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(Q). Find the first two non-zero digit (FTNZD) of \frac{{555!}}{{444!}} .
(i) 12
(ii) 72
(iii) 96
(iv) 38
Solution:- 

FTNZD(555!) = 

 

555111225545Remainder0125H = 111 + 22 +4 +0 = 13704Q = ✖Q = U(111) = 11Q = U(22×21) = 62Q = U(4×3×2×1) = 24

FTNZD(555!) = Uₜ(12ᴴ) × Uₜ(Q₁×Q₂×Q₃×Q₄)
                          =Uₜ(12¹³⁷) × Uₜ(11×62×24)

= (2×3)×62×64 = 2×3×6824+6= (2) × 2 × (81) × 3 × 68= × × 682888= 84 × 288884= 84 ×= 56

FTNZD(555!) = 56

now to find FTNZD(444!) = 

44488175535Remainder4325H = 88 + 17 + 3 + 0 = 108 03Q = U(444×443×442×441) = 24Q = U(88×87×86) = 16Q = U(17×16) = 72Q = U(3×2×1) = 06now Q= U(444 × 443 × 442 × 441)162984292 × 2224482Q = U( × 86546556 × 86380116

∴ FTNZD(444!) = Uₜ(12ᴴ) × Uₜ(Q₁ × Q₂ × Q₃ × Q₄)

= 12 × (24×16 × 72×06)= (2×3) × 84 × 32268= 2 × 3 × 88 = (2) × 2 × (81) × 88= × = 36 × 683828836= 484644

∴ FTNZD(444!) = 48
now to find FTNZD\left( {\frac{{555!}}{{444!}}} \right) 
Let 

= Quotient & Remainder = 0x000........0FTNZD×multiply

i.e. 444! × Quotient = 555!
So FTNZD(444!) × FTNZD(Quotient) = FTNZD(555!)

because we know that irrespective of the size of the number & ignoring its all trailing zeros, To find the FTNZD we only need the FTNZD of the numbers & remaining digits of the numbers are useless when finding FTNZD.
∴ 48 × x = 56
now from the given options
(i) 12
(ii) 72 
(iii) 96
(iv) 38

Uₜ(48×72) = 56
Hence option (ii) is correct

FTNZD\left( {\frac{{555!}}{{444!}}} \right) = 72 (option(ii)) Answer