1200 followers problem!

First observe that $37 \ | \ 111$ and so $37 \ | \ \overline{aaa}$ .Since $1200$ is a multiple of $3$ , $N_a$ 's decimal representation can be grouped as $400$ parts of $aaa$ .So , when $N_a$ is divided by $\overline{aaa}$ the quotient will be $\overline{1001001001\dots 001001}$ with no remainder. Since $\overline{aaa} \ | \ N_a$ and $37 \ | \ \overline{aaa}$ , we have $37 \ | \ N_a$ . Observe that :

$37 \times 3 = 111 \\ 37 \times 6 = 222 \\ 37 \times 9 = 333 \\ 37 \times 12 = 444 \\ 37 \times 15 = 555 \\ 37 \times 18 = 666 \\ 37 \times 21 = 777 \\ 37 \times 24 = 888 \\ 37 \times 27 = 999$

When $\overline{aaa}=37\delta_a$ where $\delta_a$ is a one digit number of form $3a \forall \ a \ , \ 1\leq a \leq 3$ , the quotient when $N_a$ is divided by $37$ will be of form $\overline{\delta_a00\delta_a00\delta_a\dots00\delta_a}$ where $00\delta_a$ appears $399$ times , which means $0$ appears $(399 \times 2)$ for 1 value of $a$ . Since $1\leq a \leq 3$ we have :

$\Rightarrow\Large\displaystyle\sum_{a=1}^3 \varphi_a= 399 \times 6$

When $\overline{aaa}=37\delta_a$ where $\delta_a$ is a two digit number of form $3a \forall \ a \ , \ 4\leq a \leq 9$ , the quotient when $N_a$ is divided by $37$ will be of form $\overline{\delta_a0\delta_a0\delta_a\dots0\delta_a}$ where $0\delta_a$ appears $399$ times , which means $0$ appears $(399) \ times$ for 1 value of $a$ . Since $4\leq a \leq 9$ , we have:

$\Rightarrow\Large\displaystyle\sum_{a=4}^9 \varphi_a =399 \times 6$

$\Rightarrow\Large\displaystyle\sum_{a=1}^9 \varphi_a = 2(399\times 6) = 399\times 12 = \boxed{4788}$

$aaa/37=3\cdot a$ . For $\frac{1200}{3}-1=399$ of the $aaa$ 's, the leading zeros appear. There are no trailing zeros of $3a$ for $1\le a\le9$ , so just count leading zeros. $1\le a\le3$ gives $2$ leading zeros of $3a$ ; $4\le a\le9$ gives $1$ leading zeros of $3a$ . So it's $399\cdot(2\cdot(3-1+1)+1\cdot(9-4+1))=4788$

Cheated again. I'm totally ashamed. Wait, no I'm not.

a = []

a << ('1' * 1 200).to i

a << ('2' * 1 200).to i

a << ('3' * 1 200).to i

a << ('4' * 1 200).to i

a << ('5' * 1 200).to i

a << ('6' * 1 200).to i

a << ('7' * 1 200).to i

a << ('8' * 1 200).to i

a << ('9' * 1 200).to i

b = []

a.each{|l|b << l / 37}

b.map!{|l|l.to_s.split('')}

c=0

b.each{|lob|

c += lob.inject(0){|cs,l|l == '0' ? cs+1 : cs}

}

puts c

$For ~integer~P_n=nnn, ~~f(n)=\dfrac {nnn}{37}=3*n.\\\text{ THREE member Group A, n=1 2, 3, f(n )is only one digit number,} \\\text { and has TWO leading 0's.} \\ \text{SIX member Group B, n=4, 5, . . . , 9, f(n) is two digit number,} \\ \text {and has ONE leading 0.}\\ \text{ There are 400 groups of nnn in }P_n.\\ \text {In Group A, 2 * 400 " 0s " for each member, BUT} \\ \text{first two "0s" being leading "0s" of } P_n~~itself,~~has~ to~ be~ eliminated. \\\text {So Group A contribute 3 * (2 * 400-2)= (2400-6) " 0s ".} \\ \text {SIX member Group B, in similar fashon,}\\ contribute~~ to~~ 6 * ( 1 * 400 -1)=(2400-6) " 0s ".\\Total~~ 4800-12=~~~~~~~~~~~{ \huge \color{#D61F06}{4788} }~~ " 0s ". \\~~\\$ Generalisation:-\\P_m ~and~ Q_m~\text{are m digit numbers. N, a n digit number is}\\ \text {made up by repeatedly filling P_m or its part from the left.} \\\text {Find how many X are there if n is divided by Q_m.} \\ \text{N should be big enough to be out of reach of a calculator,}\\ \text {Q_m|P_m, and X any of 0 to 9 digits.} \\\text{Example:-{m=2, n=9, X=5}..}\\\text{A 9 digit number N is made up by filled with 51 or its part.} \\ \text{ Find number of 5 in N divided by 3.}\\Solution:-\dfrac {151515151} 3 =50505050.33~\implies~FOUR~~5s