# How many 0's are in the end of this expansion?

How many $0's$ are in the end of:

$$1^1 \cdot 2^2 \cdot 3^3 \cdot 4^4.... 99^{99}$$

The answer is supposed to be $1100$ but I have absolutely NO clue how to get there.

• No. of zeroes will be equal to no. of $(5*2)$ in the above product. Also the power of $5$ will be less than the power of $2$ in the above expression(why?). So count the power of $5$ to arrive at the answer. :) Commented Feb 1, 2015 at 9:26
• Find the exponent of $10=2\cdot 5$ in this.
– AvZ
Commented Feb 1, 2015 at 9:29

To find out the number of zeroes at the end of that huge number, we just have to find out the exponent of $10$ in it.
Since $10=2\cdot 5$, we can just find out the exponent of $5$ in that number as the exponent of $2$ will be more than $5$.
All the numbers which have $5$ as their factor will be counted.
These numbers would be
$$5^5,10^{10},15^{15},\ldots,95^{95}$$
We know that all numbers except $25,50$ and $75$ will only contribute themselves as the exponent, whereas the aforementioned numbers, will contribute double of themselves.
So, we have
$$(5+10+\cdots+95)+(75+50+25)=\frac{19}{2}(5+95) + 150=1100$$

• +1. great Answer. But why only the exponent of $5$? Commented Feb 1, 2015 at 10:07
• We know that $10=2\cdot 5$. Also, for the product of these consecutive integers, there will be more even numbers than numbers divisible by $5$. Since $10$ is just basically pair of $2$ and $5$, there will be as many pairs as there are $5$.
– AvZ
Commented Feb 1, 2015 at 10:11
• Okay mmmm.. But if $10 = 2 \cdot 5$ then why cant we just use the exponent of $2$? Commented Feb 1, 2015 at 10:16
• Since there will be more $2$'s than $5$'s, the number of pairs will be equal to exponent of $5$. There will be some extra $2$'s which won't pair with anything.
– AvZ
Commented Feb 1, 2015 at 10:19
• To be sure of the answer, you should look at the number of factors of $2$ as well, but you don't have to look at them as carefully as the factors of $5$. As soon as you have found at least $1100$ factors of $2$ you can stop looking. Consider that there are $25$ even numbers in the range $50$ to $98$, and each contributes at least $50$ factors of $2$ to the product; that's $1250$ factors of $2$ already, so we're done. Commented Feb 1, 2015 at 15:38

There as many zeros as the number of the factors of 5 in the product: So we need to check the subproduct $$5^5\cdot 10^{10}\cdots 95^{95}$$ We have $5+10+15+\cdot+95=5(1+2+\cdots+19)=950$. And we have missed the case where we have two factors: $25,50,75$. So we have another $25+50+75=150$. Altogether $950+150=1100$.

• Why only number factors of 5? Commented Feb 1, 2015 at 9:55

The product is:

34553708533451189908388022423131226066233897781220888849903704212716691939238825645547711844418578690264583160130364150148833934897849726067772667131757088169872065619573621747233081114266125668525439526114504046584546639711711419992082916786355734843922298042692869987065705784829116473956530224194054061176047316074878083304704158540961227513574264554396382009198311904248670619893968827404470150033043852806874424572991439981597499280455615791403656640275929040556246647315911465550172523603238889034707801049458647782411192166268473767456816618347208926066509123852968425735479724877682799502230521659625771165864718510523569706468848664347685514105514883210642177764058825339805906895198930450007161022194799856677606014031096736032048966068191983313754191254531965089117369945463617841849193737968749188091204647671055434075889709231996667882499520868964602171064657494524859014135647581869664951789711022535686559811740615619682514219383744857570427269757531851073720831062640484299119796851971928396861185600073702280665490758760504555480528997066265261887832490026305314551644618637336459403051693795911068979447192082247542899780738178839502397036866970262632149872951406195249895070842271742809442879449015947381559538136709141451502251501217977086892951126520767573790438049580889108500562782749167708753169126765737228483327598143839454413138733502770488014211155342861905065014948032004516441371887246301904849453328363626029414435282400038326957397945045478806066852009796803048988734212719990328059456491537039512073111232019019233547758096007791109062690249696371085938368959164219590984833513033356451879493847403656495610551471411224599265059711271279207475122416482174196254517532972465464720893916847156361710641746035756874759134958809948348316339144976101922617403818050193571383203910685900832796181901759131437735663358070039012271968552739735236758819198946547646633936103377953059455751171143594811790344564729849496515305179896622108666266186579236854852057754888619887882818671372096778363931976566487950032722092865346307048044729379405388633543517078984999498821198351846375559881560541168725321280328066976438818952643356138049696435822603579787111269750240477056325776852012806387017087191979532171561093823187998612766384295611226684466326137672157698099521428014267421855917719636679974561672864388065803437951940459562105939991923814979583721897776999074236726246144999907627772928730286265500701367733597866720224180946374355544258372787974832301439673375349168036488394037166326719463512726312260185907820752691873674354948469310478774241736020973914155825825992084490790774068046152945680618966351882248015062161340544995059441721495435052184468790531312646898990036585048929448125655397054389699283516266069099793462943724795794105311183100268701298417781343661553177165529411699183594173284868850518980973319000385685902744155931045266999610831464195496273989425510393903517647346010536278295572987817525511192374610018605250021874457070711560148245645617186228799475685018235178800520273561026991329082970358239827664796950471351221206715627699154861579748833331656301060296043123421537092931093917978797816315533360860136220502469777083357928930703462039552556283603956245924590861303103647127141330056491272885090654805058438358281731820755398883482582511585690278983535863406833529478627111284867696092400332344538700703513731227628101394132500806371957784463146972689587844424673814402724314980018706795563176802080484505189823502839957622692988657721745103305437491759336467027278516685624453703277260266079713916690222978686151266194561198489750147083193124234858851728799333793531316444969377915108848689609466382623585571104693209338918373797922252030960369431007826551516651225714156576659828085196027473383052288520182275665573641224166677564615276054810142712947301873911500626906011865467515924825476948411450361439217284121921770460160221015575323624324480759619580612856651168284434093553918085678962494026587186817223959344977237039289497216097790170019665191109452741326012220351889145251830649820436696846343051960416410908131474879921477176776502421255096817434940768142894421773185648314932336576890317905314341486855330401538488888175571366088123955227454591080003222252998985584186521628570075820485849624936885828604897151383311874990515741479826837783380494197312577232602499828578359919949575593272535135825856411659690008802313196153385795744740195429333738766644873756984486540800284644453289096739943172992848235144873426708905747613070603307512521312595962127928005768886511205536410392804689241010738433254820516568957684009976378947510939736222858928033552922156391734785932669237143821740276902254166341239081596164538931905585806100807652976534344830234540078149058035628666656321324797652385305812931572709314054362754013617818669757267580268423903604626321924757517247087721729640599793734894006091791408645463930973376079083755431041387064792692348280790928326105097785917187040595407191509405086827280771516828620597657669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• Looks like Python to me...?
– AvZ
Commented Feb 1, 2015 at 17:46
• @AvZ Well, in this case Mathematica, but python works too. Commented Feb 1, 2015 at 17:59
• So how will you determine the number of zeroes in the end? Factorize? Commented Feb 1, 2015 at 19:32
• @JeppeStigNielsen just copy the terminating string of zeroes to a text editor, go to the end of line and see what column you're at :) Commented Feb 1, 2015 at 19:38
• @JeppeStigNielsen Or copy into an interpreter for the scripting language of your choice, surround in quotes, and take the length of the string! Commented Feb 2, 2015 at 8:35