19. Number Formation & Digits Based Problems

Digits based problems aur number formation ka complete concept: digit place value, forming smallest/largest numbers, digit interchange, remainder with digit-sum, divisibility with digits, SSC CGL level solved examples & practice with answers.

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1) Place value based representation (foundation)

Agar 3-digit number abc ho, to value: 100a + 10b + c

4-digit number abcd: 1000a + 100b + 10c + d

Digits based almost har question me ye representation use hoti hai.

2) Forming largest & smallest number from given digits

A) Largest number

Digits ko descending order me arrange karo.

• Digits: 3, 0, 7, 1 ⇒ largest = 7310

B) Smallest number (0 present ho to rule)

Digits ko ascending me karo, but first digit 0 nahi ho sakta. So: smallest non-zero digit pehle, phir 0, phir remaining digits ascending.

• Digits: 3, 0, 7, 1 ⇒ ascending: 0,1,3,7 ⇒ smallest = 1037
• Digits: 0,0,2,5 ⇒ smallest = 2005

3) How many numbers can be formed? (counting)

A) Without repetition (all digits distinct)

Agar n digits distinct hain, aur r-digit number banana hai: ways = nP r.

Example: digits 1,2,3,4 se 3-digit numbers (no repetition)

• Ways = 4P3 = 4×3×2 = 24

B) With repetition allowed

Ways = n^r (first place cannot be 0 if 0 included).

C) If 0 included (leading zero not allowed)

Example: digits {0,1,2,3}, 3-digit numbers (no repetition):

• Total 4P3 = 24
• Invalid (leading 0): fix 0 at first place ⇒ remaining 3P2 = 6
• Valid = 24 − 6 = 18

4) Sum of all numbers formed (SSC favorite)

Agar n distinct digits se n-digit numbers (no repetition) ban rahe hain, to: Har digit har place par equal times aata hai.

Each digit appears at each place: (n−1)! times. Sum = (sum of digits) × (n−1)! × (111...1) (n times)

111...1 means: 1 + 10 + 100 + ... (n terms)

Example: digits 1,2,3,4 se all 4-digit numbers (no repetition) ka sum

• n=4, sum digits = 10
• (n−1)! = 3! = 6
• 1111 = 1+10+100+1000
• Sum = 10 × 6 × 1111 = 60 × 1111 = 66660

5) Digit interchange / swapping problems

Trick: number ko place value me likho, then swap.

Example: 2-digit number (10a+b). Digits interchange → (10b+a)

Difference (new − old) = (10b+a) − (10a+b) = 9(b−a)

• So, interchange difference always divisible by 9.

Example (SSC): A 2-digit number differs from its reversed number by 36. Find possible pairs (a,b)

• |9(b−a)| = 36 ⇒ |b−a| = 4
• So digits differ by 4 (example: 15 & 51, 26 & 62, 37 & 73, 48 & 84, 59 & 95)

6) Remainder / divisibility linked with digits (quick)

9 (and 3) divisibility: sum of digits divisible by 9/3.

• Example: 8721 sum=18 ⇒ divisible by 9 and 3

7) Common traps

• Smallest number me 0 ko first digit mat banao.
• Counting me “repetition allowed” aur “not allowed” confuse mat karo.
• 2-digit reverse difference hamesha 9 multiple hota hai.

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8) Practice (SSC CGL) + Answers

1. Digits 3,0,7,1 se smallest and largest number banao.
2. Digits 1,2,3,4 se 3-digit numbers (no repetition) kitne?
3. Digits 0,1,2,3 se 3-digit numbers (no repetition) kitne?
4. Digits 2,4,6 se all 3-digit numbers (no repetition) ka sum?
5. 2-digit number and its reverse ka difference 45 hai. Digits ka difference?
6. How many 4-digit numbers can be formed using digits 0,1,2,3,4 without repetition?

Show Answers

1. Smallest = 1037, Largest = 7310
2. 4P3 = 24
3. 4P3 − 3P2 = 24 − 6 = 18
4. n=3, sum digits=12, (n−1)!=2, 111=111 ⇒ 12×2×111 = 2664
5. |9(b−a)|=45 ⇒ |b−a|=5
6. Total 5P4 = 120; invalid leading 0: fix 0 then 4P3=24 ⇒ valid = 120−24 = 96