Select Page

First two non-zero digit in N!

by | Aug 18, 2021 | Uncategorized | 0 comments

First two non-zero digit in N!:-

First understand some basic concept:-
To find last two digit of lets

3 6 × 5 1+×66 × 5 = 3 03 × 1 = 3+3Unit digit35 9 × 7 89×8 = 7 2 2Unit digit9×7 = 6 35×8 = 4 0Take 7 as carry7++1 004 8 3 × 5 9 8ignoreignorewe take only rightmost two digits for multiplication42741 337 8 5 6 2 × 1 5 9 8 7 ignoreignore412699we take only rightmost two digits for multiplicationnow the other concept2 = 1 0 4 8 5 7 6

& (76)ⁿ = last two digits always 76
& 3⁴ = 81
& (81)ⁿ = always 1 at unit place & last digit of (8 × last digit of n) will be at tens place.

For Example:- 

(81) 6 17 × 8 = 5 6(81)2 18 × 9 = 7 2(81)2 18 × 9 = 7 2(81)6 18 × 2 = 1 6

Now come to the point to find the first two non-zero digits of a factorial N!

find the highest power of 5 in N!. let it be H. then understand it with an example:-
182!

Remainder5555182367102211QQQQ= U[182 × 181] = 42= U[36] = 36= U[7 × 6] = 42= U[1] = 1

& Highest power of 5 in 182! is: H=36 + 7 + 1 + 0 = 44
so Uₜ(182!) = Uₜ(12ᴴ) × Uₜ(Q₁ × Q₂ × Q₃ × Q₄)

= 12 × 42 × 36 × 42 × 1

= 12⁴⁴ × 12 × 42
= 12⁴⁴ × 04
= (2²×3)⁴⁴ × 4
= 2⁸⁸ × (81)¹¹ × 4
= (2²⁰)⁴ × 2⁸ × 81 × 4
= 76 × 56 × 24
= 56 × 24
= 44 Answer

Take another example:-
Uₜ(1038!) = ?

Remainder5555103820741813123QQQQ= U[1038 × 1037 × 1036] = 16= U[207 × 206] = 42= U[41] = 41= U[8 × 7 × 6] = 36501Q = U[1] = 1

H = 207+41+8+1+0
    = 257
now Uₜ(1038!) = Uₜ(12ᴴ) × Uₜ(Q₁ × Q₂ × Q₃ × Q₄ × Q₅)

= U(12) × U(16×42×41×36)= (2×3) × 72 × 76= 2 × 3 × 72= (2) × 2 × (3) × 3 × 72= (76) × 84 × (81) × 16= 76 × 84 × 21 × 16= 84 × 36= 24 Answer

Submit a Comment

Your email address will not be published. Required fields are marked *

Captcha loading...