13. Last Two Digits / Last Digit Patterns

Last two digits (mod 100) aur last digit patterns ka complete concept: mod 100 basics, cycle building, Chinese remainder style shortcut (easy), power patterns, multiplication patterns, SSC CGL level examples & practice with answers.

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1) Last Two Digits ka matlab (mod 100)

Kisi bhi number ka last two digits = number mod 100.

• 12345 mod 100 = 45 ⇒ last two digits = 45
• 907 mod 100 = 07 ⇒ last two digits = 07 (leading zero allowed)

2) Base reduce karo (always first step)

Agar question me Aⁿ hai, to: A mod 100 le lo, phir power consider karo.

• Example: 347^k ⇒ base for last 2 digits is 47 (because 347 mod 100 = 47)

3) 2 important building blocks: mod 4 and mod 25 idea

100 = 4 × 25. Last two digits (mod 100) ka pattern often mod 4 & mod 25 ke combine se easy hota hai. SSC me mostly “pattern build” approach enough hota hai. (CRT name ya theory yaad rakhna zaroori nahi.)

4) Easy & practical method (SSC): cycle build using mod 100

Base ko 100 se reduce karo, phir 2–10 powers tak last two digits nikaal kar pattern repeat check karo.

Example 1: Find last two digits of 7²⁰²

We compute pattern mod 100:

• 7¹ = 07
• 7² = 49
• 7³ = 343 ⇒ 43
• 7⁴ = 43×7 = 301 ⇒ 01

Now 7⁴ mod 100 = 01 ⇒ after that, multiply by 7 repeats the cycle of length 20? (actually 7 and 100 are coprime, so a cycle exists; but we already got a strong shortcut:)

Since 7⁴ ≡ 01 (mod 100), then: 7²⁰² = 7^{4×50 + 2} ≡ (7⁴)⁵⁰ × 7² ≡ 1 × 49 = 49 (mod 100)

Answer: 49

Example 2: Find last two digits of 13⁵⁷

Base = 13 (already).

• 13¹ = 13
• 13² = 169 ⇒ 69
• 13⁴ last 2 digits = 69² = 4761 ⇒ 61
• 13⁸ last 2 digits = 61² = 3721 ⇒ 21
• 13¹⁶ last 2 digits = 21² = 441 ⇒ 41
• 13³² last 2 digits = 41² = 1681 ⇒ 81

Now 57 = 32 + 16 + 8 + 1 ⇒ 13⁵⁷ ≡ (13³²)(13¹⁶)(13⁸)(13¹) (mod 100)

≡ 81 × 41 × 21 × 13 (mod 100)

• 81×41 = 3321 ⇒ 21
• 21×21 = 441 ⇒ 41
• 41×13 = 533 ⇒ 33

Answer: 33

5) Special endings (very fast)

A) Last digit 0

• If number ends with 0 ⇒ any positive power ends with 00 (mod 100)
• Example: 120⁵ ends with 00

B) Last digit 5

• Any number ending in 5: for n ≥ 2, last two digits become 25
• Example: 35²=1225 ⇒ 25; 35³ also ends with 25

C) Last digit 6

• Any number ending in 6: unit digit always 6; last two digits often stabilize quickly for many bases (pattern approach use karo)

D) Last digit 1/9/3/7 (odd & coprime with 10)

In cases me mod 100 cycle exists. Practical SSC: square & multiply (Example 2) is fastest.

6) Product/Sum expressions (reduce each term)

Last two digits nikalne ke liye har term ko mod 100 reduce karo, then combine: (a+b) mod 100, (a×b) mod 100.

Example 3: Last two digits of (47¹² × 23⁵)

SSC approach: compute each power mod 100 via squaring.

• 47² = 2209 ⇒ 09
• 47⁴ ⇒ 09² = 81 ⇒ 81
• 47⁸ ⇒ 81² = 6561 ⇒ 61
• 47¹² = 47⁸×47⁴ ⇒ 61×81 = 4941 ⇒ 41
• 23² = 529 ⇒ 29
• 23⁴ ⇒ 29² = 841 ⇒ 41
• 23⁵ ⇒ 23⁴×23 ⇒ 41×23 = 943 ⇒ 43
• Now multiply: 41×43 = 1763 ⇒ 63

Answer: 63

7) Common traps

• Last two digits can have leading zero (07, 01).
• Always take mod 100 after every multiply to keep numbers small.
• 5-ending powers: for n ≥ 2 always 25 (don’t waste time).
• If base ends with 0: answer always 00 (n ≥ 1).

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8) Practice (SSC CGL) + Answers

1. Find last two digits of 7²⁰².
2. Find last two digits of 15¹⁰.
3. Find last two digits of 25⁹⁹.
4. Find last two digits of 12²⁵.
5. Find last two digits of 3²⁰.
6. Find last two digits of (19⁷ + 1).

Show Answers

1. 49
2. Ends with 5, power ≥2 ⇒ last two digits 25 ⇒ 25
3. 25ⁿ for n≥1 ends with 25
4. 12 mod 100=12; 12²=144 ⇒ 44; 12⁴ ⇒ 44²=1936 ⇒ 36; cycle length 20 for mod 100 powers of 2*3? Practical: 12⁵=248832 ⇒ 32; continuing gives 12²⁵ ends with 32 (quick pattern via repeated squaring also works)
5. 3⁴=81 ⇒ 3²⁰=(3⁴)⁵ ⇒ 81⁵ mod 100 = 01
6. 19 mod 100=19; 19²=361 ⇒ 61; 19⁴ ⇒ 61²=3721 ⇒ 21; 19⁷=19⁴×19²×19 ⇒ 21×61×19 ⇒ (21×61=1281⇒81; 81×19=1539⇒39) so 19⁷ last2=39 ⇒ +1 ⇒ 40