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Question Number 179420 by mr W last updated on 29/Oct/22

a challening question: find the number of numbers which are divisible by 9 and consist of distinct digits.

$$\underline{{a}\:{challening}\:{question}:} \\ $$$${find}\:{the}\:{number}\:{of}\:{numbers}\:{which} \\ $$$${are}\:{divisible}\:{by}\:\mathrm{9}\:{and}\:{consist}\:{of} \\ $$$${distinct}\:{digits}. \\ $$

Commented by Frix last updated on 01/Nov/22

I get 59 “unique numbers” which consist of ascending digits without 0 9 18 27 36 45 126 135 189 234 279 369 378 459 468 567 1269 1278 1359 1368 1458 1467 2349 2358 2367 2457 3456 3789 4689 5679 12348 12357 12456 12789 13689 14589 14679 15678 23589 23679 24579 24678 34569 34578 123489 123579 123678 124569 124578 134568 156789 234567 246789 345789 1236789 1245789 1345689 2345679 12345678 123456789 1 digit 1 2 digits 4 3 digits 10 4 digits 14 5 digits 14 6 digits 10 7 digits 4 8 digits 1 9 digits 1 with all permutations and 0 I get 4215386 numbers

$$\mathrm{I}\:\mathrm{get}\:\mathrm{59}\:``\mathrm{unique}\:\mathrm{numbers}''\:\mathrm{which}\:\mathrm{consist} \\ $$$$\mathrm{of}\:\mathrm{ascending}\:\mathrm{digits}\:\mathrm{without}\:\mathrm{0} \\ $$$$\mathrm{9} \\ $$$$\mathrm{18}\:\mathrm{27}\:\mathrm{36}\:\mathrm{45} \\ $$$$\mathrm{126}\:\mathrm{135}\:\mathrm{189}\:\mathrm{234}\:\mathrm{279}\:\mathrm{369}\:\mathrm{378}\:\mathrm{459}\:\mathrm{468}\:\mathrm{567} \\ $$$$\mathrm{1269}\:\mathrm{1278}\:\mathrm{1359}\:\mathrm{1368}\:\mathrm{1458}\:\mathrm{1467}\:\mathrm{2349}\:\mathrm{2358} \\ $$$$\:\:\:\mathrm{2367}\:\mathrm{2457}\:\mathrm{3456}\:\mathrm{3789}\:\mathrm{4689}\:\mathrm{5679} \\ $$$$\mathrm{12348}\:\mathrm{12357}\:\mathrm{12456}\:\mathrm{12789}\:\mathrm{13689}\:\mathrm{14589} \\ $$$$\:\:\:\mathrm{14679}\:\mathrm{15678}\:\mathrm{23589}\:\mathrm{23679}\:\mathrm{24579}\:\mathrm{24678} \\ $$$$\:\:\:\mathrm{34569}\:\mathrm{34578} \\ $$$$\mathrm{123489}\:\mathrm{123579}\:\mathrm{123678}\:\mathrm{124569}\:\mathrm{124578} \\ $$$$\:\:\:\mathrm{134568}\:\mathrm{156789}\:\mathrm{234567}\:\mathrm{246789}\:\mathrm{345789} \\ $$$$\mathrm{1236789}\:\mathrm{1245789}\:\mathrm{1345689}\:\mathrm{2345679} \\ $$$$\mathrm{12345678} \\ $$$$\mathrm{123456789} \\ $$$$ \\ $$$$\mathrm{1}\:\mathrm{digit}\:\mathrm{1} \\ $$$$\mathrm{2}\:\mathrm{digits}\:\mathrm{4} \\ $$$$\mathrm{3}\:\mathrm{digits}\:\mathrm{10} \\ $$$$\mathrm{4}\:\mathrm{digits}\:\mathrm{14} \\ $$$$\mathrm{5}\:\mathrm{digits}\:\mathrm{14} \\ $$$$\mathrm{6}\:\mathrm{digits}\:\mathrm{10} \\ $$$$\mathrm{7}\:\mathrm{digits}\:\mathrm{4} \\ $$$$\mathrm{8}\:\mathrm{digits}\:\mathrm{1} \\ $$$$\mathrm{9}\:\mathrm{digits}\:\mathrm{1} \\ $$$$ \\ $$$$\mathrm{with}\:\mathrm{all}\:\mathrm{permutations}\:\mathrm{and}\:\mathrm{0}\:\mathrm{I}\:\mathrm{get} \\ $$$$\mathrm{4215386}\:\mathrm{numbers} \\ $$

Commented by mr W last updated on 01/Nov/22

thanks sir! how did you get all these numbers? the numbers of 1 to 9 digit numbers are correct, they correspond with my table.

$${thanks}\:{sir}! \\ $$$${how}\:{did}\:{you}\:{get}\:{all}\:{these}\:{numbers}? \\ $$$${the}\:{numbers}\:{of}\:\mathrm{1}\:{to}\:\mathrm{9}\:{digit}\:{numbers} \\ $$$${are}\:{correct},\:{they}\:{correspond}\:{with}\:{my} \\ $$$${table}. \\ $$

Commented by Frix last updated on 01/Nov/22

I tried to find a system to get them: start with 9 we can “split” it: 18 27 36 45 then we can add a 9: 189 279 369 459 now sometimes we can “shift” a 1: i.e. 369 ⇒ 378 and so on if we always stay with ascending digits it′s a fast way

$$\mathrm{I}\:\mathrm{tried}\:\mathrm{to}\:\mathrm{find}\:\mathrm{a}\:\mathrm{system}\:\mathrm{to}\:\mathrm{get}\:\mathrm{them}: \\ $$$$\mathrm{start}\:\mathrm{with} \\ $$$$\mathrm{9} \\ $$$$\mathrm{we}\:\mathrm{can}\:``\mathrm{split}''\:\mathrm{it}: \\ $$$$\mathrm{18}\:\mathrm{27}\:\mathrm{36}\:\mathrm{45} \\ $$$$\mathrm{then}\:\mathrm{we}\:\mathrm{can}\:\mathrm{add}\:\mathrm{a}\:\mathrm{9}: \\ $$$$\mathrm{189}\:\mathrm{279}\:\mathrm{369}\:\mathrm{459} \\ $$$$\mathrm{now}\:\mathrm{sometimes}\:\mathrm{we}\:\mathrm{can}\:``\mathrm{shift}''\:\mathrm{a}\:\mathrm{1}: \\ $$$$\mathrm{i}.\mathrm{e}.\:\mathrm{369}\:\Rightarrow\:\mathrm{378} \\ $$$$\mathrm{and}\:\mathrm{so}\:\mathrm{on} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{always}\:\mathrm{stay}\:\mathrm{with}\:\mathrm{ascending}\:\mathrm{digits}\:\mathrm{it}'\mathrm{s} \\ $$$$\mathrm{a}\:\mathrm{fast}\:\mathrm{way} \\ $$

Commented by mr W last updated on 01/Nov/22

thanks alot for explaining!

$${thanks}\:{alot}\:{for}\:{explaining}! \\ $$

Answered by Rasheed.Sindhi last updated on 29/Oct/22

determinant (((A Try...))) C-9: Sum of digits=9 All permutations of partions of 9 consist of distinct digits which are not starting with 0 from left. a>b>c>d⇒3≤a≤9 ∧6≤ d≤0 0:1+8,2+7,3+6,4+5,... 1: C-18: Sum of digits=18 C-27: Sum of digits=27 C-36: Sum of digits=36 C-45: Sum of digits=45 ...

$$\:\begin{array}{|c|}{\mathbb{A}\:\mathbb{T}\boldsymbol{\mathrm{ry}}...}\\\hline\end{array} \\ $$$$\mathbb{C}-\mathrm{9}:\:\mathrm{Sum}\:\mathrm{of}\:\mathrm{digits}=\mathrm{9} \\ $$$${All}\:{permutations}\:{of} \\ $$$$\:{partions}\:{of}\:\mathrm{9}\:{consist}\:{of}\:{distinct}\:{digits} \\ $$$${which}\:{are}\:{not}\:{starting}\:{with}\:\mathrm{0}\:{from}\:{left}. \\ $$$${a}>{b}>{c}>{d}\Rightarrow\mathrm{3}\leqslant{a}\leqslant\mathrm{9}\:\wedge\mathrm{6}\leqslant\:{d}\leqslant\mathrm{0} \\ $$$$ \\ $$$$\mathrm{0}:\mathrm{1}+\mathrm{8},\mathrm{2}+\mathrm{7},\mathrm{3}+\mathrm{6},\mathrm{4}+\mathrm{5},... \\ $$$$ \\ $$$$\mathrm{1}: \\ $$$$ \\ $$$$\mathbb{C}-\mathrm{18}:\:\mathrm{Sum}\:\mathrm{of}\:\mathrm{digits}=\mathrm{18} \\ $$$$ \\ $$$$ \\ $$$$\mathbb{C}-\mathrm{27}:\:\mathrm{Sum}\:\mathrm{of}\:\mathrm{digits}=\mathrm{27} \\ $$$$ \\ $$$$ \\ $$$$\mathbb{C}-\mathrm{36}:\:\mathrm{Sum}\:\mathrm{of}\:\mathrm{digits}=\mathrm{36} \\ $$$$ \\ $$$$ \\ $$$$\mathbb{C}-\mathrm{45}:\:\mathrm{Sum}\:\mathrm{of}\:\mathrm{digits}=\mathrm{45} \\ $$$$ \\ $$$$... \\ $$

Commented by mr W last updated on 30/Oct/22

it seems no one takes the challenge except you. thanks for trying sir! please continue sir! i′ll alse give an attempt.

$${it}\:{seems}\:{no}\:{one}\:{takes}\:{the}\:{challenge} \\ $$$${except}\:{you}.\:{thanks}\:{for}\:{trying}\:{sir}! \\ $$$${please}\:{continue}\:{sir}! \\ $$$${i}'{ll}\:{alse}\:{give}\:{an}\:{attempt}. \\ $$

Commented by Rasheed.Sindhi last updated on 30/Oct/22

Sir actually at the moment I don′t know formulas of number of partitions/ ∼ with some restrictions! Even I am not sure of my approach to be correct! I think my approach is clear at least. (i)Digit sum of such numbers is between 9 & 45 inclusively and must be divisible by 9 i-e it may be equal to 9,18,27,36 or 45.So such numbers must be partitions of these numbers with two more restrictions: being distinct and not beginning with 0 (from left). Could you please say something about this approach.( Of course there may be better approaches) but is this a wrong approach?

$$\boldsymbol{{Sir}}\:{actually}\:{at}\:{the}\:{moment}\:{I}\:{don}'{t} \\ $$$${know}\:{formulas}\:{of}\:{number}\:{of}\:{partitions}/ \\ $$$$\sim\:{with}\:{some}\:{restrictions}!\:{Even}\:{I} \\ $$$${am}\:{not}\:{sure}\:{of}\:{my}\:{approach}\:{to}\:{be} \\ $$$${correct}! \\ $$$${I}\:{think}\:{my}\:{approach}\:{is}\:{clear}\:{at}\:{least}. \\ $$$$\left({i}\right){Digit}\:{sum}\:{of}\:{such}\:{numbers}\:{is} \\ $$$${between}\:\mathrm{9}\:\&\:\mathrm{45}\:{inclusively}\:{and}\:{must} \\ $$$${be}\:{divisible}\:{by}\:\mathrm{9}\:{i}-{e}\:{it}\:{may}\:{be}\:{equal} \\ $$$${to}\:\mathrm{9},\mathrm{18},\mathrm{27},\mathrm{36}\:{or}\:\mathrm{45}.{So}\:{such}\:{numbers} \\ $$$${must}\:{be}\:{partitions}\:{of}\:{these}\:{numbers} \\ $$$${with}\:{two}\:{more}\:{restrictions}:\:{being} \\ $$$${distinct}\:{and}\:{not}\:{beginning}\:{with}\:\mathrm{0} \\ $$$$\left({from}\:{left}\right). \\ $$$${Could}\:{you}\:{please}\:{say}\:{something} \\ $$$${about}\:{this}\:{approach}.\left(\:{Of}\:{course}\:{there}\right. \\ $$$$\left.{may}\:{be}\:{better}\:{approaches}\right)\:{but}\:{is} \\ $$$${this}\:{a}\:{wrong}\:{approach}? \\ $$$$ \\ $$

Commented by mr W last updated on 30/Oct/22

your idea is absolutely correct. i think there are no other shortcut methods than this. so the task is to find the number of ways to select r digits from [1,9] such that their sum is a multiple of 9, i.e. 9, 18, 27, 36, 45. here r=2, 3, ...,9.

$${your}\:{idea}\:{is}\:{absolutely}\:{correct}.\:{i}\:{think} \\ $$$${there}\:{are}\:{no}\:{other}\:{shortcut}\:{methods}\: \\ $$$${than}\:{this}.\:{so}\:{the}\:{task}\:{is}\:{to}\:{find}\:{the}\: \\ $$$${number}\:{of}\:{ways}\:{to}\:{select}\:{r}\:{digits}\:{from} \\ $$$$\left[\mathrm{1},\mathrm{9}\right]\:{such}\:{that}\:{their}\:{sum}\:{is}\:{a}\:{multiple}\: \\ $$$${of}\:\mathrm{9},\:{i}.{e}.\:\mathrm{9},\:\mathrm{18},\:\mathrm{27},\:\mathrm{36},\:\mathrm{45}.\: \\ $$$${here}\:{r}=\mathrm{2},\:\mathrm{3},\:...,\mathrm{9}. \\ $$

Commented by Rasheed.Sindhi last updated on 30/Oct/22

Grateful sir!

$$\mathbb{G}\boldsymbol{\mathrm{rateful}}\:\boldsymbol{\mathrm{sir}}! \\ $$

Commented by Rasheed.Sindhi last updated on 30/Oct/22

Sorry sir,I′ve only rough idea. Complete solution is beyond my very limited capacity!

$${Sorry}\:{sir},{I}'{ve}\:{only}\:{rough}\:{idea}. \\ $$$${Complete}\:{solution}\:{is}\:{beyond}\:{my} \\ $$$${very}\:{limited}\:{capacity}! \\ $$

Commented by mr W last updated on 31/Oct/22

thanks for till now! please check my answer sir!

$${thanks}\:{for}\:{till}\:{now}!\:{please}\:{check}\:{my} \\ $$$${answer}\:{sir}! \\ $$

Answered by mr W last updated on 31/Oct/22

an attempt since such a number should consist of distinct digits, it must have at least two digits and at most 10 digits. let′s consider at first only numbers without the digit “0”, since if we have a r−digit−number which is divisible by 9, but without “0”, then we have automatically r (r+1)−digit−numbers which are also divisible by 9. as example: say d_1 d_2 ...d_r is divisible by 9, with d_i ≠0, then d_1 0d_2 ...d_r ,d_1 d_2 0...d_r ,...,d_1 d_2 ...d_r 0 are also divisible by 9. we know a number is divisible by 9, if the sum of its digits is divisible by 9. the sum of 9 different digits can at most be 1+2+3+4+5+6+7+8+9=45. therefore the sum of the digits from a number with distinct digits and divisible by 9 must be 9 or 18 or 27 or 36 or 45. now we should find the number of ways to select r digits from [1,9] such that their sum is 9 or 18 or 27 or 36 or 45. here r=1,2,...,9. say the r distinct digits are d_1 ,d_2 ,...,d_r with d_1 <d_2 <...<d_r and 1≤d_i ≤9 with them we can form r! r−digit−numbers. d_1 +d_2 +...+d_r =9 or 18 or 27 or 36 or 45. the number of integer solutions of this equation represents the number of ways to select d_1 ,d_2 ,...,d_r from [1,9].

$$\boldsymbol{{an}}\:\boldsymbol{{attempt}} \\ $$$${since}\:{such}\:{a}\:{number}\:{should}\:{consist} \\ $$$${of}\:{distinct}\:{digits},\:{it}\:{must}\:{have}\:{at} \\ $$$${least}\:{two}\:{digits}\:{and}\:{at}\:{most}\:\mathrm{10}\:{digits}. \\ $$$${let}'{s}\:{consider}\:{at}\:{first}\:{only}\:\:{numbers} \\ $$$${without}\:{the}\:{digit}\:``\mathrm{0}'',\:{since}\:{if}\:{we}\:{have} \\ $$$${a}\:{r}−{digit}−{number}\:{which}\:{is}\:{divisible} \\ $$$${by}\:\mathrm{9},\:{but}\:{without}\:``\mathrm{0}'',\:\:{then}\:{we}\:{have}\: \\ $$$${automatically}\:{r}\: \\ $$$$\left({r}+\mathrm{1}\right)−{digit}−{numbers}\:{which}\:{are} \\ $$$${also}\:{divisible}\:{by}\:\mathrm{9}.\:{as}\:{example}: \\ $$$${say}\:{d}_{\mathrm{1}} {d}_{\mathrm{2}} ...{d}_{{r}} \:{is}\:{divisible}\:{by}\:\mathrm{9},\:{with}\:{d}_{{i}} \neq\mathrm{0}, \\ $$$${then}\:{d}_{\mathrm{1}} \mathrm{0}{d}_{\mathrm{2}} ...{d}_{{r}} ,{d}_{\mathrm{1}} {d}_{\mathrm{2}} \mathrm{0}...{d}_{{r}} ,...,{d}_{\mathrm{1}} {d}_{\mathrm{2}} ...{d}_{{r}} \mathrm{0}\: \\ $$$${are}\:{also}\:{divisible}\:{by}\:\mathrm{9}. \\ $$$${we}\:{know}\:{a}\:{number}\:{is}\:{divisible}\:{by}\:\mathrm{9}, \\ $$$${if}\:{the}\:{sum}\:{of}\:{its}\:{digits}\:{is}\:{divisible}\:{by} \\ $$$$\mathrm{9}.\:{the}\:{sum}\:{of}\:\mathrm{9}\:{different}\:{digits}\:{can} \\ $$$${at}\:{most}\:{be}\: \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}+\mathrm{7}+\mathrm{8}+\mathrm{9}=\mathrm{45}. \\ $$$${therefore}\:{the}\:{sum}\:{of}\:{the}\:{digits}\:{from} \\ $$$${a}\:{number}\:{with}\:{distinct}\:{digits}\:{and} \\ $$$${divisible}\:{by}\:\mathrm{9}\:{must}\:{be} \\ $$$$\mathrm{9}\:{or}\:\mathrm{18}\:{or}\:\mathrm{27}\:{or}\:\mathrm{36}\:{or}\:\mathrm{45}. \\ $$$${now}\:{we}\:{should}\:{find}\:{the}\:{number}\:{of} \\ $$$${ways}\:{to}\:{select}\:{r}\:{digits}\:{from}\:\left[\mathrm{1},\mathrm{9}\right]\: \\ $$$${such}\:{that}\:{their}\:{sum}\:{is}\:\mathrm{9}\:{or}\:\mathrm{18}\:{or}\:\mathrm{27}\:{or} \\ $$$$\mathrm{36}\:{or}\:\mathrm{45}.\:{here}\:{r}=\mathrm{1},\mathrm{2},...,\mathrm{9}. \\ $$$${say}\:{the}\:{r}\:{distinct}\:{digits}\:{are} \\ $$$${d}_{\mathrm{1}} ,{d}_{\mathrm{2}} ,...,{d}_{{r}} \:{with}\: \\ $$$${d}_{\mathrm{1}} <{d}_{\mathrm{2}} <...<{d}_{{r}} \:{and}\:\mathrm{1}\leqslant{d}_{{i}} \leqslant\mathrm{9} \\ $$$${with}\:{them}\:{we}\:{can}\:{form}\:{r}!\: \\ $$$${r}−{digit}−{numbers}. \\ $$$${d}_{\mathrm{1}} +{d}_{\mathrm{2}} +...+{d}_{{r}} =\mathrm{9}\:{or}\:\mathrm{18}\:{or}\:\mathrm{27}\:{or}\:\mathrm{36}\:{or}\:\mathrm{45}. \\ $$$${the}\:{number}\:{of}\:{integer}\:{solutions}\:{of} \\ $$$${this}\:{equation}\:{represents}\:{the}\:{number} \\ $$$${of}\:{ways}\:{to}\:{select}\:{d}_{\mathrm{1}} ,{d}_{\mathrm{2}} ,...,{d}_{{r}} \:{from}\:\left[\mathrm{1},\mathrm{9}\right]. \\ $$

Commented by mr W last updated on 31/Oct/22

number of ways to select r digits from [1,9] such that their sum is equal to...

$${number}\:{of}\:{ways}\:{to}\:{select}\:{r}\:{digits}\:{from} \\ $$$$\left[\mathrm{1},\mathrm{9}\right]\:{such}\:{that}\:{their}\:{sum}\:{is}\:{equal}\:{to}... \\ $$

Commented by mr W last updated on 01/Nov/22

Commented by mr W last updated on 01/Nov/22

2−digit numbers: without “0”: 4×2!=8 with “0”: 1×1=1 3−digit numbers: without “0”: (3+7)×3!=60 with “0”: 8×2=16 4−digit numbers: without “0”: (11+3)×4!=336 with “0”: 60×3=180 5−digit numbers: without “0”: (3+11)×5!=1 680 with “0”: 336×4=1 344 6−digit numbers: without “0”: (7+3)×6!=7 200 with “0”: 1680×5=8 400 7−digit numbers: without “0”: 4×7!=20 160 with “0”: 7200×6=43 200 8−digit numbers: without “0”: 1×8!=40 320 with “0”: 20160×7=141 120 9−digit numbers: without “0”: 1×9!=362 880 with “0”: 40320×8=322 560 10−digit numbers: with “0”: 362880×9=3 265 920 totally: 4 215 385 numbers

$$\mathrm{2}−{digit}\:{numbers}: \\ $$$$\:\:{without}\:``\mathrm{0}'':\:\:\:\mathrm{4}×\mathrm{2}!=\mathrm{8} \\ $$$$\:\:{with}\:``\mathrm{0}'':\:\:\:\:\:\:\:\:\:\:\mathrm{1}×\mathrm{1}=\mathrm{1} \\ $$$$\mathrm{3}−{digit}\:{numbers}: \\ $$$$\:\:{without}\:``\mathrm{0}'':\:\:\:\left(\mathrm{3}+\mathrm{7}\right)×\mathrm{3}!=\mathrm{60} \\ $$$$\:\:{with}\:``\mathrm{0}'':\:\:\:\:\:\:\:\:\:\:\mathrm{8}×\mathrm{2}=\mathrm{16} \\ $$$$\mathrm{4}−{digit}\:{numbers}: \\ $$$$\:\:{without}\:``\mathrm{0}'':\:\:\:\left(\mathrm{11}+\mathrm{3}\right)×\mathrm{4}!=\mathrm{336} \\ $$$$\:\:{with}\:``\mathrm{0}'':\:\:\:\:\:\:\:\:\:\:\mathrm{60}×\mathrm{3}=\mathrm{180} \\ $$$$\mathrm{5}−{digit}\:{numbers}: \\ $$$$\:\:{without}\:``\mathrm{0}'':\:\:\:\left(\mathrm{3}+\mathrm{11}\right)×\mathrm{5}!=\mathrm{1}\:\mathrm{680} \\ $$$$\:\:{with}\:``\mathrm{0}'':\:\:\:\:\:\:\:\:\:\:\mathrm{336}×\mathrm{4}=\mathrm{1}\:\mathrm{344} \\ $$$$\mathrm{6}−{digit}\:{numbers}: \\ $$$$\:\:{without}\:``\mathrm{0}'':\:\:\:\left(\mathrm{7}+\mathrm{3}\right)×\mathrm{6}!=\mathrm{7}\:\mathrm{200} \\ $$$$\:\:{with}\:``\mathrm{0}'':\:\:\:\:\:\:\:\:\:\:\mathrm{1680}×\mathrm{5}=\mathrm{8}\:\mathrm{400} \\ $$$$\mathrm{7}−{digit}\:{numbers}: \\ $$$$\:\:{without}\:``\mathrm{0}'':\:\:\:\mathrm{4}×\mathrm{7}!=\mathrm{20}\:\mathrm{160} \\ $$$$\:\:{with}\:``\mathrm{0}'':\:\:\:\:\:\:\:\:\:\:\mathrm{7200}×\mathrm{6}=\mathrm{43}\:\mathrm{200} \\ $$$$\mathrm{8}−{digit}\:{numbers}: \\ $$$$\:\:{without}\:``\mathrm{0}'':\:\:\:\mathrm{1}×\mathrm{8}!=\mathrm{40}\:\mathrm{320} \\ $$$$\:\:{with}\:``\mathrm{0}'':\:\:\:\:\:\:\:\:\:\:\mathrm{20160}×\mathrm{7}=\mathrm{141}\:\mathrm{120} \\ $$$$\mathrm{9}−{digit}\:{numbers}: \\ $$$$\:\:{without}\:``\mathrm{0}'':\:\:\:\mathrm{1}×\mathrm{9}!=\mathrm{362}\:\mathrm{880} \\ $$$$\:\:{with}\:``\mathrm{0}'':\:\:\:\:\:\:\:\:\:\:\mathrm{40320}×\mathrm{8}=\mathrm{322}\:\mathrm{560} \\ $$$$\mathrm{10}−{digit}\:{numbers}: \\ $$$$\:\:{with}\:``\mathrm{0}'':\:\:\:\:\:\:\:\:\:\:\mathrm{362880}×\mathrm{9}=\mathrm{3}\:\mathrm{265}\:\mathrm{920} \\ $$$$ \\ $$$${totally}:\:\:\mathrm{4}\:\mathrm{215}\:\mathrm{385}\:{numbers} \\ $$

Commented by mr W last updated on 31/Oct/22

to check the table above we can calculate the total number (Σ) of ways to select 1,2,...,9 to get the sum k. it is the coefficient of x^k in the expansion of (1+x)(1+x^2 )(1+x^3 )...(1+x^9 ). here we only need to look at the values for k=9, 18, 27, 36, 45.

$${to}\:{check}\:{the}\:{table}\:{above}\:{we}\:{can}\: \\ $$$${calculate}\:{the}\:{total}\:{number}\:\left(\Sigma\right)\:{of} \\ $$$${ways}\:{to}\:{select}\:\mathrm{1},\mathrm{2},...,\mathrm{9}\:{to}\:{get}\:{the}\:{sum}\:{k}. \\ $$$${it}\:{is}\:{the}\:{coefficient}\:{of}\:{x}^{{k}} \:{in}\:{the} \\ $$$${expansion}\:{of}\: \\ $$$$\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{3}} \right)...\left(\mathrm{1}+{x}^{\mathrm{9}} \right). \\ $$$${here}\:{we}\:{only}\:{need}\:{to}\:{look}\:{at}\:{the}\:{values} \\ $$$${for}\:{k}=\mathrm{9},\:\mathrm{18},\:\mathrm{27},\:\mathrm{36},\:\mathrm{45}. \\ $$

Commented by mr W last updated on 31/Oct/22

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