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Question Number 71534 by mr W last updated on 16/Oct/19
a number consists of digits 1 and 2.  the sum of its digits is 2018.  if the number is multiplied with 5,   the sum of the digits will be 10000.  find how many digits this number  has.
$${a}\:{number}\:{consists}\:{of}\:{digits}\:\mathrm{1}\:{and}\:\mathrm{2}. \\ $$$${the}\:{sum}\:{of}\:{its}\:{digits}\:{is}\:\mathrm{2018}. \\ $$$${if}\:{the}\:{number}\:{is}\:{multiplied}\:{with}\:\mathrm{5},\: \\ $$$${the}\:{sum}\:{of}\:{the}\:{digits}\:{will}\:{be}\:\mathrm{10000}. \\ $$$${find}\:{how}\:{many}\:{digits}\:{this}\:{number} \\ $$$${has}. \\ $$
Answered by Rasheed.Sindhi last updated on 17/Oct/19
Let 1 comes  x times & 2  y times in  the number.  The sum of the digits=2018       x+2y=2018........................A  The number will equal:  1(10^u_1  +10^u_2  +...+10^u_x  )                +2(10^v_1  +10^v_2  +...+10^v_y  )  5 times the number:     5(10^u_1  +10^u_2  +...+10^u_x  )                +10(10^v_1  +10^v_2  +...+10^v_y  )    = 5(10^u_1  +10^u_2  +...+10^u_x  )                +1(10^(v_1 +1) +10^(v_2 +1) +...+10^(v_y +1) )  The sum of the digits now:             5x+y=10000...................B  From A & B:            x=1998 &  y=10  ∴    The number of digits in the number          =x+y=1998+10=2008
$${Let}\:\mathrm{1}\:{comes}\:\:{x}\:{times}\:\&\:\mathrm{2}\:\:{y}\:{times}\:{in} \\ $$$${the}\:{number}. \\ $$$${The}\:{sum}\:{of}\:{the}\:{digits}=\mathrm{2018} \\ $$$$\:\:\:\:\:{x}+\mathrm{2}{y}=\mathrm{2018}……………………{A} \\ $$$${The}\:{number}\:{will}\:{equal}: \\ $$$$\mathrm{1}\left(\mathrm{10}^{{u}_{\mathrm{1}} } +\mathrm{10}^{{u}_{\mathrm{2}} } +…+\mathrm{10}^{{u}_{{x}} } \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{2}\left(\mathrm{10}^{{v}_{\mathrm{1}} } +\mathrm{10}^{{v}_{\mathrm{2}} } +…+\mathrm{10}^{{v}_{{y}} } \right) \\ $$$$\mathrm{5}\:{times}\:{the}\:{number}: \\ $$$$\:\:\:\mathrm{5}\left(\mathrm{10}^{{u}_{\mathrm{1}} } +\mathrm{10}^{{u}_{\mathrm{2}} } +…+\mathrm{10}^{{u}_{{x}} } \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{10}\left(\mathrm{10}^{{v}_{\mathrm{1}} } +\mathrm{10}^{{v}_{\mathrm{2}} } +…+\mathrm{10}^{{v}_{{y}} } \right) \\ $$$$\:\:=\:\mathrm{5}\left(\mathrm{10}^{{u}_{\mathrm{1}} } +\mathrm{10}^{{u}_{\mathrm{2}} } +…+\mathrm{10}^{{u}_{{x}} } \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{1}\left(\mathrm{10}^{{v}_{\mathrm{1}} +\mathrm{1}} +\mathrm{10}^{{v}_{\mathrm{2}} +\mathrm{1}} +…+\mathrm{10}^{{v}_{{y}} +\mathrm{1}} \right) \\ $$$${The}\:{sum}\:{of}\:{the}\:{digits}\:{now}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{5}{x}+{y}=\mathrm{10000}……………….{B} \\ $$$${From}\:{A}\:\&\:{B}: \\ $$$$\:\:\:\:\:\:\:\:\:\:{x}=\mathrm{1998}\:\&\:\:{y}=\mathrm{10} \\ $$$$\therefore\:\:\:\:{The}\:{number}\:{of}\:{digits}\:{in}\:{the}\:{number} \\ $$$$\:\:\:\:\:\:\:\:={x}+{y}=\mathrm{1998}+\mathrm{10}=\mathrm{2008} \\ $$
Commented by mr W last updated on 17/Oct/19
thanks!
$${thanks}! \\ $$
Answered by MJS last updated on 17/Oct/19
m+2n=2018  (1)  1×5=5  2×5=10 ⇒  5m+n=10000  (2)  ⇒ m=1998 ∧ n=10  ⇒ it has 2008 digits
$${m}+\mathrm{2}{n}=\mathrm{2018}\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{1}×\mathrm{5}=\mathrm{5}\:\:\mathrm{2}×\mathrm{5}=\mathrm{10}\:\Rightarrow \\ $$$$\mathrm{5}{m}+{n}=\mathrm{10000}\:\:\left(\mathrm{2}\right) \\ $$$$\Rightarrow\:{m}=\mathrm{1998}\:\wedge\:{n}=\mathrm{10} \\ $$$$\Rightarrow\:\mathrm{it}\:\mathrm{has}\:\mathrm{2008}\:\mathrm{digits} \\ $$
Commented by mr W last updated on 17/Oct/19
thanks!
$${thanks}! \\ $$

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