Light-version-of-Q-13724-Expansion-of-100-has-24-0-s-at-the-end-Find-the-first-non-zero-digit-from-right-100-d000-00-What-is-the-value-of-d-

Question Number 15292 by RasheedSoomro last updated on 09/Jun/17

You can't use 'macro parameter character #' in math mode

E xpansion of 100! has 24, 0^{'} s at the end .

Find the first non - zero digit from right .

100! = \dots . . d 000 \dots 00

What is the value of d ?

Answered by RasheedSoomro last updated on 11/Jun/17

100! has 24 0′s from right. p=((100!)/(10^(24) )) has no zero at the end. ∴ First non-zero digit of 100! =unit digit of p ∴ p≡d(mod 10) Simplifying p=((100!)/(10^(24) ))=((100!)/(2^(24) ×5^(24) )) 10^(24) =2^(24) ×5^(24) =2^(6•) (5.10.15...100) All the 24 5′s have come in 5.10...100 ^• Also some 2′s have come: One 2 in each multiple of 10 10 2′s One 2 extra in each multiple of 20 5 2′s One 2 extra in each multiple of 40 2 2′s One 2 extra in each multiple of 80 1 2′s _(−) Total(included in 5.10...100) 18 2′s Remaining 24−18=6 So now p=((100!)/(10^(24) ))=((100!)/(2^(24) ×5^(24) ))=((100!)/(2^6 (5.10....100))) p=(1/(2.2^5 )){(1.2.3.4)(6.7.8.9)...(96.97.98.99} p={(((1.2.3.4)/2))(6.7.8.9).(11.12.13.14)(((16.17.18.19)/2))} ×{(21.22.23.24)(26.27.28.29).(31.32.33.34)(((36.37.38.39)/2))} ...×{(81.82.83.84)(86.87.88.89).(91.92.93.94)(((96.97.98.99)/2))} Account of 2^6 : One 2 comes in read bracket 5 2′s come in five blue brackets (one blue bracket is per each 20) Now we can verify/prove the following congruences: ((1.2.3.4)/2)≡2(mod 10) {: ((6.7.8.9)),(⋮),((91.92.93.94)) }≡4(mod 10) [14 congruencs] (5k+1)(5k+2)(5k+3)(5k+4)≡4(mod 10) { ((^• See Answer of mrW1 ⌉)),((to Q#14614 or an)),(( alternative proof in)),((the comment below.)) :} {: (((16.17.18.19)/2)),(⋮),(((96.97.98.99)/2)) }≡2 [5 congruences] {(5k+1)(5k+2)(5k+3)(5k+4)}/2≡4(mod 10) { ((^∗ Fo proof see the)),((comment below.)) :} Multiplying all the above congruences p≡2.4^(14) .2^5 (mod 10) 4^(14) ≡6(mod 10) [4^(2k) ≡6(mod 10)] 2^5 ≡32≡2(mod 10) p≡2.6.2(mod 10) p≡4(mod 10) Hence the require non-zero digit is 4 It′s only accedent that the answer is correct! Actually there′s a fault in the logic. I′ll try to correct my logic. Please stay with me!

100! has 24 0^{'} s from right .

p = \frac{100!}{10^{24}} has no zero at the end .

∴ First non - zero digit of 100!

= unit digit of p

∴ p \equiv d (\mod 10)

S i m p l i f y i n g p = \frac{100!}{10^{24}} = \frac{100!}{2^{24} \times 5^{24}}

10^{24} = 2^{24} \times 5^{24} = 2^{6 ∙} (5.10.15 \dots 100)

All the 24 5^{'} s have come in 5.10 \dots 100

^{∙} Also some 2^{'} s have come :

One 2 in each multiple of 10 10 2^{'} s

One 2 extra in each multiple of 20 5 2^{'} s

One 2 extra in each multiple of 40 2 2^{'} s

One 2 extra in each mult \underline{iple of 80 1 2^{'} s}

Total (included in 5.10 \dots 100) 18 2^{'} s

Remaining 24 - 18 = 6

So now p = \frac{100!}{10^{24}} = \frac{100!}{2^{24} \times 5^{24}} = \frac{100!}{2^{6} (5.10 \dots . 100)}

p = \frac{1}{2 . 2^{5}} {(1.2.3.4) (6.7.8.9) \dots (96.97.98.99}

p = {(\frac{1.2.3.4}{2}) (6.7.8.9) . (11.12.13.14) (\frac{16.17.18.19}{2})}

\times {(21.22.23.24) (26.27.28.29) . (31.32.33.34) (\frac{36.37.38.39}{2})}

\dots \times {(81.82.83.84) (86.87.88.89) . (91.92.93.94) (\frac{96.97.98.99}{2})}

Account of 2^{6} : One 2 comes in read bracket

5 2^{'} s come in five blue brackets

(one blue bracket is per each 20)

Now we can verify / prove the following congruences :

\frac{1.2.3.4}{2} \equiv 2 (\mod 10)

\begin{matrix} 6.7.8.9 \\ ⋮ \\ 91.92.93.94 \end{matrix}} \equiv 4 (\mod 10) [14 congruencs]

You can't use 'macro parameter character #' in math mode

\begin{matrix} (16.17.18.19) / 2 \\ ⋮ \\ (96.97.98.99) / 2 \end{matrix}} \equiv 2 [5 congruences]

{(5 k + 1) (5 k + 2) (5 k + 3) (5 k + 4)} / 2 \equiv 4 (\mod 10) {\begin{cases} ^{*} Fo proof see the \\ comment below . \end{cases}

Multiplying all the above congruences

p \equiv 2 . 4^{14} . 2^{5} (\mod 10)

4^{14} \equiv 6 (\mod 10) [4^{2 k} \equiv 6 (\mod 10)]

2^{5} \equiv 32 \equiv 2 (\mod 10)

p \equiv 2.6.2 (\mod 10)

p \equiv 4 (\mod 10)

Hence the require non - zero digit is 4

\boldsymbol {It}^{'} \boldsymbol s \boldsymbol only \boldsymbol accedent \boldsymbol that \boldsymbol the \boldsymbol answer

\boldsymbol is \boldsymbol correct! \boldsymbol Actually \boldsymbol {there}^{'} \boldsymbol s \boldsymbol a \boldsymbol fault \boldsymbol in

\boldsymbol the \boldsymbol logic . \boldsymbol I^{'} \boldsymbol ll \boldsymbol try \boldsymbol to \boldsymbol correct \boldsymbol my \boldsymbol logic .

\boldsymbol Please \boldsymbol stay \boldsymbol with \boldsymbol me!

Commented by RasheedSoomro last updated on 09/Jun/17

•(5k+1)(5k+2)(5k+3)(5k+4)(mod 10)=4 Case-1: k is even.let it′s equal to 2m ∀m∈Z =(5(2m)+1)(5(2m)+2)(5(2m)+3)(5(2m)+4)(mod 10) =10000m^4 +10000m^3 +3500m^2 +500m+24(mod 10) =(10000m^4 +10000m^3 +3500m^2 +500m+20)+4(mod 10) =10(1000m^4 +1000m^3 +350m^2 +50m+2)+4(mod 10) =4(mod 10) Case-2:k is odd.Let it′s equal to 2m+1 (5(2m+1)+1)(5(2m+1)+2)(5(2m+1)+3)(5(2m+1)+4)(mod 10) (10m+6)(10m+7)(10m+8)(10m+9)(mod 10) 10000m^4 +30000m^3 +33500m^2 +16500m+3024(mod 10) (10000m^4 +30000m^3 +33500m^2 +16500m+3020)+4(mod 10) 10(1000m^4 +3000m^3 +3350m^2 +1650m+302)+4(mod 10) =4(mod 10 Hence for all k∈Z ,the proposition itrue. ∗{(5k+1)(5k+2)(5k+3)(5k+4)/2} (mod 10)=2 Case-1:k ieven.Let it′s equal to 2m ={(5(2m)+1)(5(2m)+2)(5(2m)+3)(5(2m)+4)/2}(mod 10) =5000m^4 +5000m^3 +1750m^2 +250m+12(mod 10) =10(500m^4 +500m^3 +175m^2 +25m+1)+2(mod 10) =2(mod 10) Case-2:k is odd.Let it′s equal to 2m+1 =(5(2m+1)+1)(5(2m+1)+2)(5(2m+1)+3)(5(2m+1)+4)/2(mod 10) =(10m+6)(10m+7)(10m+8)(10m+9)/2(mod 10) =(10000m^4 +30000m^3 +33500m^2 +16500m+3024)/2(mod 10) =5000m^4 +15000m^3 +16750m^2 +8250m+1512(mod 10) =(5000m^4 +15000m^3 +16750m^2 +8250m+1510)+2(mod 2) =2(mod 2)

∙ (5 k + 1) (5 k + 2) (5 k + 3) (5 k + 4) (\mod 10) = 4

Case - 1 : k is even . let {it}^{'} s equal to 2 m \forall m \in Z

= (5 (2 m) + 1) (5 (2 m) + 2) (5 (2 m) + 3) (5 (2 m) + 4) (\mod 10)

= {10000 m}^{4} + {10000 m}^{3} + {3500 m}^{2} + 500 m + 24 (\mod 10)

= ({10000 m}^{4} + {10000 m}^{3} + {3500 m}^{2} + 500 m + 20) + 4 (\mod 10)

= 10 ({1000 m}^{4} + {1000 m}^{3} + {350 m}^{2} + 50 m + 2) + 4 (\mod 10)

= 4 (\mod 10)

Case - 2 : k is odd . Let {it}^{'} s equal to 2 m + 1

(5 (2 m + 1) + 1) (5 (2 m + 1) + 2) (5 (2 m + 1) + 3) (5 (2 m + 1) + 4) (\mod 10)

(10 m + 6) (10 m + 7) (10 m + 8) (10 m + 9) (\mod 10)

{10000 m}^{4} + {30000 m}^{3} + {33500 m}^{2} + 16500 m + 3024 (\mod 10)

({10000 m}^{4} + {30000 m}^{3} + {33500 m}^{2} + 16500 m + 3020) + 4 (\mod 10)

10 ({1000 m}^{4} + {3000 m}^{3} + {3350 m}^{2} + 1650 m + 302) + 4 (\mod 10)

= 4 (\mod 10

Hence for all k \in Z, the proposition itrue .

* {(5 k + 1) (5 k + 2) (5 k + 3) (5 k + 4) / 2} (\mod 10) = 2

Case - 1 : k ieven . Let {it}^{'} s equal to 2 m

= {(5 (2 m) + 1) (5 (2 m) + 2) (5 (2 m) + 3) (5 (2 m) + 4) / 2} (\mod 10)

= {5000 m}^{4} + {5000 m}^{3} + {1750 m}^{2} + 250 m + 12 (\mod 10)

= 10 ({500 m}^{4} + {500 m}^{3} + {175 m}^{2} + 25 m + 1) + 2 (\mod 10)

= 2 (\mod 10)

Case - 2 : k is odd . Let {it}^{'} s equal to 2 m + 1

= (5 (2 m + 1) + 1) (5 (2 m + 1) + 2) (5 (2 m + 1) + 3) (5 (2 m + 1) + 4) / 2 (\mod 10)

= (10 m + 6) (10 m + 7) (10 m + 8) (10 m + 9) / 2 (\mod 10)

= ({10000 m}^{4} + {30000 m}^{3} + {33500 m}^{2} + 16500 m + 3024) / 2 (\mod 10)

= {5000 m}^{4} + {15000 m}^{3} + {16750 m}^{2} + 8250 m + 1512 (\mod 10)

= ({5000 m}^{4} + {15000 m}^{3} + {16750 m}^{2} + 8250 m + 1510) + 2 (\mod 2)

= 2 (\mod 2)

Commented by mrW1 last updated on 09/Jun/17

(5k+1)(5k+2)(5k+3)(5k+4) =(5k+1)(5k+4)(5k+3)(5k+2) = [(5k)^2 +5(5k)+4][(5k)^2 +5(5k)+6] =[(5k)^2 +5(5k)]^2 +10[(5k)^2 +5(5k)]+24 =625[k(k+1)]^2 +250[k(k+1)]+24 =2500[((k(k+1))/2)]^2 +500[((k(k+1))/2)]+24 since one of k and k+1 is even, ((k(k+1))/2) is integer ⇒4 (mod 10)

(5 k + 1) (5 k + 2) (5 k + 3) (5 k + 4)

= (5 k + 1) (5 k + 4) (5 k + 3) (5 k + 2)

= [{(5 k)}^{2} + 5 (5 k) + 4] [{(5 k)}^{2} + 5 (5 k) + 6]

= {[{(5 k)}^{2} + 5 (5 k)]}^{2} + 10 [{(5 k)}^{2} + 5 (5 k)] + 24

= 625 {[k (k + 1)]}^{2} + 250 [k (k + 1)] + 24

= 2500 {[\frac{k (k + 1)}{2}]}^{2} + 500 [\frac{k (k + 1)}{2}] + 24

since one of k and k + 1 is even,

\frac{k (k + 1)}{2} is integer

\Rightarrow 4 (\mod 10)

Commented by RasheedSoomro last updated on 09/Jun/17

Your method is better Sir! It was also used in your answer of Q#14614. I merely tried to use different approach!

Your method is better Sir! It was also used

You can't use 'macro parameter character #' in math mode

I merely tried to use different approach!

Commented by mrW1 last updated on 09/Jun/17

congratulation to you for solving this light version! the last non−zero digit of 100! is indeed 4. i hope you will crack 1000! soon!

congratulation to you for solving this

light version! the last non - zero digit

of 100! is indeed 4 .

i hope you will crack 1000! soon!

Commented by RasheedSoomro last updated on 09/Jun/17

Please help me in finding error in answer of Q#13724. In one of its answers I have used same logic. Bu the answer is wrong! Mr prakash jain & mrW1 Do you know any standard method of such questions? If you do please do share.

You can't use 'macro parameter character #' in math mode

In one of its answers I have used same logic .

Bu the answer is wrong!

Mr prakash jain & mrW 1

Do you know any standard method of such questions ?

If you do please do share .

Commented by mrW1 last updated on 09/Jun/17

I definitively don′t know any standard methods for solving such questions. I am learning, also from posts such like yours.

I definitively {don}^{'} t know any standard

methods for solving such questions .

I am learning, also from posts such

like yours .

Commented by prakash jain last updated on 10/Jun/17

Actually I don′t know a short cut method to this question. I posted this question to get some ideas.

Actually I {don}^{'} t know a short cut

method to this question .

I posted this question to get some

ideas .

Commented by RasheedSoomro last updated on 10/Jun/17

T h a n k s of you both!

Answered by RasheedSoomro last updated on 11/Jun/17

Easy way 100! has 24 0′s from right. p=((100!)/(10^(24) )) has no zeo from right ∴ first non-zero digit in 100! from right = unit digit in p ∴ p≡d(mod 10) p=((100!)/(10^(24) ))=((100!)/(2^(24) ×5^(24) ))=((100!)/(2^6 (5.10....100))) [18 2′s from 2^(24) have come in 5.10....100] =(1/2^6 ){(1.2.3.4)(6.7.8.9)...(96.97.98.99} p≡d(mod 10)⇒ (1/2^6 ){(1.2.3.4)(6.7.8.9)...(96.97.98.99}≡d(mod 10) (1/2^6 )(4)^(20) ≡d(mod 10) [∵ Each bracket≡4(mod 10) Total brackets=20] (2^(40) /2^6 )≡d(mod 10) d≡2^(34) (mod 10) Verify that: 2^4 ≡6(mod 10 (2^4 )^8 ≡6^8 ≡6(mod 10) 2^(32) .2^2 ≡6.2^2 ≡4(mod 10) 2^(34) ≡4(mod 10) Hence d≡2^(34) ≡4(mod 10) d≡4(mod 10) Wrong logic! Accedently correct answer! I′ll try to correct my logic. Please stay with me!

Easy way

100! has 24 0^{'} s from right .

p = \frac{100!}{10^{24}} has no zeo from right

∴ first non - zero digit in 100! from right

= unit digit in p

∴ p \equiv d (\mod 10)

p = \frac{100!}{10^{24}} = \frac{100!}{2^{24} \times 5^{24}} = \frac{100!}{2^{6} (5.10 \dots . 100)}

[18 2^{'} s from 2^{24} have come in 5.10 \dots . 100]

= \frac{1}{2^{6}} {(1.2.3.4) (6.7.8.9) \dots (96.97.98.99}

p \equiv d (\mod 10) \Rightarrow

\frac{1}{2^{6}} {(1.2.3.4) (6.7.8.9) \dots (96.97.98.99} \equiv d (\mod 10)

\frac{1}{2^{6}} {(4)}^{20} \equiv d (\mod 10) [∵ Each bracket \equiv 4 (\mod 10) Total brackets = 20]

\frac{2^{40}}{2^{6}} \equiv d (\mod 10)

d \equiv 2^{34} (\mod 10)

Verify that :

2^{4} \equiv 6 (\mod 10

{(2^{4})}^{8} \equiv 6^{8} \equiv 6 (\mod 10)

2^{32} . 2^{2} \equiv 6 . 2^{2} \equiv 4 (\mod 10)

2^{34} \equiv 4 (\mod 10)

Hence

d \equiv 2^{34} \equiv 4 (\mod 10)

d \equiv 4 (\mod 10)

\boldsymbol Wrong \boldsymbol logic! \boldsymbol Accedently \boldsymbol correct \boldsymbol answer!

\boldsymbol I^{'} \boldsymbol ll \boldsymbol try \boldsymbol to \boldsymbol correct \boldsymbol my \boldsymbol logic .

\boldsymbol Please \boldsymbol stay \boldsymbol with \boldsymbol me!

Answered by RasheedSoomro last updated on 11/Jun/17

According to Corrected Logic. Right logic & Right answer Expansion of 100! have 24 0′s in the right. Let p=((100!)/(10^(24) )) p has no zero from right. First non-zero digit in 100! =unit digit in p d≡p(mod 10) p=((100!)/(10^(24) ))=((100!)/(2^(24) ×5^(24) )) =((100!)/(2^(24) ×(5×((10)/2)×((15)/3)×((20)/2^2 )×5^2 ×...×((100)/2^2 )))) =((100!)/(2^(24) ×(((5.10.15.....100)/(2^(18) ×3^8 ×7^2 ×11×13×17×19))))) =3^8 ×7^2 ×11×13×17×19×((100!)/(2^6 (5.10....100))) =3^8 ×7^2 ×11×13×17×19{(((1...4)/2))(6...9)(11...14)(((16....19)/2))} ×...×{(81....84)(86...89)(91...94)(((96...99)/2))} Number of red bracket= 1 Number of black brackets= 14 3 per each 20 except first. Number of blue brackets= 5 1 per each 20 We can verify/prove the following congruences 3^8 ≡1(mod 10) 7^2 ≡9(mod 10) 11≡1(mod 10) 13≡3(mod 10) 17≡7(mod 10) 19≡9(mod 10) ((1.2.3.4)/2)≡2 Any of Black-bracket≡4(mod 10) [14 congruences] Any of Blue-bracket≡2(mod 10) [ 5 congruences] Multiplying all above congruences p≡1.9.1.3.7.9.2.4^(14) .2^5 (mod 10) p≡1.9.1.3.7.9.2.6.2(mod 10) [∵4^(even) ≡6(mod 10] p≡40824≡4(mod 10) [∵4^(even) ≡6(mod 10]

According to Corrected Logic .

Right logic & Right answer

E xpansion of 100! have 24 0^{'} s in the right .

Let p = \frac{100!}{10^{24}}

p has no zero from right .

First non - zero digit in 100!

= unit digit in p

d \equiv p (\mod 10)

p = \frac{100!}{10^{24}} = \frac{100!}{2^{24} \times 5^{24}}

= \frac{100!}{2^{24} \times (5 \times \frac{10}{2} \times \frac{15}{3} \times \frac{20}{2^{2}} \times 5^{2} \times \dots \times \frac{100}{2^{2}})}

= \frac{100!}{2^{24} \times (\frac{5.10.15 \dots . . 100}{2^{18} \times 3^{8} \times 7^{2} \times 11 \times 13 \times 17 \times 19})}

= 3^{8} \times 7^{2} \times 11 \times 13 \times 17 \times 19 \times \frac{100!}{2^{6} (5.10 \dots . 100)}

= 3^{8} \times 7^{2} \times 11 \times 13 \times 17 \times 19 {(\frac{1 \dots 4}{2}) (6 \dots 9) (11 \dots 14) (\frac{16 \dots . 19}{2})}

\times \dots \times {(81 \dots . 84) (86 \dots 89) (91 \dots 94) (\frac{96 \dots 99}{2})}

Number of red bracket = 1

Number of black brackets = 14 3 per each 20 except first .

Number of blue brackets = 5 1 per each 20

We can verify / prove the following congruences

3^{8} \equiv 1 (\mod 10)

7^{2} \equiv 9 (\mod 10)

11 \equiv 1 (\mod 10)

13 \equiv 3 (\mod 10)

17 \equiv 7 (\mod 10)

19 \equiv 9 (\mod 10)

\frac{1.2.3.4}{2} \equiv 2

Any of \boldsymbol Black - \boldsymbol bracket \equiv 4 (\mod 10) [14 congruences]

Any of \boldsymbol Blue - \boldsymbol bracket \equiv 2 (\mod 10) [5 congruences]

Multiplying all above congruences

p \equiv 1.9.1.3.7.9.2 . 4^{14} . 2^{5} (\mod 10)

p \equiv 1.9.1.3.7.9.2.6.2 (\mod 10) [∵ 4^{even} \equiv 6 (\mod 10]

p \equiv 40824 \equiv 4 (\mod 10) [∵ 4^{even} \equiv 6 (\mod 10]

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Light-version-of-Q-13724-Expansion-of-100-has-24-0-s-at-the-end-Find-the-first-non-zero-digit-from-right-100-d000-00-What-is-the-value-of-d-

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