noteswale
Noteclerical

Number system

Lesson 62 of 113

1. Ye topic hai kya?

LDC CBT: digit ki jagah (place), decimal ↔ binary, last digit of 723, 258 ko 3 kat-ta hai kya, missing digit. Ye counting-speed hai, number-line philosophy nahi.

Yahan lock

Place vs face. Even/odd/prime (2 only even prime). Binary powers of 2. Cyclicity 4. Rules 2,3,4,5,6,8,9,10,11. Remainder = last piece after divide.

Yahan kya nahi

Terminate/repeat p/q, Euclid algorithm, √2 proof — Real numbers. Bells, tiles, rooms — HCF/LCM. BODMAS / 15² table — Simplification. Agla term — Number series. P(E) — Probability.

Digit sirf 0–9. Binary mein sirf 0,1. Unit digit = remainder when ÷10. Divisibility 3/9 = digits ka sum, poori number ko divide mat — unless chhoti ho.

2. Place value, even–odd, prime

Face value = digit khud (8 hamesha 8). Place value = digit × uski jagah. Indian: ones, tens, hundreds, thousands, ten-thousands, lakh, ten-lakh, crore.

5 8 2 3 4 7 lakh 10-th thous hund tens ones

5,82,347. Leftmost 5: place 5,00,000; face 5. 8 ki place 80,000, face 8. Expanded: 5×100000 + 8×10000 + 2×1000 + 3×100 + 4×10 + 7.

Even: last digit 0,2,4,6,8 (0 even hai). Odd: 1,3,5,7,9. Even+even=even, odd+odd=even, even+odd=odd. Even×anything=even except 0-talk later. Prime: sirf 1 aur khud — 2,3,5,7,11… 1 na prime na composite. 2 akela even prime. Negative integers prime nahi (Class-X/LDC).
Example

5,82,347 mein 8 ka place value aur face value? 2+3 even ya odd?

Solution

  1. Place

    8 ten-thousands = 80000. Face = 8. Trap: 8×1000=8000 (thousands 2 ki jagah).

  2. 2+3

    Even+odd=odd (5). Classification “odd man” nahi — yahan parity.

Answer80000, 8 ; odd

3. Binary — base 2

Decimal base 10: 1,10,100,1000. Binary: 1,2,4,8,16,32,… Digit sirf 0 ya 1. Paper: (1101)2 = ?₁₀ ya 38 = ?₂.

32 16 8 4 2 1 1 0 1 1 0 1 32+8+4+1 = 45

Bits = 1 wale power. 0 wale skip. (101101)2 = 45. 2 ki jagah 10 mat padhna.

Binary → decimal: har 1 ko uski power se jod. Decimal → binary: baar-baar ÷2, remainder neeche se upar padho (LSB last remainder pehle likha jata hai jab upar se padho — remainders ulta).
Example

(1101)2 decimal? 38 decimal → binary?

Solution

  1. 1101

    8+4+0+1 = 13. (Pehle 8, phir 4, 2, 1.) 1+1+0+1=3 galat (binary add nahi, weights).

  2. 38÷2

    19 r0, 9 r1, 4 r1, 2 r0, 1 r0, 0 r1. Remainders ulta: 100110. Check: 32+4+2=38.

Answer13 ; 1001102
Binary 10 = ten nahi, do (2). 100₂ = 4. Leading zeros matter nahi: 01101 = 1101. Digit 2 binary mein illegal.

4. Unit digit — cycle

Last digit of 723 poori power expand nahi. Last digits repeat. ÷10 ka remainder = unit. Product ka unit = units ka product ka unit.

2 4 8 6 2ⁿ

2n: 2 → 4 → 8 → 6 → 2… period 4. Exponent ko 4 se modulo. Remainder 0 matlab cycle ki 4th = 6 (24, 28…).

Base unitCycleLength
0,1,5,6khud hi1
44, 6 (odd exp → 4, even → 6)2
99, 1 (odd → 9, even → 1)2
22,4,8,64
33,9,7,14
77,9,3,14
88,4,2,64
Period 4: n mod 4. Rem 1 → cycle 1st, rem 2 → 2nd, rem 3 → 3rd, rem 0 → 4th. Base ka sirf unit use karo: 27k = 7k ka unit.
Example

Unit digit: 723, 253, 314, 421, 23×47?

Solution

  1. 723

    23÷4 rem 3. Cycle 7,9,3,1 → 3rd = 3.

  2. 253

    53÷4 rem 1 → 2. (Rem 0 hota to 6.)

  3. 314

    14 rem 2 → 9.

  4. 421

    Odd exponent → 4. (42=6, 43=4.)

  5. 23×47

    3×7=21 → unit 1. Poora 1081 check, last 1. AP 7,9,3,1 agla term nahi — yahan cycle se power.

Answer3 ; 2 ; 9 ; 4 ; 1
210 = 1024, unit 4: 10 rem 2 → cycle 2nd = 4. 10 se divide remainder bhi 4. 5 se divide: 4 rem 4. Mean of digits Statistics nahi.

5. Divisibility 2–11, missing digit

Poori lambi number ko divide karne se pehle rule. 7 ka rule (last digit double, minus) LDC mein kam; 2–6, 8–11 roz.

Se kat-taRule
2Last digit even
3Digits ka sum ÷3
4Last 2 digits ÷4
5Last 0 ya 5
62 aur 3 dono
8Last 3 digits ÷8
9Digits ka sum ÷9 (3 se tight)
10Last 0
11Alternate sum (jagah-jagah + −) ÷11, including 0
3 vs 9: 258 → 2+5+8=15, 15÷3 yes, 15÷9 no. 4 vs 8: 4 ke liye last do, 8 ke liye last teen — 12 last two 12÷4 yes, 012=12÷8 nahi. 11: 2178 → 2−1+7−8=0, 0÷11 yes.
Example

258 ko 3 aur 9? 1324 ko 4? 3512 ko 8? 918082 ko 11?

Solution

  1. 258

    Sum 15: 3 haan, 9 nahi.

  2. 1324

    Last 24÷4=6. Haan. Poora 1324/4 sochne ki zaroorat nahi.

  3. 3512

    512÷8=64. Haan. Last two 12÷8 nahi — galat rule 4 wala.

  4. 918082

    9−1+8−0+8−2=22, 22÷11. Haan. Euclid HCF nahi.

Answer3 yes / 9 no ; 4 yes ; 8 yes ; 11 yes
Example

5a2 ko 9 kat-ta. a (0–9)? 2x4 ko 9?

Solution

  1. 5a2

    5+a+2=7+a = 9,18,27… a=2 (9) ya a=11 invalid. a=2. (7+a=18 → a=11 nahi.)

  2. 2x4

    6+x=9 → x=3. (18 → x=12 nahi.)

Answera=2 ; x=3
Example

Number 4a6b 4 se kat-ta, last digit b=0. 9 se bhi kat-ta. a?

Solution

  1. Rule 4

    Last two 6b=60. 60÷4=15, already OK for b=0.

  2. Rule 9

    4+a+6+0=10+a ÷9. a=8 → 18. a=−1 nahi. a=8.

Answer8
Bells 12,18,24 saath — LCM, agla chapter. 22/7 terminate test — Real numbers. 15² BODMAS — Simplification. Number series 2,4,8,6 agla? Cycle yahan power ke liye, series ke liye nahi.

6. Sawal — basic se pro

Q1 · Basic

5,82,347. 8 ka place value?

  1. 80000
  2. 8
  3. 8000
  4. 800000

Solution

  1. Ten-thousand

    80000. B face. C thousands.

Answer80000
Q2 · Basic

Kaun even prime?

  1. 2
  2. 1
  3. 4
  4. 0

Solution

  1. Sirf 2

    1 prime nahi. 0 even lekin prime nahi.

Answer2
Q3 · Basic

(1101)2 = ?

  1. 13
  2. 11
  3. 8
  4. 1101

Solution

  1. 8+4+1

    13. B 8+2+1. D as decimal padh diya.

Answer13
Q4 · Basic

38 in binary?

  1. 100110
  2. 100101
  3. 110100
  4. 38

Solution

  1. 32+4+2

    100110. B 32+4+1=37.

Answer100110
Q5 · Medium

Unit digit 723?

  1. 3
  2. 7
  3. 9
  4. 1

Solution

  1. 23 rem 3

    7,9,3,1 → 3. B rem 1. D rem 0.

Answer3
Q6 · Medium

Unit digit 253?

  1. 2
  2. 4
  3. 8
  4. 6

Solution

  1. 53 rem 1

    2. D agar rem 0 samjha.

Answer2
Q7 · Medium

258 divisible by 3? by 9?

  1. 3 yes, 9 no
  2. Dono yes
  3. Dono no
  4. 3 no, 9 yes

Solution

  1. Sum 15

    A. 9 ke liye 18 chahiye.

AnswerA
Q8 · Medium

2178 divisible by 11?

  1. Haan
  2. Nahi
  3. Sirf 3 se
  4. Sirf 2 se

Solution

  1. 2−1+7−8=0

    Haan. D even last 8, 11 alag.

AnswerHaan
Q9 · Pro

5a2 ÷9. a?

  1. 2
  2. 5
  3. 7
  4. 9

Solution

  1. 7+a=9

    a=2. C 7+7=14 not ÷9.

Answer2
Q10 · Pro

12, 18, 24 bells saath. Is chapter?

  1. Nahi — HCF / LCM
  2. Haan, divisible by 3
  3. Binary
  4. Unit digit

Solution

  1. Together

    LCM. A. 24÷3 yahan ho sakta, bells nahi.

AnswerA

Practice

Pehle khud, phir neeche kholo.

12345 678910

P1

8 ka face value in 5,82,347?

  1. 8
  2. 80000
  3. 80
  4. 800

Solution

  1. Face

    8. B place.

Answer8
P2

(101101)2 = ?

  1. 45
  2. 44
  3. 53
  4. 32

Solution

  1. 32+8+4+1

    45. C 32+16+4+1.

Answer45
P3

Unit 314?

  1. 9
  2. 3
  3. 7
  4. 1

Solution

  1. 14 rem 2

    3,9,7,1 → 9.

Answer9
P4

Unit 421?

  1. 4
  2. 6
  3. 2
  4. 0

Solution

  1. Odd

    4. B even exp.

Answer4
P5

23×47 unit?

  1. 1
  2. 21
  3. 7
  4. 3

Solution

  1. 3×7=21

    1. B poora 21.

Answer1
P6

1324 ÷4?

  1. Haan
  2. Nahi
  3. Sirf 2 se
  4. Sirf 8 se

Solution

  1. 24÷4

    Haan. D last 324÷8 nahi zaroori.

AnswerHaan
P7

3512 ÷8?

  1. Haan
  2. Nahi
  3. Last 12 se 4
  4. Sum digits

Solution

  1. 512÷8

    Haan.

AnswerHaan
P8

2x4 ÷9. x?

  1. 3
  2. 2
  3. 6
  4. 9

Solution

  1. 6+x=9

    x=3.

Answer3
P9

4a60 ÷9, a?

  1. 8
  2. 0
  3. 4
  4. 6

Solution

  1. 10+a÷9

    a=8 → 18.

Answer8
P10

7/8 decimal terminate — is chapter?

  1. Nahi — Real numbers
  2. Haan, ÷8 last 3
  3. Binary
  4. Unit 8

Solution

  1. p/q 2a5b

    A. Rule 8 yahan integer 3512 pe, fraction 7/8 nahi.

AnswerA

← Index · Probability · Real numbers

Agle topic: HCF / LCM.