SOLUTION TO PROBLEM 1

Since

a₁=2001²⁰⁰¹<(10⁴)²⁰⁰¹=10⁸⁰⁰⁴,

we plainly have $a_2\leq 8004\cdot 9=72036$. Therefore, $a_3=S(a_2)\leq 9\cdot 5 = 45$, and hence $a_4\leq 3+9=12$. Consequently $a_5\leq 9$. Also clearly $a_5\geq 1$.

Now we note that

Claim:        For every positive integer m, the difference m-S(m) is divisible by 9.

Proof of the claim:         Write the integer m as the sum

$$m=a_0+a_1\cdot 10+a_2\cdot 10^2+\ldots+a_{k-1}\cdot 10^{k-1}+a_k\cdot 10^k.$$

Then

$$\begin{array}{r} m-S(m)=(a_0+a_1\cdot 10+\ldots+a_k\cdot 10^k... ...-1)+a_{k-1}\cdot (10^{k-1}-1)+ a_k\cdot (10^k-1). \end{array}$$

Since, for a positive integer $\ell$, $10^\ell-1=9\cdot (1+10+\ldots+ 10^{\ell-1})$, we conclude that m-S(m) is divisible by 9.

Next we note that the number 2001 is divisible by 3, so its power 2001²⁰⁰¹ is divisible by 9 - just note that $2001^{2001}=(3\cdot 667)^2 \cdot 2001^{1999}$. (As a matter of fact the power is divisible by 3²⁰⁰¹ - clearly). By the Claim, a₂ is divisible by 9, and also a₃, a₄ and a₅ are.

Putting things together we get that a₅ is a divisible by 9 integer between 1 and 9. Therefore a₅=9

Questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified:   Fri Jan 12 21:59:53 CST 2001