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Sum of Digits / Digital Root

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Introduction

Sum of Digits और Digital Root patterns बड़े numbers को एक single-digit representative में बदलने की तेज़ technique देते हैं। ये divisibility checks (खासकर 3 और 9), checksum problems और repeated-sum tasks को simplify करने में बहुत उपयोगी हैं।

Pattern: Sum of Digits / Digital Root

Pattern: Sum of Digits / Digital Root

Digital root वह single-digit value है जो किसी number की digits को बार-बार जोड़ने पर अंत में बचती है। इसे divisibility (3 और 9) test करने और fast calculation के लिए modular formula से निकालें।

  • Sum of digits (S): अगर n के digits d_k...d_1d_0 हैं, तो S(n) = d_0 + d_1 + ... + d_k.
  • Digital root (DR): S को तब तक apply करें जब तक एक digit न बचे (0-9)।
  • Fast modular formula:
    n > 0 के लिए: DR(n) = 1 + ((n - 1) mod 9) n = 0 के लिए: DR(0) = 0
  • Digit-sum divisibility tests:
    • S(n) अगर 3 से divisible → n, 3 से divisible।
    • S(n) अगर 9 से divisible → n, 9 से divisible।
  • Casting out 9s: Digit-sum में 9 subtract करते रहना mod 9 remainder नहीं बदलता-mental calculation आसान होती है।
  • Relation to mod 9: Digital root (9 आने पर छोड़कर) n mod 9 के बराबर होता है; DR(n) = 0 जब n ≡ 0 (mod 9), वरना DR = n mod 9 (या formula 1+((n-1) mod 9) use करें)।

Step-by-Step Example

Question

(a) 987654 का digital root निकालें।
(b) Sum-of-digits से जाँचें कि 987654, 3 और 9 से divisible है या नहीं।

Solution

  1. Step 1: Sum of digits निकालें

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

  2. Step 2: Digital root (iterative)

    S(39) = 3 + 9 = 12.
    S(12) = 1 + 2 = 3.
    इसलिए DR(987654) = 3.

  3. Step 3: Modular formula से quick check

    DR = 1 + ((987654 - 1) mod 9). 987654 mod 9 = (S = 39) → 39 mod 9 = 3, इसलिए DR = 3 → दोनों methods match।

  4. Step 4: Divisibility conclusions

    DR = 3 (digit-sum 39, 3 से divisible लेकिन 9 से नहीं), इसलिए 987654 → 3 से divisible, पर 9 से नहीं।

  5. Final Answer:

    (a) Digital root = 3.
    (b) 3 से divisible: Yes. 9 से divisible: No.

  6. Quick Check:

    987654 ÷ 3 = 329218 (integer) 987654 ÷ 9 = 109739.333... (integer नहीं) → result confirm। ✅

Quick Variations

1. Missing digit problems: किसी number में unknown digit x हो और उसे 9 से divisible होना हो, तो (S + x) ≡ 0 (mod 9) से x निकालें।

2. Negative numbers: Digit-sum और digital root के लिए absolute value लें।

3. Base-b version: Digital root का concept दूसरे bases में भी चलता है-modulus (b-1) का उपयोग करें।

Trick to Always Use

  • Step 1 → Digit-sum निकालते समय 9s cast out करें-बड़ा sum जल्दी छोटा हो जाता है।
  • Step 2 → Comfortable हों तो DR = 1 + ((n - 1) mod 9) से direct एक-step result लें।
  • Step 3 → Missing-digit problems में (known sum + x) ≡ 0 (mod 3 या 9) से x तुरंत मिल जाता है।

Summary

  • Digit-sum से 3 और 9 की divisibility जल्दी check होती है।
  • Digital root digits को बार-बार जोड़कर single digit तक पहुँचता है।
  • Formula DR(n) = 1 + ((n - 1) mod 9) बहुत fast है।
  • Missing-digit problems के लिए digit-sum से congruence बनाएं।

Example:
987654 का digit-sum 39 → digital root 3 → 3 से divisible, 9 से नहीं।

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.