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Q58885-reposted-How-many-4-digit-numbers-abcd-exist-which-are-divisible-by-3-and-satisfy-a-b-c-d-




Question Number 59021 by mr W last updated on 03/May/19
[Q58885 reposted]  How many 4−digit numbers abcd exist  which are divisible by 3 and satisfy  a≤b≤c≤d?
$$\left[{Q}\mathrm{58885}\:{reposted}\right] \\ $$$${How}\:{many}\:\mathrm{4}−{digit}\:{numbers}\:{abcd}\:{exist} \\ $$$${which}\:{are}\:{divisible}\:{by}\:\mathrm{3}\:{and}\:{satisfy} \\ $$$${a}\leqslant{b}\leqslant{c}\leqslant{d}? \\ $$
Commented by tanmay last updated on 03/May/19
is there any co−relation between this problem  and Euler enumeration partition ...
$${is}\:{there}\:{any}\:{co}−{relation}\:{between}\:{this}\:{problem} \\ $$$${and}\:{Euler}\:{enumeration}\:{partition}\:… \\ $$
Commented by mr W last updated on 03/May/19
It′s not as easy as I thought. Please  give a try.
$${It}'{s}\:{not}\:{as}\:{easy}\:{as}\:{I}\:{thought}.\:{Please} \\ $$$${give}\:{a}\:{try}. \\ $$
Commented by Kunal12588 last updated on 03/May/19
Commented by Kunal12588 last updated on 03/May/19
I think it may help someone who is trying to solve the problem.
Commented by mr W last updated on 03/May/19
thank you sir!  can you give a formula how to calculate the  number without knowing the numbers  themselves?
$${thank}\:{you}\:{sir}! \\ $$$${can}\:{you}\:{give}\:{a}\:{formula}\:{how}\:{to}\:{calculate}\:{the} \\ $$$${number}\:{without}\:{knowing}\:{the}\:{numbers} \\ $$$${themselves}? \\ $$
Commented by mr W last updated on 03/May/19
my formula (not exactly proved yet) is  (C_(9−1) ^(4+9−1) /3)=(C_8 ^(12) /3)=((495)/3)=165    in case of 5−digit numbers it is  (C_(9−1) ^(5+9−1) /3)=(C_8 ^(13) /3)=((1287)/3)=429    in case of 6−digit numbers it is  (C_(9−1) ^(6+9−1) /3)=(C_8 ^(14) /3)=((3003)/3)=1001    can you check them using your method?
$${my}\:{formula}\:\left({not}\:{exactly}\:{proved}\:{yet}\right)\:{is} \\ $$$$\frac{{C}_{\mathrm{9}−\mathrm{1}} ^{\mathrm{4}+\mathrm{9}−\mathrm{1}} }{\mathrm{3}}=\frac{{C}_{\mathrm{8}} ^{\mathrm{12}} }{\mathrm{3}}=\frac{\mathrm{495}}{\mathrm{3}}=\mathrm{165} \\ $$$$ \\ $$$${in}\:{case}\:{of}\:\mathrm{5}−{digit}\:{numbers}\:{it}\:{is} \\ $$$$\frac{{C}_{\mathrm{9}−\mathrm{1}} ^{\mathrm{5}+\mathrm{9}−\mathrm{1}} }{\mathrm{3}}=\frac{{C}_{\mathrm{8}} ^{\mathrm{13}} }{\mathrm{3}}=\frac{\mathrm{1287}}{\mathrm{3}}=\mathrm{429} \\ $$$$ \\ $$$${in}\:{case}\:{of}\:\mathrm{6}−{digit}\:{numbers}\:{it}\:{is} \\ $$$$\frac{{C}_{\mathrm{9}−\mathrm{1}} ^{\mathrm{6}+\mathrm{9}−\mathrm{1}} }{\mathrm{3}}=\frac{{C}_{\mathrm{8}} ^{\mathrm{14}} }{\mathrm{3}}=\frac{\mathrm{3003}}{\mathrm{3}}=\mathrm{1001} \\ $$$$ \\ $$$${can}\:{you}\:{check}\:{them}\:{using}\:{your}\:{method}? \\ $$
Commented by mr W last updated on 03/May/19
ajfour sir, MJS sir, tanmay sir:  can you please also have a look at this  problem?
$${ajfour}\:{sir},\:{MJS}\:{sir},\:{tanmay}\:{sir}: \\ $$$${can}\:{you}\:{please}\:{also}\:{have}\:{a}\:{look}\:{at}\:{this} \\ $$$${problem}? \\ $$
Commented by Kunal12588 last updated on 03/May/19
Commented by Kunal12588 last updated on 03/May/19
Commented by Kunal12588 last updated on 03/May/19
Sir MrW you are correct with 5-digit numbers There are 429 such numbers. but for 6-digit number they are total 1005.
Answered by Rasheed.Sindhi last updated on 03/May/19
I don′t know any formula.I want to  think about ′process′ of determining  required number.Although I′m not  sure of my process being  useful at  this stage.  I think upto three digits we need not  be anxious about divisibility of 3.  It′s only fixation of one digit which  makes the number divisible by 3.  So upto abc we should consider only one  condition that is a≤b≤c.  The fourth digit (unit) d may be fixed  in such a way that it makes the number  multiple of  3 along with d≥c.  Particularily if a_0 b_0 c_(0 ) is any of such  permutations and d is unit digit then  for 3m>a_0 +b_0 +c_0  &1≤3m−(a_0 +b_0 +c_0  )≤9:           d≥c   ∧  d=3m−(a_0 +b_0 +c_0  )    For example if abc=112  According to the condition d≥c , the  candidates for d are 2,3,4,...,9 but  only 2,5 & 8 are such which make  the number divisible by 3. So three  numbers (1122,1125 & 1128) can be   made from 112  The limitation of the process is this  that we have to compose all the three  digit numbers abc with condition  a≤b≤c   So this is not a good method to determine  number of the required numbers.  However it may be used to construct  numbers under the given conditions
$${I}\:{don}'{t}\:{know}\:{any}\:{formula}.{I}\:{want}\:{to} \\ $$$${think}\:{about}\:'{process}'\:{of}\:{determining} \\ $$$${required}\:{number}.{Although}\:{I}'{m}\:{not} \\ $$$${sure}\:{of}\:{my}\:{process}\:{being}\:\:{useful}\:{at} \\ $$$${this}\:{stage}. \\ $$$${I}\:{think}\:{upto}\:{three}\:{digits}\:{we}\:{need}\:{not} \\ $$$${be}\:{anxious}\:{about}\:{divisibility}\:{of}\:\mathrm{3}. \\ $$$${It}'{s}\:{only}\:{fixation}\:{of}\:{one}\:{digit}\:{which} \\ $$$${makes}\:{the}\:{number}\:{divisible}\:{by}\:\mathrm{3}. \\ $$$${So}\:{upto}\:{abc}\:{we}\:{should}\:{consider}\:{only}\:{one} \\ $$$${condition}\:{that}\:{is}\:{a}\leqslant{b}\leqslant{c}. \\ $$$${The}\:{fourth}\:{digit}\:\left({unit}\right)\:{d}\:{may}\:{be}\:{fixed} \\ $$$${in}\:{such}\:{a}\:{way}\:{that}\:{it}\:{makes}\:{the}\:{number} \\ $$$${multiple}\:{of}\:\:\mathrm{3}\:{along}\:{with}\:{d}\geqslant{c}. \\ $$$${Particularily}\:{if}\:{a}_{\mathrm{0}} {b}_{\mathrm{0}} {c}_{\mathrm{0}\:} {is}\:{any}\:{of}\:{such} \\ $$$${permutations}\:{and}\:{d}\:{is}\:{unit}\:{digit}\:{then} \\ $$$${for}\:\mathrm{3}{m}>{a}_{\mathrm{0}} +{b}_{\mathrm{0}} +{c}_{\mathrm{0}} \:\&\mathrm{1}\leqslant\mathrm{3}{m}−\left({a}_{\mathrm{0}} +{b}_{\mathrm{0}} +{c}_{\mathrm{0}} \:\right)\leqslant\mathrm{9}: \\ $$$$\:\:\:\:\:\:\:\:\:{d}\geqslant{c}\:\:\:\wedge\:\:{d}=\mathrm{3}{m}−\left({a}_{\mathrm{0}} +{b}_{\mathrm{0}} +{c}_{\mathrm{0}} \:\right) \\ $$$$ \\ $$$${For}\:{example}\:{if}\:{abc}=\mathrm{112} \\ $$$${According}\:{to}\:{the}\:{condition}\:{d}\geqslant{c}\:,\:{the} \\ $$$${candidates}\:{for}\:{d}\:{are}\:\mathrm{2},\mathrm{3},\mathrm{4},…,\mathrm{9}\:{but} \\ $$$${only}\:\mathrm{2},\mathrm{5}\:\&\:\mathrm{8}\:{are}\:{such}\:{which}\:{make} \\ $$$${the}\:{number}\:{divisible}\:{by}\:\mathrm{3}.\:{So}\:{three} \\ $$$${numbers}\:\left(\mathrm{1122},\mathrm{1125}\:\&\:\mathrm{1128}\right)\:{can}\:{be}\: \\ $$$${made}\:{from}\:\mathrm{112} \\ $$$${The}\:{limitation}\:{of}\:{the}\:{process}\:{is}\:{this} \\ $$$${that}\:{we}\:{have}\:{to}\:{compose}\:{all}\:{the}\:{three} \\ $$$${digit}\:{numbers}\:{abc}\:{with}\:{condition} \\ $$$${a}\leqslant{b}\leqslant{c}\: \\ $$$${So}\:{this}\:{is}\:{not}\:{a}\:{good}\:{method}\:{to}\:{determine} \\ $$$${number}\:{of}\:{the}\:{required}\:{numbers}. \\ $$$${However}\:{it}\:{may}\:{be}\:{used}\:{to}\:{construct} \\ $$$${numbers}\:{under}\:{the}\:{given}\:{conditions} \\ $$
Answered by Rasheed.Sindhi last updated on 04/May/19
Case-1: When a+b+c+d=3  No possibility because 1111 is the  smallest number under condition  a≤b≤c≤d    Case-2: When a+b+c+d=6    Case-3: When a+b+c+d=9    Case-12: When a+b+c+d=36     ........      .......
$${Case}-\mathrm{1}:\:{When}\:{a}+{b}+{c}+{d}=\mathrm{3} \\ $$$${No}\:{possibility}\:{because}\:\mathrm{1111}\:{is}\:{the} \\ $$$${smallest}\:{number}\:{under}\:{condition} \\ $$$${a}\leqslant{b}\leqslant{c}\leqslant{d} \\ $$$$ \\ $$$${Case}-\mathrm{2}:\:{When}\:{a}+{b}+{c}+{d}=\mathrm{6} \\ $$$$ \\ $$$${Case}-\mathrm{3}:\:{When}\:{a}+{b}+{c}+{d}=\mathrm{9} \\ $$$$ \\ $$$${Case}-\mathrm{12}:\:{When}\:{a}+{b}+{c}+{d}=\mathrm{36} \\ $$$$\:\:\:…….. \\ $$$$\:\:\:\:……. \\ $$$$ \\ $$
Commented by mr W last updated on 03/May/19
thank you Rasheed sir for sharing  your thoughts!
$${thank}\:{you}\:{Rasheed}\:{sir}\:{for}\:{sharing} \\ $$$${your}\:{thoughts}! \\ $$
Commented by mr W last updated on 03/May/19
the question is indeed to find how  many solutions we have for the  equation  a+b+c+d=3k with k=2...12 and  1≤a≤b≤c≤d≤9
$${the}\:{question}\:{is}\:{indeed}\:{to}\:{find}\:{how} \\ $$$${many}\:{solutions}\:{we}\:{have}\:{for}\:{the} \\ $$$${equation} \\ $$$${a}+{b}+{c}+{d}=\mathrm{3}{k}\:{with}\:{k}=\mathrm{2}…\mathrm{12}\:{and} \\ $$$$\mathrm{1}\leqslant{a}\leqslant{b}\leqslant{c}\leqslant{d}\leqslant\mathrm{9} \\ $$
Commented by Rasheed.Sindhi last updated on 03/May/19
The problem goes to “number of  partitions” under some conditions.  For example: In how many ways  6 can be partitioned in four parts  a,b,c,d such that        1≤a,b,c,d≤9 & a≤b≤c≤d
$${The}\:{problem}\:{goes}\:{to}\:“{number}\:{of} \\ $$$${partitions}''\:{under}\:{some}\:{conditions}. \\ $$$${For}\:{example}:\:{In}\:{how}\:{many}\:{ways} \\ $$$$\mathrm{6}\:{can}\:{be}\:{partitioned}\:{in}\:{four}\:{parts} \\ $$$${a},{b},{c},{d}\:{such}\:{that} \\ $$$$\:\:\:\:\:\:\mathrm{1}\leqslant{a},{b},{c},{d}\leqslant\mathrm{9}\:\&\:{a}\leqslant{b}\leqslant{c}\leqslant{d} \\ $$
Commented by Rasheed.Sindhi last updated on 04/May/19
Sir I want to do the same.I thought  that if we be able to determine the  number of partitions of an integer n  (in this case  n=6,9,...36) in k parts  (in our case k=4) we can solve our  problem.
$${Sir}\:{I}\:{want}\:{to}\:{do}\:{the}\:{same}.{I}\:{thought} \\ $$$${that}\:{if}\:{we}\:{be}\:{able}\:{to}\:{determine}\:{the} \\ $$$${number}\:{of}\:{partitions}\:{of}\:{an}\:{integer}\:{n} \\ $$$$\left({in}\:{this}\:{case}\:\:{n}=\mathrm{6},\mathrm{9},…\mathrm{36}\right)\:{in}\:{k}\:{parts} \\ $$$$\left({in}\:{our}\:{case}\:{k}=\mathrm{4}\right)\:{we}\:{can}\:{solve}\:{our} \\ $$$${problem}. \\ $$

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