Find the smallest number greater than zero which can be written with ones and zeroes and is evenly divisble by 225.


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Question Number 515 by 112358 last updated on 25/Jan/15

$F i n d t h e s m a l l e s t n u m b e r g r e a t e r$ $t h a n z e r o w h i c h c a n b e w r i t t e n$ $w i t h o n e s a n d z e r o e s a n d i s e v e n l y d i v i s b l e$ $b y 225 .$
	Answered by prakash jain last updated on 22/Jan/15
	$225=25×9 For 25, 10^2 number is of form n =Σ_(i=1) ^j 10^k_i , k_i ∈N∪{0} n is divisible 9. Let say say k_i >k_(i−1) . It is clear that for smallest n k_1 =0 ((10^k_j +10^k_(j−1) +10^k_1 )/9)=(((10^k_j −1)+(10^k_(j−1) −1)+...+(1+j−1))/9) RHS is divisible by 9 only for j=9 So 9 terms are needed in the sum. For smallest value for n n=Σ_(k=0) ^8 10^k =111111111 Smallest n of the given form for 225. 11111111100$
	$225 = 25 \times 9$ $For 25, 10^{2}$ $number is of form n = \underset{i = 1}{\sum^{j}} 10^{k_{i}}, k_{i} \in N \cup {0}$ $n is divisible 9 .$ $Let say say k_{i} > k_{i - 1} .$ $It is clear that for smallest n k_{1} = 0$ $\frac{10^{k_{j}} + 10^{k_{j - 1}} + 10^{k_{1}}}{9} = \frac{(10^{k_{j}} - 1) + (10^{k_{j - 1}} - 1) + . . . + (1 + j - 1)}{9}$ $RHS is divisible by 9 only for j = 9$ $So 9 terms are needed in the sum .$ $For smallest value for n$ $n = \underset{k = 0}{\sum^{8}} 10^{k} = 111111111$ $Smallest n of the given form for 225 .$ $11111111100$
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