littletush wrote: Decimal digits $a,b,c$ satisfy \[ 37\mid (a0a0\ldots a0b0c0c\ldots 0c)_{10} \] where there are $1001$ a's and $1001$ c's. Prove that $b=a+c$. Just write the number as $N=a\frac{10^{4006}-10^{2004}}{99}$ $+b10^{2002}$ $+c\frac{10^{2002}-1}{99}$ Then notice that $10^3\equiv 1\pmod {37}$ and that $3\times 99\equiv 1\pmod{37}$ So $N\equiv a\times 3\times(10-1)+b\times 10+c\times 3(10-1)\pmod{37}$ So $N\equiv 27(a+c-b)\pmod{37}$ Hence the result.