SSC CGL Maths · Number System

Missing Digits

A hidden digit is found by writing the divisibility test as a condition on the unknown and keeping only the digits 0 to 9 that meet it.

All Number System notes

Why this matters

Nineteen questions in the bank, fifteen EASY and four MODERATE, fifteen of them from the 2024 shifts. The unknown is a single digit, so the test leaves at most a handful of values, and the word 'least' or 'greatest' in the stem picks one.

Concept 1 of 3: A missing digit from a digit-sum test

For 3 and 9 only the digit sum matters. Add the digits you can see, and the hidden digit must top that sum up to the next multiple of 3 or 9.

Definition

Add the known digits; call it $S$ . The digit $x$ must make $S + x$ a multiple of 9 (or of 3), with $0 \le x \le 9$ .
For 9 there is one answer, except when $S$ is already a multiple of 9: then $x$ is 0 or 9.
For 3 there are three or four answers, 3 apart (for example 1, 4, 7). The stem's 'least' or 'greatest' decides.
For 6, 18 or 36 combine two tests: 6 is even and 3, 18 is even and 9, 36 is 4 and 9. Run the last-digit test first; it often fixes the digit or its parity.
Two unknowns $x$ and $y$ : the digit sum gives $x + y$ as one of two values 9 apart. A second condition picks one: $x + y$ and $x - y$ are always both even or both odd.
The same letter in two places is the same digit. A digit can never be 10 or more, and the first digit of a number is never 0.

Hidden digit from a digit sum

S + x = 9k \ (\text{or } 3k), \quad 0 \le x \le 9

Sthe sum of the digits you can see
xthe hidden digit

Worked example

Find the digit k so that 5k4632 is divisible by 9. What is the least k that makes it divisible by 3?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2024 · Tier I, 17 Sep 2024, Shift 2 · Q51Easy

Example 1 · Number System · Missing Digits

Find the least value of x for which 57x716 is divisible by 9.

Accepting a value that is not a digit

Options sometimes include 10 or 12. A missing digit lies between 0 and 9, so throw those out before testing anything.

Missing the second answer for 9

If the visible digits already add to a multiple of 9, the hidden digit can be 0 or 9. Read whether the stem wants the least or the greatest.

Drill 10 more on a missing digit from a digit-sum test

Concept 2 of 3: A missing digit from a last-digits test

For 4, 5 and 8 only the tail of the number matters, so the hidden digit is found by trying the ten possible tails. A small fact about the hundreds digit cuts the trying down for 8.

Definition

5 and 8 together force a last digit of 0: 5 needs 0 or 5, and a multiple of 8 is even.
8: the last three digits $\overline{abc}$ must be a multiple of 8. Since 200 is a multiple of 8 and 100 leaves 4, look at the hundreds digit $a$ : if $a$ is even, $\overline{bc}$ must be a multiple of 8; if $a$ is odd, $\overline{bc} + 4$ must be.
4: the last two digits $\overline{bc}$ must be a multiple of 4, which happens exactly when $2b + c$ is a multiple of 4.
40 is $8 \times 5$ : the number ends in 0, and then the 8 test runs on the last three digits.
To count numbers of a pattern, fix one digit at a time and list the others that pass.

Last-digit tests for 8 and 4

\overline{abc} \text{ is a multiple of } 8: \ a \text{ even} \Rightarrow 8 \text{ divides } \overline{bc}; \ a \text{ odd} \Rightarrow 8 \text{ divides } \overline{bc} + 4 \qquad 4 \text{ divides } \overline{bc} \text{ exactly when } 4 \text{ divides } 2b + c

$\overline{abc}$ the last three digits, with a in the hundreds place
$\overline{bc}$ the last two digits read as a two-digit number

Worked example

Find the smallest digit q for which 591q6 is divisible by 8.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2024 · Tier I, 26 Sep 2024, Shift 2 · Q55Moderate

Example 2 · Number System · Missing Digits

The five-digit number 45yz0 is divisible by 40. What is the maximum possible value of (y + z)?

Letting a multiple of 8 and 5 end in 5

A number ending in 5 is odd, and no odd number is a multiple of 8. Divisible by both 5 and 8 means the last digit is 0.

Running the 8 test on two digits

The tail for 8 is three digits long. 136 and 236 share the last two digits 36, but only 136 is a multiple of 8; the hundreds digit decides.

Drill 5 more on a missing digit from a last-digits test

Concept 3 of 3: A missing digit from the test for 11

Write the alternating difference with the unknown digit in it. Since the digit is only 0 to 9, the difference can reach just one or two multiples of 11, and each gives one value.

Definition

Count places from the right. Put the hidden digit in the odd-place sum or the even-place sum, wherever it sits.
Write $D$ = (odd-place sum) $-$ (even-place sum) in terms of the digit, then set $D$ equal to 0, 11 or $-11$ .
Keep only the value from 0 to 9. When the known part of $D$ is large, the answer comes from $D = 11$ or $D = -11$ , not from 0.

Worked example

Find the digit t so that 52t817 is divisible by 11.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2024 · Tier I, 9 Sep 2024, Shift 3 · Q59Easy

Example 3 · Number System · Missing Digits

If 7845K854 is divisible by 11, then what is the value of K?

Setting the difference to 0 and stopping

D = 0

needs a digit of 10 or more, try

D = 11

and

D = -11

. A digit that makes the difference 11 is just as valid.

Drill 1 more on a missing digit from the test for 11

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 (2)

A missing digit from a digit-sum test
Hidden digit from a digit sum
$S + x = 9k \ (\text{or } 3k), \quad 0 \le x \le 9$
A missing digit from a last-digits test
Last-digit tests for 8 and 4
$\overline{abc} \text{ is a multiple of } 8: \ a \text{ even} \Rightarrow 8 \text{ divides } \overline{bc}; \ a \text{ odd} \Rightarrow 8 \text{ divides } \overline{bc} + 4 \qquad 4 \text{ divides } \overline{bc} \text{ exactly when } 4 \text{ divides } 2b + c$

Watch out for (5)

Accepting a value that is not a digit
→ A missing digit from a digit-sum test
Missing the second answer for 9
→ A missing digit from a digit-sum test
Letting a multiple of 8 and 5 end in 5
→ A missing digit from a last-digits test
Running the 8 test on two digits
→ A missing digit from a last-digits test
Setting the difference to 0 and stopping
→ A missing digit from the test for 11

Test yourself on Number System

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