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CAT|ByAshankDubey(100%ilerofCAT15)
TESTCRACKERCATTOPPER’sPROGRAM:QUANT
Factorial
Theproductofnconsecutivenaturalnumbersstartingfrom1toniscalledasthefactorial‘n’.
n!=1x2x3x4x5x6x7x……x(n–2)x(n–1)xn
e.g.5!=1x2x3x4x5=120;6!=1x2x3x4x5x6=720
F 0!=1and1!=1
F n!alwaysendswithzeroifn≥5
F Theproductofnconsecutivenaturalnumbersisalwaysdivisiblebyn!,wheren!=1×2×3×4×
5….×n
1. Given𝒇 𝒙 = 𝒙 ×𝒇 𝒙 − 𝟏 foranynaturalnumber′𝒙′:if 𝒇(𝒙 + 𝟐) = 𝟐𝟎𝒇 𝒙 ,thenwhatisthevalue
of𝒙?
(a)4 (b)5 (c)2 (d)Noneofthese
Solution:Asweknow,n!=n(n–1)!.Thegivenfunctioncanbeassumedtobethefactorialfunction.
𝑓 𝑥 = 𝑥 ×𝑓 𝑥 − 1 𝑓 𝑥 + 2 = (𝑥 + 2) ×𝑓 𝑥 + 1 𝑓 𝑥 + 2 = (𝑥 + 2) (𝑥 + 1)×𝑓 𝑥 ………….(1)
𝑓(𝑥 + 2) = 20𝑓 𝑥 {Giveninthequestion,comparethisequationwithequation(1)}
(𝑥 + 2) 𝑥 + 1 =20⇒ (𝑥 + 2) 𝑥 + 1 =5x4⇒𝑥=3.Thereforetheansweris(d).
2. Ifnisanoddnaturalnumber,whatisthehighestnumberthatalwaysdividesn×(n2–1)?
(a)6 (b)12 (c)24 (d)Noneofthese
2
Solution:n×(n –1)=(n–1)×n×(n+1),whichisaproductofthreeconsecutivenumbers.Sincenis
odd,thenumbers(n–1)and(n+1)arebotheven.Astheyaretwoconsecutiveevennumbersoneof
these numbers will be a multiple of 2 and the other will be a multiple of 4. Hence, their product is a
multiple of 8. Since one out of every three consecutive numbers is a multiple of 3, one of the three
numberswillbeamultipleofthree.Hence,theproductofthreenumberswillbeamultipleof8×3=
24.
Hence,thehighestnumberthatalwaysdividesn×(n2–1)is24.
3. Findthesumofallvaluesofnwhichsatisfy
𝒏! 𝟐
𝟒!
+
𝟕! × 𝟓!
𝟒 × 𝟒!
=240
𝒏!
𝟒!
Solution:Letn!=k.Taking4!offbothsidesandsimplifying,weget
k2+(7!x5!)(4!/4)=240x(4!)k
⇒k2–5!x2x4!k+7!x5!x3!
⇒k2–5!x48k+7!x5!x6
⇒k2–5!x48k+7!x6!
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⇒k2–5!x(6+42)k+7!x6!
[Attention:Whyarewesplitting48into6+42above?Becausewewanttheabovequadraticequationto
beoftheformx2-(a+b)x+ab]
⇒k2–5!x6k–5!x6x7k+7!x6!
⇒k2–6!k–7!k+7!x6!
⇒k2–(6!+7!)k+7!x6!
⇒k=6!and7!⇒n=6and7
⇒Sum=13
4. Whatistheremainderif(2n)!isdividedby(n!)2?
(a)0 (b)2 (c)4 (d)1
Solution:(2n)!=1×2×3×4×…×(n–1)×n×(n+1)×…×2n=(n)!×(n+1)×(n+2)×…×2n.Since(n
+1)×(n+2)×…×2nisaproductofnconsecutivenumbers,itisdivisiblebyn!.Hence,theproduct(n)!
×(n+1)×(n+2)×…×2nisdivisiblebyn!×n!=(n!)2.Theremainderthereforeis0.
Type1:HighestPowerofaNumberinaFactorial
Supposewehavetofindthehighestpowerofkthatcanexactlydividen!,wedividenbyK,nbyK ! ,n
!
byK ! …..andsoontillweget
tox)andthenaddupas
!
!
+
!!
!
!!
equalto1(where,[x]meansthegreatestintegerlessthanorequal
+
!
!!
+……..+
!
!!
Examples
1. Whatisthelargestpowerof5in121!?
Solution:
!"!
!
+
`"#"
!!
=24+4=28.(Wecannotdoitfurthersince121islessthan5! =125)
Hence,5isinvolved28timesasafactorin121!
2. Whatisthelargestpowerof2thatcandivide269!?
Solution:
!"#
!
+
!"#
!!
+
!"#
!!
+
!"#
!!
+
!"#
!!
+
!"#
!!
+
!"#
!!
+
!"#
!!
=134+67+33+16+8+4+2+1=265.
Thusthegreatestpowerof2is265thatcandivideexactly268!.
TestCrackerTrick:Now,youmusthavenoticedthatweconsideronlyintegralvaluesofthequotient
as[q]whichgivesanintegerjustlessthanorequaltoq.Sowecandothiscalculationinaslightly
easiermanner.Herewejustdividethegivennumberandthensucceedingquotientswillbedivided
by the same remainder as in case of successive division. So, we solve previous problem in this
manner.
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CAT|ByAshankDubey(100%ilerofCAT15)
Sowecanseethatthismethodofcalculationiseasiersinceweneednotknowthevaluesof2! ,2! ,2! ,
……,2! etc.Alsothedivisionby2iseasierthanthedivisionby2 ! ,where𝑥isanylargeinteger.
3. Whatisthelargestpowerof7thatcanexactlydivide780!?
Solution:
Thusthehighestpowerof7is128bywhich777!canbedivided.
4. Whatisthehighestpowerof72in100!?
Solution: Since our method is applicable only for the prime factors, we solve it by breaking 72 in its
primefactors.
72=2! × 3! .Therefore,weneedtofindthehighestpowerof2! and 3! in72!.
72=8x9.Sincetomakea8weneed2x2x2i.e.threetimes.Soweneedtodivide97by3andget32.
And,for9weneed3x3i.e.twotimes.Soweneedtodivide48by2andget24.
So
2!" x3!" ⇒8!" x9!" Thusthelargestpowerof72is24.Thatcanexactlydivide100!
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CAT|ByAshankDubey(100%ilerofCAT15)
5. Whatisthehighestpowerof81thatcandivide1802!?
Solution:81=3! So,
Thusweget3 .Butweneedtofindoutfor 3! .
Therefore3!"# = 3! !!" x 3! = 81 !!" x3
Thereforethelargestpowerof81is224whichcandivideexactly1802!
!"#
Type2:Numberofzeroesattheendoftheproductofnumbers
Weknowthat10=2x5,100=2! x5! ……etc.Sowecansaythat"𝑛"numberofzeroesattheendofthe
productweneedexactly"𝑛"combinationsof“5x2”
F Thenumberofzeroesattheendoftheproductdependsupon2x5,buttheconditionisthat
(a) 𝟐𝒙 x𝟓𝒚 gives𝒙numberofzeroesif𝒙<y
(b) 𝟐𝒙 x𝟓𝒚 gives𝒚numberofzeroesify<𝒙
Examples
1. Whatisthenumberofzeroesattheendof10!?
Solution:10!=1x2x3x4x5x6x7x8x9x10.
Itisobviousfromtheaboveexpressionthatthereareonlytwo5’sandeight2’s.Sincethenumberof5’s
arelessthannumberof2’s,thenumberof5’swillbeeffectivetoformthecombinationof‘2x5’.Thus
thereareonly2zeroesattheendof10!
2. Findthenumberofzeroesattheendof100!?
(a)24 (b)97 (d)23 (d)Noneofthese
Solution:Sinceweknowthatthezeroesattheendofanyproductareduetothepresenceof10asthe
factoroftheproductandthenumberofzeroesdependsuponthenumberoftimes10isinvolved.
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CAT|ByAshankDubey(100%ilerofCAT15)
Sothereare24combinationsof2x5:itmeanstherewillbe24zeroesattheendof100!
(Needtocheckfor5only,sinceithassmallerpower.)
3. Findthenumberofzeroesattheendoftheproductoffirst100multiplesof10.
(a)1000
(b)24 (b)100 (c)124
Solution:(10x20x30x40x……………………..x1000)
=10!"" [1x2x3x4x…………………x100]
=10!"" [100!]
Asweknowfrompreviousexample,highestpowerof5in100!is24.
Numberofzeroes=100+24=124.
4. Howmanynaturalnumbers‘n’arethere,suchthat‘n!’endswithexactly30zeroes?
(a)0 (b)1 (c)3 (d)4
Solution:
Accordingtoquestion,n!shouldhave30zeroesintheend.Ifn=100(takingrandomlykeepinginview
30isthenumberofzeroes]
100!has
!""
!
+
!""
!!
=24,Sonshouldbegreaterthan100.Nextmultipleof5is105.But105=5x21,
hasonlyoneextra5.Numberofzeroeswillincreaseby1only.Similarly110,115and120alsohaveone
extra5.Numberofzeroes(from120!to124!)=28.Now,thenextmultipleof5is125and125contains
three5’s.So,numberofzeroeswillincreaseby3.Numberofzeroesin125!=28+3=31.So,thereisno
factorialofanumberwhichendswith30zeroes.
5. n!hasxnumberofzeroesattheendand(n+1)!has(x+3)zeroesattheend.1≤n≤1000.Howmany
solutionsarepossiblefor‘n’?
(a)8 (b)7 (c)1 (d)4
Solution: We can see that increasing the natural number by 1, we are gathering 3 more powers of 5.
Therefore, (n + 1) is a multiple of 125 but not a multiple of 625 as it would result in 4 powers of 5.
Therefore,(n+1)willbeequaltoallthemultiplesof125minus625.
Totalnumberofmultiplesof125lessthan1000=8
Totalnumberofmultiplesof625lessthan1000=1
Therequiredansweris(8–1)=7
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CAT|ByAshankDubey(100%ilerofCAT15)
6. Ifn!and(4n)!endwith25zeroesand106zeroesrespectively,thenwhichofthefollowingisafactor
ofn?
(a)25 (b)26 (c)27 (d)28
Solution:
Step1:
Accordingtothequestionn!has25zeroesattheend.
Hereyouhavetotryabittoknowtherangeofn.Ifn=100(takingrandomlykeepinginview26isthe
numberofzeroes]
100!has
!""
!
+
!""
!!
=24,Sonshouldbegreaterthan100.
Next multiple of 5 is 105. But 105 = 5 x 21, has only one 5. Number of zeroes will increase by 1 only.
From105to109,thereisnomultipleof5.So,therangeofnisinbetween105to109.
Step2:
Asthenextconditionis(4n)!has106zeroesintheend.Forgettingtheactualvaluejustverifythevalue
ofnwithintherange{105to109}.Youwillgetn=108.108=4x27.Therequiredanswer=27.
7. If146!Isdivisibleby𝟔𝒏 thenfindthemaximumvalueofn.
(a)74 (b)70 (c)76 (d)75
Solution:Initscurrentformthequestionlooksintimidating.Butisit?
Wesimplyneedtofindthehighestpowerof6containedin146!
6=2x3.As3isthelargestprimefactorof6,wewillfindthehighestpowerof3in146!
Thehighestpowerof3in146!=48+16+5+1=70
8. Thenumber2006!iswritteninbase22.Howmanyzeroesarethereattheend?
(a)450 (b)500 (c)199 (d)200
Solution:Beforewecometotheexactquestion,letusquicklygetafeelofhowtochangebasesquickly
TestCrackerTrick:Anumberwritteninbase10canbeconvertedtoanybasebbyfirstdividingthe
numberbybandthendividingthesuccessivequotientsbyb.Theremainderthusobtained,written
inreverseorder,giventheequivalentnumberinbaseb.
Let’sconvert 63 !" tobase7
63
=
120
!"
!
There is exactly one zero at the end of the number converted to base 7 because the number 63 is
divisibleonceby7i.e.thehighestpowerof7containedinthenumber63isexactlyequalto1.63=3! 7! [email protected]+919035001996
CAT|ByAshankDubey(100%ilerofCAT15)
Similarly, if we convert 56 to base 2 then there will be exactly three zeroes at the end of converted
number in base 2 because the highest power of 2 contained in 56 is exactly equal to three i.e. the
number56isdivisiblethriceby2.
Comingbacktoourquestion:Here,thenumberofzeroespresentattheendof2006!inbase22willbe
equal to the number of times 22 divides 2006! completely. Therefore, we need to find the highest
powerof22containedin2006!
22=2x11.As11isthelargestprimefactorof22.Wewillfindthehighestpowerof11containedin
2006!
Therefore,ouranswerhastobe[2006/11]+[2006/112]+[2006/113]=199
9. 30!whenexpressedinbase12endswithkzeroes.Findk
(a)12 (b)13 (c)14 (d)Noneofthese
!
Solution:12=2 x3
Thehighestpowerof2! in30!=
!"!!!!!!
!
=
!"
!
=13
Thehighestpowerof3in30!=10+3+1=14
Asthehighestpowerof2! islessthanthatof3,thehighestpowerof12in100!Isequaltothatof2! .
Hence,k=13.
Type3:FactorialOfaNumberContainingHighestPower(DifferentBase)
Let,z(n)bethenumberofzeroespresentattheendofn!inbaseb.Therefore,thehighestpowerofb
contained in n! is z(n). The base b = 2! x 3! x 5! x…… where a, b, c, …denote the power of prime
numbers contained in base b. Let p be the largest prime factor of base b. So, the highest power of p
containedinn!isz(n).
z(n)=
!
!!
!
+
z(n)= 𝑛 +
!
z(n)xp=𝑛 +
!
!!
!
+
!!
!
!!
!
!!
+
+
+
!
!!
!
!!
!
!!
+
+
+…….+
!
!!
!
!!
+
+
!
!!
!
!!
!
!!
+ …….+
+ …….+
!
!(!!!)
!
!(!!!)
(approximately)
(approximately)
z(n)xp–n=z(n)
n=(p–1)xz(n)
TestCrackerTrick:
Foragivenz(n),thesmallestvalueofanaturalnumbernsuchthattherearez(n)zeroespresentat
theendofn!inbasebcanbecalculatedapproximatelyasn=(p–1)xz(n),wherez(n)denotesthe
numberofzeroespresentattheendofn!inbasebandpisthelargestprimefactorofbaseb.
10. Findtheleastvalueofnsuchthatn!hasexactly2394zeroes
Solution:Here,z(n)=2394.n=(5-1)x2394=9576(approximately)
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CAT|ByAshankDubey(100%ilerofCAT15)
11.
12.
13.
14.
[Herethelargestprimefactorofthebase(10)is5]
Let’scountthenumberofzeroespresentattheendof9576!=1915+383+76+15+3=2392
To get more zeroes, we will have to find the second least multiple of five greater than 9576 and it is
9585.Thenumberofzeroespresentattheendof9585!=1917+383+76+15+3=2394.Hence,n=
9585.
Note:Increaseinnumberofzeroesdependsonwhethergivennumberismultipleof5,25,125,625……
Findthesmallestpositiveintegernsuchthatn!isamultipleof𝟏𝟎𝟐𝟎𝟎𝟗 .
Solution:Itmeansthatfindthesmallestnaturalnumber“n”suchthatn!hasatleast2009zeroesatthe
end.n=(5-1)x2009=8036(approximately)
Thenumberofzeroespresentattheendof8036!=1607+321+64+12+2=2006
Here,numberofzeroesarethreelessthanrequired.So,wehavetolookforthirdmultipleof5greater
than8036.
Therequiredmultipleof5is8050.Thenumberofzeroespresentattheendof8050!=1610+322+64
+12+2=2010.Therefore,8050!has2010trailingzeroes.Hence,8050!istheleastmultipleof10!""# n!has100trailingzeroesinbase35.Findtheminimumvalueofn.
Solution:Here,Base=35=5x7.As7isthelargestprimefactorof35.z(n)=100.
n=(p–1)xz(n)=6x100=600(approximately)
Let’scountthenumberofzeroespresentattheendof600!inbase35.
Thenumberofzeroespresentattheendof600!Inbase35=Numberoffactorsof7in600!
=85+12+1=98
Togettwomorezeroes,wewillhavetofindthesecondleastmultipleof7greaterthan600.
600=85x7+5.Therefore,therequiredmultipleof7is{600+(7–5)+7}=609.
What is the smallest n such that n! ends in exactly 91 zeroes in base 91
Solution:Here,Base=91=7x13.As13isthelargestprimefactorof91.z(n)=91.
n=(p–1)xz(n)=12x91=1092(approximately)
Let’scountthenumberofzeroespresentattheendof1092!inbase91.
Thenumberofzeroespresentatendof1092!Inbase91=84+6=90
Togetmorezeroeswewillhavetofindtheleastmultipleof13greaterthan1092.1092=84x13
Therefore,thenextmultipleof13is1092+(13)=1105.
The number of zeroes present at the end of 1105! In base 91 = 85 + 6 = 91
Hence,n=1105.
Find the number of possible values of n = 125 x m if n is less than or equal to 1000 and m is not
divisibleby5.
Solution:n=5! xm,wheremisnotdivisibleby5i.e.thehighestpowerof5inthenumbernisexactly
equaltothree.
Thehighestpowerof5containedin1000!=200+40+8+1.
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CAT|ByAshankDubey(100%ilerofCAT15)
8 and 1 are representing here number of numbers less than 1000 and divisible by 5! and 5! respectively.
Hence,therequiredanswer=(8–1)=7.
[Sincemisnotamultipleof5,weknowthatnisnotdivisibleby54]
15. Thequotientobtainedbydividingnby121andthenumber11areco–primetoeachother.Findthe
numberofpossiblevaluesofnifnislessthanorequalto10,000.
Solution: It means that the highest power of 11 contained in the number n is exactly equal to two.
Thehighestpowerof11containedin10,000!=909+82+7
82–7=75.
Hence,thenumberofpossiblevaluesofnis75.
16. n!has𝒙numberofzeroesattheendand(n+1)!has(𝒙+3)zeroesattheend.Findthenumberof
possiblevaluesofnifnisathreedigitnumber.
Solution:Thehighestpowerof5inn!is𝑥.Howmanyzeroeswe’llgetattheendof(n+1)!dependson
whether(n+1)isamultipleof5! ,5! , 5! ,5! ,……andsoon.Here(n+1)!has(𝑥+3)zeroesattheend.
Thereforethehighestpowerof5inthenumber(n+1)isexactlyequalto3.
5! x1=125
5! x2=250
5! x3=375
5! x4=500
5! x6=750
5! x7=875
5! x8=1000
Thehighestpowerof5in5! x5is4,that’swhy,weignoreit.Hence,thepossiblevaluesofn=7.
[Note:1000isn+1,thereforen,whichis999willbeathreedigitnumber]
Type4:LastNon–Zerodigit/digitsinafactorial
Findthelastnon–zerodigitofn!:Lastnon–zerodigitof(tenconsecutiveintegers)
Let’stakea10–tupleoftenconsecutiveintegersstartingfrom(10a+1)to(10a+10).Lastnon–zero
digitoftheirproductcanbewrittenas:
=(10a+1)(10a+2)(10a+3)(10a+4)(10a+5)(10a+6)(10a+7)(10a+8)(10a+9)(10a+10)
=(10a+1)(10a+2)(10a+3)(10a+4)5(2a+1)(10a+6)(10a+7)(10a+8)(10a+9)5(2a+2)
=(10a+1)(10a+2)(10a+3)2(5a+2)5(2a+1)2(5a+3)(10a+7)(10a+8)(10a+9)5(2a+2)
=100(10a+1)(10a+2)(10a+3)(5a+2)(2a+1)(5a+3)(10a+7)(10a+8)(10a+9)(2a+2)
=100(2a+1)(2a+2)(10a+1)(10a+2)(10a+3)(5a+2)(5a+3)(10a+7)(10a+8)(10a+9)
{Now,wecanremovebothzeroes}
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CAT|ByAshankDubey(100%ilerofCAT15)
=Lastnon–zerodigitof(2a+1)(2a+2)(1)(2)(3)(7)(8)(9)(5a+2)(5a+3)
=Lastnon–zerodigitof4{25a(a+1)+6}(2a+1)(2a+2)
=4xLastnon–zerodigitof(2a+1)(2a+2)
Examples
1. Whatisthelastnon–zerodigitof20!
Solution:Usingabovepropertywecanwrite:
1x2x3x4x5x6x7x8x9x10=4x remainder
! ×!
!"
[Here,a=0]
11x12x13x14x15x16x17x18x19x20=4x remainder
!
!×!
!"
[Here,a=1]
Solastnon–zerodigitof20!=4 xlastnon–zerodigitof4!=lastnon–zerodigitof6x4=4
TestCrackerTrick:
F LNZ(10n!)=LNZ(𝟒𝒏 )xLNZ(2n!)
2. Whatisthelastnon–zerodigitof100!
Solution:Lastnon–zerodigitof100!=LNZ(10x10!)=LNZ(4!" )xLNZ(20!)
LNZof410isthesameasLNZof42,whichis6
Fromtheaboveexample,weknowlastnon–zerodigitof20!,whichis4.LNZ(100!)=LNZ(6x4)=4
3. Findthelastnon–zerodigitof125!
Solution:125isnotamultipleof10;hencewecanapplytheabovesamemethodupto120!andthen
multiplytheresultwithlastnon–zerodigitoftheproductofremainingnumbersi.e.121to125.
So,thelastnon–zerodigitof125!=LNZ(125!)=LNZ(4!" )xLNZ(24!)xLNZ(121x122x123x124x125)
=LNX(6x4! )xLNZ(4!)xLNZ(21x22x23x24)xLNZ(121x61x123x31)=LNZ(6x4x4x3)=8.
LASTTWODIGITSINAFACTORIAL
Withtheuseofaboveconcept,wecandirectlyderiveformulaforthesetypesofquestions.
Lasttwonon–zerodigitsof(10n!)=Lasttwonon–zerodigitsof(44! )xLasttwonon–zerodigitsof
(2n!)
TestCrackerTrick:
F LTNZ(10n!)=LTNZ(𝟒𝟒𝒏 )xLTNZ(2n!)
Examples
1. Whatisthelasttwonon-zerodigitsof100!
(a)44(b)36(c)76(d)64
Solution:Lasttwodigitsof100!=4410xlasttwonon-zeroof20!
=220x1110x442xlasttwonon-zerodigitsof4!
=76x01x36x24=76x01x64=64;Option(d)
(Note:if210powerisoddweget24asthelasttwodigits,ifevenweget76).
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CAT|ByAshankDubey(100%ilerofCAT15)
2. Whatisthelasttwonon–zerodigitsof500!
(a)64 (b)44 (c)36 (d)Noneofthese
!"
Solution:Lasttwonon–zerodigitsof500!=44 xlasttwonon–zerodigitsof100!
=Lasttwonon–zerodigitsof76x01x64=64
3. Howmanyzeroesarepresentattheendof25!+26!+27!+28!+30!
Solution:Let’sfindlasttwonon-zerodigitsof20!=442xlasttwonon-zerodigitsof4!
=36x24=64
Lasttwonon-zerodigitsof25!=25x24x23x22x21xlasttwonon-zerodigitsof20!
=25x24x23x22x21x64
Lasttwonon–zerodigitsof25!=84
25!+26!+27!+28!+30!=25!(1+26+26x27+26x27x28+26x27x28x30)
=25!(1+26+x02++y56+z80)
=84000000xx65
=a0,000,000
Hence,givenexpressionhas7zeroes.
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