CDS Mathematics · Number System

Divisibility Rules & Missing Digits

The tests that decide divisibility from a number's written form alone — digit sums for 3 and 9, alternating sums for 11, the last few digits for the powers of 2 and 5 — and how to run them backwards to recover a digit the paper has hidden.

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Why this matters

Nine PYQs, two of them HARD, and the smallest unit in the chapter by count — but it is pure mechanism, so it is close to guaranteed marks. Every question is one of two shapes: apply a test, or apply a test in reverse to find a hidden digit. The traps are all in the wording, not the mathematics.

Concept 1 of 4: The divisibility test table

Every one of these tests exists because of what

10

does modulo the divisor. For 3 and 9,

10 \equiv 1

, so each digit contributes its face value and the digit sum decides. For 11,

10 \equiv -1

, so digits alternate in sign. For the powers of 2 and 5,

10^k

is itself a multiple, so only the last few digits survive. Knowing the reason means you never mix up which test belongs to which divisor.

Definition

Learn the table below cold. Two structural facts make it much shorter than it looks:

For a composite divisor, test its coprime prime-power parts separately: $6 = 2\times3$ , $12 = 4\times3$ , $33 = 3\times11$ , $45 = 9\times5$ .
Testing $2$ and $3$ is not enough for $12$ ; you need $4$ and $3$ , because $12$ carries two factors of 2.

Divisor	Test	Reason
2	last digit is even	$10 \equiv 0 \pmod 2$
3	digit sum divisible by 3	$10 \equiv 1 \pmod 3$
4	last two digits divisible by 4	$100 \equiv 0 \pmod 4$
5	last digit is 0 or 5	$10 \equiv 0 \pmod 5$
6	passes both the 2 and 3 tests	$6 = 2\times 3$ , coprime parts
8	last three digits divisible by 8	$1000 \equiv 0 \pmod 8$
9	digit sum divisible by 9	$10 \equiv 1 \pmod 9$
10	last digit is 0	$10 \equiv 0 \pmod{10}$
11	alternating digit sum divisible by 11	$10 \equiv -1 \pmod{11}$ Alternate the signs from the units digit leftwards. A result of $0$ counts as divisible.
16	last four digits divisible by 16	$10^4 \equiv 0 \pmod{16}$
25	last two digits are 00, 25, 50 or 75	$100 \equiv 0 \pmod{25}$
12	passes the 4 and 3 tests	$12 = 4\times 3$ , not $2\times 6$ Testing 2 and 6 is wrong: 18 passes both and is not a multiple of 12.
7 and 13	no short test worth learning	$10$ has order 6 modulo both

Thirteen rows. The reason column is not decoration — it is what tells you how many trailing digits a power-of-2 test needs.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2020 · CDS (II) 2020 — Elementary Mathematics · Q6Moderate

Example 1 · Number System · Divisibility Rules and Missing Digits

If the number 23P62971335 is divisible by the smallest odd composite number, then what is the value of P ?

The smallest odd composite number is 9, not 1, 3 or 15

CDS hides the divisor behind a description.

1

is neither prime nor composite, and

3, 5, 7

are all prime — so the smallest odd composite is

9

. Get that wrong and you run a perfectly correct digit-sum test against the wrong divisor.

For a composite divisor, split into COPRIME parts

To test 12, use 4 and 3 — not 2 and 6. The parts must be coprime and must multiply to the divisor, otherwise you lose a factor: 18 passes the 2-test and the 6-test yet is not divisible by 12. Same trap for 8 (use 8 directly, not 2 and 4).

Drill 2 more on the divisibility test table

Concept 2 of 4: Recovering a hidden digit from a divisibility condition

A test that turns a number into a digit sum can be run in reverse: if the sum must be a multiple of 9 and you know all the digits but one, the missing digit is forced. The only subtlety is that sometimes two values of the digit work, and the question expects you to notice.

Definition

Procedure for a hidden digit $P$ under a 3- or 9-condition:

Add the known digits to get a partial sum $s$ .
Require $s + P \equiv 0$ modulo 3 or 9, so $P \equiv -s$ .
Solve within $0 \le P \le 9$ and list every solution; modulo 3 there are usually three, modulo 9 usually one or two.

With two hidden digits $A$ and $B$ , the condition fixes only $A+B$ , so the answer is a count of digit pairs — and any extra condition (B is odd, $A+B \le 5$ ) prunes that list.

Hidden-digit condition

P \equiv -\!\!\sum(\text{known digits}) \pmod{9}, \qquad 0 \le P \le 9

Worked example

The number

58P41

is divisible by 9. Find

P

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2019 · CDS (II) 2019 — Elementary Mathematics · Q3Moderate

Example 2 · Number System · Divisibility Rules and Missing Digits

How many pairs (A, B) are possible in the number 479865AB if the number is divisible by 9 and it is given that the last digit of the number is odd ?

A mod-9 condition often has TWO digit solutions, not one

If the required residue is

0

, both

P=0

and

P=9

satisfy it. A question asking "the value of

P

" when two exist is asking you to notice; one asking for a count is counting both. Always solve the congruence and then enumerate the range rather than stopping at the first hit.

With two hidden digits the condition fixes only their SUM

9 \mid N

pins

A+B

to a residue, never

A

and

B

individually — so the answer is a count of pairs. Remember that

A+B=0

is a legitimate total (both digits zero) and is divisible by 9, which is the case students drop.

Drill 1 more on recovering a hidden digit from a divisibility condition

Concept 3 of 4: Divisors for which only the tail of the number matters

1000

is a multiple of 8, so every digit above the hundreds place contributes nothing to divisibility by 8. The same logic scales: to test

2^k

5^k

you may delete all but the last

k

digits, however monstrous the number is. A 190-digit number can be settled by looking at four digits.

Definition

Because $10^k = 2^k 5^k$ , we have $10^k \equiv 0$ modulo $2^k$ and modulo $5^k$ . So for these divisors, $N$ and its last $k$ digits leave the same remainder:

$4$ and $25$ : last two digits;
$8$ and $125$ : last three digits;
$16$ and $625$ : last four digits.

This works for remainders, not just for a yes-or-no answer — the remainder of $N$ on division by 16 equals the remainder of its last four digits.

Tail rule

N \equiv \left(N \bmod 10^{k}\right) \pmod{2^{k}} \quad\text{and}\quad \pmod{5^{k}}

Worked example

What is the remainder when the 9-digit number

473816952

is divided by 8?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2024 · CDS (I) 2024 — Elementary Mathematics · Q19Moderate

Example 3 · Number System · Divisibility Rules and Missing Digits

When the number (12345678910111213 ... 99100) is divided by 16, what will be the remainder ?

The tail rule works only for divisors built from 2s and 5s

It fails the moment a factor of 3 or 7 appears, because

10^k

is then not a multiple of the divisor. There is no "last two digits" test for 12 or 24. If the divisor is

2^a 5^b m

with

m>1

, split it: handle

2^a5^b

by the tail and

m

by its own rule.

Reading the tail of a described number is where this goes wrong

When the number is described rather than printed — "write 1 to 100 in order" — the hard part is working out what the last four digits actually are. The string ends

\ldots 99\,100

, so the final four characters are

9100

, not

0099

9910

. Write out the tail explicitly before dividing.

Drill 1 more on divisors for which only the tail of the number matters

Concept 4 of 4: What to do when the divisor has no usable test

There is no digit-sum test for 7 or 13 worth memorising, and CDS knows it. When one of those turns up with a hidden digit, the intended method is not a clever rule — it is structured trial: reduce the problem to one unknown digit and test at most ten candidates, or exploit a structural factorisation instead.

Definition

Two workable routes when the divisor is awkward:

Bounded trial. If the condition leaves a single unknown digit, testing $0$ through $9$ is only ten divisions and the answer is unique. This is a legitimate exam method, not a fallback.
Structural factorisation. Rewrite the number in a form whose factors you can see. $1001 = 7\times 11 \times 13$ is the most useful such fact: any six-digit repeated block is automatically divisible by 7 and 13, and grouping digits in threes reduces a long number modulo 7, 11 and 13 at once.

For a counting question over a whole range, look for the constraint that shrinks the search before enumerating.

The 1001 grouping fact

1001 = 7 \times 11 \times 13

Worked example

861

divisible by 7, and use the 1001 fact to decide whether

861861

is divisible by 13.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2021 · CDS (I) 2021 — Elementary Mathematics · Q1Hard

Example 4 · Number System · Divisibility Rules and Missing Digits

If the number 413283P759387 is divisible by 13, then what is the value of P ?

Trial is the intended method here, so do not hunt for a rule

On a 13-divisibility question with one hidden digit, students lose two or three minutes trying to recall a test that does not exist. Ten divisions is faster and certain. The exam-craft point is recognising immediately that there is no rule to recall, which is why 7 and 13 have a row of their own in the table above.

Drill 1 more on what to do when the divisor has no usable test

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Recovering a hidden digit from a divisibility condition
Hidden-digit condition
$P \equiv -\!\!\sum(\text{known digits}) \pmod{9}, \qquad 0 \le P \le 9$
Divisors for which only the tail of the number matters
Tail rule
$N \equiv \left(N \bmod 10^{k}\right) \pmod{2^{k}} \quad\text{and}\quad \pmod{5^{k}}$
What to do when the divisor has no usable test
The 1001 grouping fact
$1001 = 7 \times 11 \times 13$

Reference tables (1)

The divisibility test table13 rows

Divisor	Test	Reason
2	last digit is even	$10 \equiv 0 \pmod 2$
3	digit sum divisible by 3	$10 \equiv 1 \pmod 3$
4	last two digits divisible by 4	$100 \equiv 0 \pmod 4$
5	last digit is 0 or 5	$10 \equiv 0 \pmod 5$
6	passes both the 2 and 3 tests	$6 = 2\times 3$ , coprime parts
8	last three digits divisible by 8	$1000 \equiv 0 \pmod 8$
9	digit sum divisible by 9	$10 \equiv 1 \pmod 9$
10	last digit is 0	$10 \equiv 0 \pmod{10}$
11	alternating digit sum divisible by 11	$10 \equiv -1 \pmod{11}$ Alternate the signs from the units digit leftwards. A result of $0$ counts as divisible.
16	last four digits divisible by 16	$10^4 \equiv 0 \pmod{16}$
25	last two digits are 00, 25, 50 or 75	$100 \equiv 0 \pmod{25}$
12	passes the 4 and 3 tests	$12 = 4\times 3$ , not $2\times 6$ Testing 2 and 6 is wrong: 18 passes both and is not a multiple of 12.
7 and 13	no short test worth learning	$10$ has order 6 modulo both

Thirteen rows. The reason column is not decoration — it is what tells you how many trailing digits a power-of-2 test needs.

Watch out for (7)

The smallest odd composite number is 9, not 1, 3 or 15
→ The divisibility test table
For a composite divisor, split into COPRIME parts
→ The divisibility test table
A mod-9 condition often has TWO digit solutions, not one
→ Recovering a hidden digit from a divisibility condition
With two hidden digits the condition fixes only their SUM
→ Recovering a hidden digit from a divisibility condition
The tail rule works only for divisors built from 2s and 5s
→ Divisors for which only the tail of the number matters
Reading the tail of a described number is where this goes wrong
→ Divisors for which only the tail of the number matters
Trial is the intended method here, so do not hunt for a rule
→ What to do when the divisor has no usable test

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Related notes

Division, parity and consecutive integers →