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find-the-number-of-5-digit-natural-numbers-with-strictly-ascending-digits-whose-sum-is-20-example-12458-is-such-a-number-

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Question Number 188362 by mr W last updated on 28/Feb/23
find the number of 5 digit natural numbers with strictly ascending digits whose sum is 20. example: 12458 is such a number
$${find}\:{the}\:{number}\:{of}\:\mathrm{5}\:{digit}\:{natural} \\ $$$${numbers}\:{with}\:{strictly}\:{ascending}\: \\ $$$${digits}\:{whose}\:{sum}\:{is}\:\mathrm{20}. \\ $$$${example}:\:\mathrm{12458}\:{is}\:{such}\:{a}\:{number} \\ $$
Answered by ARUNG_Brandon_MBU last updated on 28/Feb/23
6
$$\mathrm{6} \\ $$
Commented by ARUNG_Brandon_MBU last updated on 28/Feb/23
#include <iostream> using namespace std; int main(void) { for (int i0, i1, i2, i3, i4, i=10000; i<100000; i++) { i0 = i/10000; i1 = i%10000/1000; i2 = i%1000/100; i3 = i%100/10; i4 = i%10; if (i0<i1 && i1<i2 && i2<i3 && i3<i4) if ((i0+i1+i2+i3+i4) == 20) cout << i <<", "; } return 0; }
Commented by ARUNG_Brandon_MBU last updated on 28/Feb/23
Output: 12359, 12368, 12458, 12467, 13457, 23456,
$$\mathrm{Output}:\:\mathrm{12359},\:\mathrm{12368},\:\mathrm{12458}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{12467},\:\mathrm{13457},\:\mathrm{23456}, \\ $$
Commented by mr W last updated on 28/Feb/23
yes!
$${yes}! \\ $$
Commented by ARUNG_Brandon_MBU last updated on 28/Feb/23
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Commented by mr W last updated on 01/Mar/23
in this case, i mean when the sum of digits is ≤20, we can still have a smart solution, which gives us the answer 7−1=6. 7 is the coef. of term x^(20) in the expansion Π_(k=1) ^5 (x^k /(1−x^k )). but it includes also the case that a digit is 10, which is certainly not true, therefore the correct answer is 7−1=6. if the sum of digits is >20, i think we can only enumerate all possible possibilities as your program does. there is no smarter approach.
$${in}\:{this}\:{case},\:{i}\:{mean}\:{when}\:{the}\:{sum}\:{of}\: \\ $$$${digits}\:{is}\:\leqslant\mathrm{20},\:{we}\:{can}\:{still}\:{have}\:{a}\: \\ $$$${smart}\:{solution},\:{which}\:{gives}\:{us}\:{the} \\ $$$${answer}\:\mathrm{7}−\mathrm{1}=\mathrm{6}. \\ $$$$\mathrm{7}\:{is}\:{the}\:{coef}.\:{of}\:{term}\:{x}^{\mathrm{20}} \:{in}\:{the} \\ $$$${expansion}\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{5}} {\prod}}\frac{{x}^{{k}} }{\mathrm{1}−{x}^{{k}} }.\:{but}\:{it}\:{includes} \\ $$$${also}\:{the}\:{case}\:{that}\:{a}\:{digit}\:{is}\:\mathrm{10},\:{which} \\ $$$${is}\:{certainly}\:{not}\:{true},\:{therefore}\:{the} \\ $$$${correct}\:{answer}\:{is}\:\mathrm{7}−\mathrm{1}=\mathrm{6}. \\ $$$${if}\:{the}\:{sum}\:{of}\:{digits}\:{is}\:>\mathrm{20},\:{i}\:{think} \\ $$$${we}\:{can}\:{only}\:{enumerate}\:{all}\:{possible} \\ $$$${possibilities}\:{as}\:{your}\:{program}\:{does}. \\ $$$${there}\:{is}\:{no}\:{smarter}\:{approach}. \\ $$
Commented by ARUNG_Brandon_MBU last updated on 28/Feb/23
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Commented by mr W last updated on 04/Mar/23
i found a new way even for n>20, see below. can you please check with your program?
$${i}\:{found}\:{a}\:{new}\:{way}\:{even}\:{for}\:{n}>\mathrm{20}, \\ $$$${see}\:{below}.\:{can}\:{you}\:{please}\:{check}\:{with} \\ $$$${your}\:{program}? \\ $$
Commented by ARUNG_Brandon_MBU last updated on 07/Mar/23
// for n > 20 #include <iostream> using namespace std; int main(void) { for(int i0,i1,i2,i3,i4,i=10000;i<100000;i++) { i0 = i/10000; i1 = i%10000/1000; i2 = i%1000/100; i3 = i%100/10; i4 = i%10; if (i0<i1 && i1<i2 && i2<i3 && i3<i4) if (i0+i1+i2+i3+i4 > 20) cout << i <<" "; } return 0; }
Commented by ARUNG_Brandon_MBU last updated on 07/Mar/23
Commented by ARUNG_Brandon_MBU last updated on 07/Mar/23
We have 108 5-digit numbers such that the digits are different and their sum > 20
$$\mathrm{We}\:\mathrm{have}\:\mathrm{108}\:\mathrm{5}-\mathrm{digit}\:\mathrm{numbers}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{digits}\:\mathrm{are}\:\mathrm{different}\:\mathrm{and}\:\mathrm{their}\:\mathrm{sum}\:>\:\mathrm{20} \\ $$
Commented by mr W last updated on 07/Mar/23
yes, that′s correct! as we can see below there are 8 numbers with sum 21, 9 numbers with sum 22, 11 numbers with sum 23, ... 2 numbers with sum 33, 1 number with sum 34, 1 number with sum 35. so totally 108 numbers with sum>20.
$${yes},\:{that}'{s}\:{correct}! \\ $$$${as}\:{we}\:{can}\:{see}\:{below}\:{there}\:{are} \\ $$$$\mathrm{8}\:{numbers}\:{with}\:{sum}\:\mathrm{21}, \\ $$$$\mathrm{9}\:{numbers}\:{with}\:{sum}\:\mathrm{22}, \\ $$$$\mathrm{11}\:{numbers}\:{with}\:{sum}\:\mathrm{23}, \\ $$$$… \\ $$$$\mathrm{2}\:{numbers}\:{with}\:{sum}\:\mathrm{33}, \\ $$$$\mathrm{1}\:{number}\:{with}\:{sum}\:\mathrm{34}, \\ $$$$\mathrm{1}\:{number}\:{with}\:{sum}\:\mathrm{35}. \\ $$$${so}\:{totally}\:\mathrm{108}\:{numbers}\:{with}\:{sum}>\mathrm{20}. \\ $$
Answered by mr W last updated on 04/Mar/23
a new try say the number is d_1 d_2 d_3 d_4 d_5 with d_1 +d_2 +d_3 +d_4 +d_5 =n and 1≤d_1 <d_2 <d_3 <d_4 <d_5 ≤9 let d_1 =1+k_1 with 0≤k_1 ≤8 d_2 =d_1 +1+k_2 =2+k_1 +k_2 with 0≤k_2 ≤7 d_3 =d_2 +1+k_3 =3+k_1 +k_2 +k_3 with 0≤k_3 ≤6 ...... d_5 =d_4 +1+k_4 =5+k_1 +k_2 +k_3 +...+k_5 with 0≤k_3 ≤4 (1+2+3+...+5)+5k_1 +4k_2 +3k_3 +2k_4 +k_5 =n number of solutions is the coef. of term x^n in the expansion of x^(15) Π_(r=1) ^5 ((1−x^(10−r) )/(1−x^r )) for n=20 we have 6 numbers and for n=25 we have 12 numbers etc.
$${a}\:{new}\:{try} \\ $$$${say}\:{the}\:{number}\:{is}\:{d}_{\mathrm{1}} {d}_{\mathrm{2}} {d}_{\mathrm{3}} {d}_{\mathrm{4}} {d}_{\mathrm{5}} \:{with} \\ $$$${d}_{\mathrm{1}} +{d}_{\mathrm{2}} +{d}_{\mathrm{3}} +{d}_{\mathrm{4}} +{d}_{\mathrm{5}} ={n}\:{and} \\ $$$$\mathrm{1}\leqslant{d}_{\mathrm{1}} <{d}_{\mathrm{2}} <{d}_{\mathrm{3}} <{d}_{\mathrm{4}} <{d}_{\mathrm{5}} \leqslant\mathrm{9} \\ $$$${let}\: \\ $$$${d}_{\mathrm{1}} =\mathrm{1}+{k}_{\mathrm{1}} \:{with}\:\mathrm{0}\leqslant{k}_{\mathrm{1}} \leqslant\mathrm{8} \\ $$$${d}_{\mathrm{2}} ={d}_{\mathrm{1}} +\mathrm{1}+{k}_{\mathrm{2}} =\mathrm{2}+{k}_{\mathrm{1}} +{k}_{\mathrm{2}} \:{with}\:\mathrm{0}\leqslant{k}_{\mathrm{2}} \leqslant\mathrm{7} \\ $$$${d}_{\mathrm{3}} ={d}_{\mathrm{2}} +\mathrm{1}+{k}_{\mathrm{3}} =\mathrm{3}+{k}_{\mathrm{1}} +{k}_{\mathrm{2}} +{k}_{\mathrm{3}} \:{with}\:\mathrm{0}\leqslant{k}_{\mathrm{3}} \leqslant\mathrm{6} \\ $$$$…… \\ $$$${d}_{\mathrm{5}} ={d}_{\mathrm{4}} +\mathrm{1}+{k}_{\mathrm{4}} =\mathrm{5}+{k}_{\mathrm{1}} +{k}_{\mathrm{2}} +{k}_{\mathrm{3}} +…+{k}_{\mathrm{5}} \:{with}\:\mathrm{0}\leqslant{k}_{\mathrm{3}} \leqslant\mathrm{4} \\ $$$$\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+…+\mathrm{5}\right)+\mathrm{5}{k}_{\mathrm{1}} +\mathrm{4}{k}_{\mathrm{2}} +\mathrm{3}{k}_{\mathrm{3}} +\mathrm{2}{k}_{\mathrm{4}} +{k}_{\mathrm{5}} ={n} \\ $$$${number}\:{of}\:{solutions}\:{is}\:{the}\:{coef}.\:{of} \\ $$$${term}\:{x}^{{n}} \:{in}\:{the}\:{expansion}\:{of} \\ $$$${x}^{\mathrm{15}} \underset{{r}=\mathrm{1}} {\overset{\mathrm{5}} {\prod}}\frac{\mathrm{1}−{x}^{\mathrm{10}−{r}} }{\mathrm{1}−{x}^{{r}} } \\ $$$${for}\:{n}=\mathrm{20}\:{we}\:{have}\:\mathrm{6}\:{numbers}\:{and} \\ $$$${for}\:{n}=\mathrm{25}\:{we}\:{have}\:\mathrm{12}\:{numbers}\:{etc}. \\ $$
Commented by mr W last updated on 04/Mar/23
Commented by mr W last updated on 04/Mar/23
in case of 6 digit numbers we need to find the coef. of term x^n in x^(21) Π_(r=1) ^6 ((1−x^(10−r) )/(1−x^r )) for n=25 we have 4 numbers and for n=30 we have 8 numbers etc.
$${in}\:{case}\:{of}\:\mathrm{6}\:{digit}\:{numbers}\:{we}\:{need} \\ $$$${to}\:{find}\:{the}\:{coef}.\:{of}\:{term}\:{x}^{{n}} \:{in} \\ $$$${x}^{\mathrm{21}} \underset{{r}=\mathrm{1}} {\overset{\mathrm{6}} {\prod}}\frac{\mathrm{1}−{x}^{\mathrm{10}−{r}} }{\mathrm{1}−{x}^{{r}} } \\ $$$${for}\:{n}=\mathrm{25}\:{we}\:{have}\:\mathrm{4}\:{numbers}\:{and} \\ $$$${for}\:{n}=\mathrm{30}\:{we}\:{have}\:\mathrm{8}\:{numbers}\:{etc}. \\ $$
Commented by mr W last updated on 04/Mar/23

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