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.


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Question Number 71534 by mr W last updated on 16/Oct/19

$a n u m b e r c o n s i s t s o f d i g i t s 1 a n d 2 .$ $t h e s u m o f i t s d i g i t s i s 2018 .$ $i f t h e n u m b e r i s m u l t i p l i e d w i t h 5,$ $t h e s u m o f t h e d i g i t s w i l l b e 10000 .$ $f i n d h o w m a n y d i g i t s t h i s n u m b e r$ $h a s .$
	Answered by Rasheed.Sindhi last updated on 17/Oct/19

	$L e t 1 c o m e s x t i m e s & 2 y t i m e s i n$ $t h e n u m b e r .$ $T h e s u m o f t h e d i g i t s = 2018$ $x + 2 y = 2018 . . . . . . . . . . . . . . . . . . . . . . . . A$ $T h e n u m b e r w i l l e q u a l :$ $1 (10^{u_{1}} + 10^{u_{2}} + . . . + 10^{u_{x}})$ $+ 2 (10^{v_{1}} + 10^{v_{2}} + . . . + 10^{v_{y}})$ $5 t i m e s t h e n u m b e r :$ $5 (10^{u_{1}} + 10^{u_{2}} + . . . + 10^{u_{x}})$ $+ 10 (10^{v_{1}} + 10^{v_{2}} + . . . + 10^{v_{y}})$ $= 5 (10^{u_{1}} + 10^{u_{2}} + . . . + 10^{u_{x}})$ $+ 1 (10^{v_{1} + 1} + 10^{v_{2} + 1} + . . . + 10^{v_{y} + 1})$ $T h e s u m o f t h e d i g i t s n o w :$ $5 x + y = 10000 . . . . . . . . . . . . . . . . . . . B$ $F r o m A & B :$ $x = 1998 & y = 10$ $∴ T h e n u m b e r o f d i g i t s i n t h e n u m b e r$ $= x + y = 1998 + 10 = 2008$
	Commented by mr W last updated on 17/Oct/19

	$t h a n k s!$
	Answered by MJS last updated on 17/Oct/19

	$m + 2 n = 2018 (1)$ $1 \times 5 = 5 2 \times 5 = 10 \Rightarrow$ $5 m + n = 10000 (2)$ $\Rightarrow m = 1998 \land n = 10$ $\Rightarrow it has 2008 digits$
	Commented by mr W last updated on 17/Oct/19

	$t h a n k s!$
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