Menu Close

Q-how-many-natural-numbers-less-than-6000-exist-such-as-the-sum-of-their-digits-equal-to-8-choices-155-165-164-158-

Tinku Tara 0 Comments
Pin




Question Number 172802 by mnjuly1970 last updated on 01/Jul/22
Q: how many natural numbers less than 6000 , exist such as the sum of their digits equal to 8 ? choices: 155 165 164 158
$$ \\ $$$$\:\:\:\: \\ $$$$\:\mathrm{Q}:\:\:\:\:\mathrm{how}\:\mathrm{many}\:\mathrm{natural}\:\mathrm{numbers}\:\mathrm{less}\:\mathrm{than} \\ $$$$\:\:\:\:\:\:\mathrm{6000}\:\:,\:\mathrm{exist}\:\mathrm{such}\:\mathrm{as}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{their}\:\mathrm{digits}\:\:\mathrm{equal}\:\mathrm{to}\:\:\:\mathrm{8}\:? \\ $$$$\:\mathrm{choices}:\:\:\:\:\mathrm{155}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{165}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{164}\:\:\:\:\:\:\:\:\:\:\mathrm{158} \\ $$$$\:\:\:\:\:\:\: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by mr W last updated on 02/Jul/22
1−digit numbers: 8 ⇒1 number 2−digit numbers: 17/71,26/62,35/53,44,80 ⇒8 numbers 3−digit numbers: abc with a+b+c=8 and a≥1 and b, c≥0 (x+x^2 +...)(1+x+x^2 +...)^2 =xΣ_(k=0) ^∞ C_2 ^(k+2) x^k ⇒C_2 ^(7+2) =36 numbers 4−digit numbers: abcd with a+b+c+d=8 and 1≤a≤5 and b,c,d≥0 (x+x^2 +x^3 +x^4 +x^5 )(1+x+x^2 +...)^3 =x(1−x^5 )Σ_(k=0) ^∞ C_3 ^(k+3) x^k ⇒C_3 ^(7+3) −C_3 ^(2+3) =110 numbers totally: 1+8+36+110=155 numbers exist. ✓ or (1+x+x^2 +x^3 +x^4 +x^5 )(1+x+x^2 +...)^3 =(1−x^6 )Σ_(k=0) ^∞ C_3 ^(k+3) x^k ⇒C_3 ^(8+3) −C_3 ^(2+3) =155 numbers
$$\underline{\mathrm{1}−{digit}\:{numbers}:} \\ $$$$\mathrm{8}\: \\ $$$$\Rightarrow\mathrm{1}\:{number} \\ $$$$ \\ $$$$\underline{\mathrm{2}−{digit}\:{numbers}:} \\ $$$$\mathrm{17}/\mathrm{71},\mathrm{26}/\mathrm{62},\mathrm{35}/\mathrm{53},\mathrm{44},\mathrm{80}\: \\ $$$$\Rightarrow\mathrm{8}\:{numbers} \\ $$$$ \\ $$$$\underline{\mathrm{3}−{digit}\:{numbers}:} \\ $$$${abc}\:{with}\:{a}+{b}+{c}=\mathrm{8}\:{and} \\ $$$${a}\geqslant\mathrm{1}\:{and}\:{b},\:{c}\geqslant\mathrm{0} \\ $$$$\left({x}+{x}^{\mathrm{2}} +…\right)\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +…\right)^{\mathrm{2}} ={x}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{2}} ^{{k}+\mathrm{2}} {x}^{{k}} \\ $$$$\Rightarrow{C}_{\mathrm{2}} ^{\mathrm{7}+\mathrm{2}} =\mathrm{36}\:{numbers} \\ $$$$ \\ $$$$\underline{\mathrm{4}−{digit}\:{numbers}:} \\ $$$${abcd}\:{with}\:{a}+{b}+{c}+{d}=\mathrm{8}\:{and} \\ $$$$\mathrm{1}\leqslant{a}\leqslant\mathrm{5}\:{and}\:{b},{c},{d}\geqslant\mathrm{0} \\ $$$$\left({x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +{x}^{\mathrm{4}} +{x}^{\mathrm{5}} \right)\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +…\right)^{\mathrm{3}} \\ $$$$={x}\left(\mathrm{1}−{x}^{\mathrm{5}} \right)\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{3}} ^{{k}+\mathrm{3}} {x}^{{k}} \\ $$$$\Rightarrow{C}_{\mathrm{3}} ^{\mathrm{7}+\mathrm{3}} −{C}_{\mathrm{3}} ^{\mathrm{2}+\mathrm{3}} =\mathrm{110}\:{numbers} \\ $$$$ \\ $$$$\underline{{totally}:}\: \\ $$$$\mathrm{1}+\mathrm{8}+\mathrm{36}+\mathrm{110}=\mathrm{155}\:{numbers}\:{exist}.\:\checkmark \\ $$$${or} \\ $$$$\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +{x}^{\mathrm{4}} +{x}^{\mathrm{5}} \right)\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +…\right)^{\mathrm{3}} \\ $$$$=\left(\mathrm{1}−{x}^{\mathrm{6}} \right)\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{3}} ^{{k}+\mathrm{3}} {x}^{{k}} \\ $$$$\Rightarrow{C}_{\mathrm{3}} ^{\mathrm{8}+\mathrm{3}} −{C}_{\mathrm{3}} ^{\mathrm{2}+\mathrm{3}} =\mathrm{155}\:{numbers} \\ $$
Commented by Tawa11 last updated on 01/Jul/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by mr W last updated on 02/Jul/22
you mean the answer is correct Miss?
$${you}\:{mean}\:{the}\:{answer}\:{is}\:{correct}\:{Miss}? \\ $$
Commented by mahdipoor last updated on 02/Jul/22
abcd with a+b+c+d=8 and a,b,c,d≥0 ⇒C_(k−1) ^( n+k−1) =C_3 ^( 11) =165 abcd with a+b+c+d=8 and a,b,c,d≥0 and bigger than 6000 : { ((a=6 ⇒ b+c+d=2 ⇒C_2 ^( 4) =6)),((a=7 ⇒ b+c+d=1 ⇒C_2 ^( 3) =3)),((a=8 ⇒ b+c+d=0 ⇒C_2 ^( 2) =1)) :} ⇒6+3+1=10 165−10=155
$${abcd}\:{with}\:{a}+{b}+{c}+{d}=\mathrm{8}\:{and}\:{a},{b},{c},{d}\geqslant\mathrm{0} \\ $$$$\Rightarrow{C}_{{k}−\mathrm{1}} ^{\:{n}+{k}−\mathrm{1}} ={C}_{\mathrm{3}} ^{\:\mathrm{11}} =\mathrm{165} \\ $$$${abcd}\:{with}\:{a}+{b}+{c}+{d}=\mathrm{8}\:{and}\:{a},{b},{c},{d}\geqslant\mathrm{0} \\ $$$${and}\:{bigger}\:{than}\:\mathrm{6000}\:: \\ $$$$\begin{cases}{{a}=\mathrm{6}\:\Rightarrow\:{b}+{c}+{d}=\mathrm{2}\:\Rightarrow{C}_{\mathrm{2}} ^{\:\mathrm{4}} =\mathrm{6}}\\{{a}=\mathrm{7}\:\Rightarrow\:{b}+{c}+{d}=\mathrm{1}\:\Rightarrow{C}_{\mathrm{2}} ^{\:\mathrm{3}} =\mathrm{3}}\\{{a}=\mathrm{8}\:\Rightarrow\:{b}+{c}+{d}=\mathrm{0}\:\Rightarrow{C}_{\mathrm{2}} ^{\:\mathrm{2}} =\mathrm{1}}\end{cases} \\ $$$$\Rightarrow\mathrm{6}+\mathrm{3}+\mathrm{1}=\mathrm{10} \\ $$$$\mathrm{165}−\mathrm{10}=\mathrm{155} \\ $$
Commented by peter frank last updated on 02/Jul/22
thanks
$$\mathrm{thanks} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *