Mathematical-Reasoning · Unit 29 · Video 5 · Interactive Practice

Why the Divisibility Test for 9 Works: Powers of Ten Are 1 Mod 9

IKey Formulas

Formula	Meaning
$n = \sum_{j=0}^{k} d_j \cdot 10^j$ , $\ s(n) = \sum_{j=0}^{k} d_j$	place value and the digit sum of $n$ , with digits $d_k \ldots d_1 d_0$
$10^j \equiv 1 \pmod 9$ for every integer $j \ge 0$	proved by induction from $10 \equiv 1 \pmod 9$ and the product property
$[n] = [s(n)]$ in $\mathbb{Z}_9$ , $\ n \equiv s(n) \pmod 9$	a number and its digit sum leave the same remainder on division by $9$
$9 \mid n \iff 9 \mid s(n)$	the divisibility test for $9$

Key Insight: The whole proof rests on $10 = 9 + 1$ : every power of ten has class $[1]$ in $\mathbb{Z}_9$ , so each digit counts with weight $1$ .

IIFrom Place Value to the Digit Sum

Well-defined $\oplus$ and $\odot$ carry the place-value equation into $\mathbb{Z}_9$ , where every power of ten becomes $[1]$ .

n Step 1

IIIPowers of Ten, One Induction Step at a Time

Each power of ten is the previous one times $10$ , so its remainder is the previous remainder times that of $10$ .

j 3 Modulus

💡 Modulo $11$ the same induction gives $10^j \equiv (-1)^j$ , which is why the test for $11$ uses the alternating digit sum instead.

IVQuiz Questions

Problem 1 · A Remainder Without Long Division

What remainder does $84\,512$ leave on division by $9$ ?

✅ Correct!

s(84\,512) = 20 = 9 \cdot 2 + 2

, and

84\,512 \equiv s(84\,512) \pmod 9

❌ Not quite.

20

is the digit sum; a remainder on division by

9

lies between

0

and

8

. Reduce

20

❌ Not quite. The remainder modulo

9

depends on every digit, not on one digit alone.

❌ Not quite. Add the digits, then reduce the sum modulo

9

Show solution

By the theorem, $n \equiv s(n) \pmod 9$ . Here

s(84\,512) = 8 + 4 + 5 + 1 + 2 = 20 = 9 \cdot 2 + 2,

so the remainder is $2$ . Check: $84\,512 = 9 \cdot 9390 + 2$ .

Problem 2 · The Inductive Step

Assume $10^j \equiv 1 \pmod 9$ . We also know $10 \equiv 1 \pmod 9$ . Which property of congruence gives $10^{j+1} \equiv 1 \pmod 9$ directly?

✅ Correct!

10^{j+1} = 10^j \cdot 10 \equiv 1 \cdot 1 = 1 \pmod 9

❌ Not quite. Adding gives

10^j + 10 \equiv 2

, which is not a statement about

10^{j+1}

❌ Not quite. Symmetry only reverses one congruence; it never reaches the exponent

j + 1

❌ Not quite. The Division Algorithm finds remainders, but the step needs a rule for combining two congruences.

Show solution

The product property says: if $x \equiv y$ and $z \equiv w \pmod 9$ , then $xz \equiv yw \pmod 9$ . With $x = 10^j$ , $y = 1$ , $z = 10$ , $w = 1$ :

10^{j+1} = 10^j \cdot 10 \equiv 1 \cdot 1 = 1 \pmod 9.

Together with the base case $10^0 = 1$ , induction gives $10^j \equiv 1 \pmod 9$ for every $j \ge 0$ .

Problem 3 · A Missing Digit

The four-digit number $4a72$ (thousands digit $4$ , hundreds digit $a$ , tens digit $7$ , units digit $2$ ) is divisible by $9$ . Find the digit $a$ .

✅ Correct!

4 + 5 + 7 + 2 = 18

, a multiple of

9

, and

4572 = 9 \cdot 508

❌ Not quite.

4 + 4 + 7 + 2 = 17

is not a multiple of

9

❌ Not quite.

4 + 0 + 7 + 2 = 13

is not a multiple of

9

❌ Not quite. The digit sum

13 + a

must be a multiple of

9

Show solution

By the test, $9 \mid 4a72$ exactly when $9 \mid s(4a72) = 13 + a$ . Since $0 \le a \le 9$ , the sum lies between $13$ and $22$ , and the only multiple of $9$ there is $18$ . So $a = 5$ , and indeed $4572 = 9 \cdot 508$ .

Problem 4 · The Same Proof Modulo 3

Since $10 \equiv 1 \pmod 3$ , the same argument shows $n \equiv s(n) \pmod 3$ . What remainder does $7319$ leave on division by $3$ ?

✅ Correct!

s(7319) = 20 = 3 \cdot 6 + 2

, and

7319 = 3 \cdot 2439 + 2

❌ Not quite.

20

is not a multiple of

3

, so

3 \nmid 7319

❌ Not quite.

20 = 3 \cdot 6 + 2

; recheck the reduction.

❌ Not quite.

20

is the digit sum, not a remainder on division by

3

Show solution

Every power of ten is $\equiv 1 \pmod 3$ (the same induction, using $10 \equiv 1 \pmod 3$ ), so $7319 \equiv 7 + 3 + 1 + 9 = 20 \pmod 3$ . Since $20 = 3 \cdot 6 + 2$ , the remainder is $2$ .

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