How many zeros are at the end of (100!)(200!)(300!)(400!) when multiplied out?
Interesting question.....
(100!)(200!)(300!)(400!) = 1442479056324507163768893362866374541125068265912031567020824850448396469814026991181151652398154401541353464981477804509478856204278416549336628890146780976281683298427997873623746540442050679206085277326775360168386468184548624892343793368862825704818174747645075063532813500322891615066320436567447212842607606293109245006405295449732628272986750937282771864100385199443420968795575293569180503401121014435681414313275736191855467863035962032568894421715621146306564676422680709900518687334479973626949500873235740316649255777793329076639420929731916483051908521236671467390110069785923169501164235461529586345929472222503947655023289076765675238170904925218929225801890448324145753637944201802244045417264004007204583067972725036082931233426080724851974424262593295013905463972161399227793721153609694179457044777898025008952275073740256060905440111916106718592627044497725728405361593568500681751058962740871337127284659955038182836051496929123260838351489482906412906032943425567260090866970031170258315250587356162840155331390366390698079916889440864984562288324437179582205537787971988525395021688799056671002702545340093275531313526186216009546381977210100661240469406105064828245906873687558347833624728922353218924863303126031244140866164960244762850115007099281081081244333509123163125813006827832203023291440985894586661357908836055254246045246323827232970169430354238034489964080952862039618928520262540129431155766740472566502657750873133642586952453162349781265017638816538525942642507929766075639108412013121787890140349483123712622866947061772553669141079633818944714793149907930746924876619393371934377643785232381422468384288914016804204053492139104721578098319855724550051050227046736683546014057180011576857934173338161252656855583817970457346310144000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
That's a total of 246 zeros!!!
To have a trailing zero, you need a 5 and 2. Note that there will always be more 2s than 5s, the number of trailing zeros is basically the number of factors of 5.
Let's count the number of 5s in each term separately, then add them up.
100!: 20 (every multiple of 5) + 4 (every multiple of 25) = 24
200!: 40 (every multiple of 5) + 8 (every multiple of 25) + 1( every multiple of 125) = 49
300!: 60 (every multiple of 5) + 12 (every multiple of 25) + 2(every multiple of 125) = 74
400!: 80 (every multiple of 5) + 16 (every multiple of 25) + 3 (every multiple of 125) = 99
So, the total number of trailing zeros is: 24+49+74+99=246, just as Vin found