ≡ 4 (mod 9), DR = 4 (not 9). ✅
Hint: Compute base mod 9 and use exponent cycles to find result mod 9 quickly.
Common Mistakes: Trying to compute full power instead of using modular reduction.
5. What is the digital root of 2^100?
medium
A. 1
B. 8
C. 9
D. 7

Solution

  1. Step 1: Use mod 9 property and cycle length:

    Compute 2^100 (mod 9). Note 2^6 ≡ 64 ≡ 1 (mod 9), so powers of 2 cycle every 6 in mod 9.

  2. Step 2: Reduce exponent modulo cycle:

    100 mod 6 = 4 → 2^100 ≡ 2^4 ≡ 16 ≡ 7 (mod 9).

  3. Final Answer:

    Digital root = 7 → Option D.

  4. Quick Check:

    Because 2^4 = 16 and 16 mod 9 = 7, and cycle repeats every 6, 2^100 has same DR as 2^4 → 7 ✅
Hint: Find exponent mod cycle length (6) for base 2 under mod 9, then compute small power.
Common Mistakes: Forgetting Euler/cycle behaviour and trying to compute huge powers directly.

Mock Test

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