Divisibility Rules and Missing Digits

Divisibility Rules from 2 to 125

A divisibility rule tells you whether one number divides another without doing the long division. Many CGL Tier 1 number questions from 2022 to 2024 were built on these rules, so learn them until you can use them without thinking.

Why the end-digit rules work. 10 = 2 × 5, 100 = 4 × 25, 1000 = 8 × 125 and 10000 = 16 × 625. Everything in front of the last three digits is a multiple of 1000, so it is already a multiple of 8. Only the tail needs checking. That gives an easy pattern to remember: for 2, 4, 8 and 16 look at the last 1, 2, 3 and 4 digits, and for 5, 25 and 125 look at the last 1, 2 and 3 digits.

The rule for 11 in detail. Number the places from the right: the units digit is in place 1, the tens digit in place 2, and so on. Add the digits in the odd places, add the digits in the even places, and subtract. For 9,28,147 the odd places hold 7, 1 and 2 (sum 10) and the even places hold 4, 8 and 9 (sum 21). The difference is 21 − 10 = 11, so 9,28,147 is divisible by 11. It equals 11 × 84,377. A difference of 0 counts too.

[SSC Tip] A faster check for 8. If the hundreds digit is even, the last two digits on their own must be divisible by 8. If the hundreds digit is odd, the last two digits plus 4 must be divisible by 8. For 59,136 the hundreds digit 1 is odd, and 36 + 4 = 40 = 8 × 5, so it passes. This works because 200 is a multiple of 8 and 100 leaves remainder 4.

Key Point: The rules for 2, 4, 8, 16, 5, 25 and 125 look only at the end of the number. The rules for 3, 9 and 11 need every digit.

Working Rules for 7, 13, 17 and 19

These four have no end-digit shortcut, but one method handles all of them. Cut off the last digit, multiply it by a fixed number, and add it to or subtract it from the number that is left. Repeat until the number is small enough to recognise. The new number is divisible by the divisor exactly when the old one is.

Test 3,39,213 for 7.

• 33921 − 2 × 3 = 33915
• 3391 − 2 × 5 = 3381
• 338 − 2 × 1 = 336
• 33 − 2 × 6 = 21

21 = 7 × 3, so 3,39,213 is divisible by 7.

Test 2,47,741 for 13.

• 24774 + 4 × 1 = 24778
• 2477 + 4 × 8 = 2509
• 250 + 4 × 9 = 286
• 28 + 4 × 6 = 52

52 = 13 × 4, so 2,47,741 is divisible by 13.

Test 39,865 for 17. 3986 − 5 × 5 = 3961, then 396 − 5 × 1 = 391, then 39 − 5 × 1 = 34 = 17 × 2. Divisible.

Test 86,773 for 19. 8677 + 2 × 3 = 8683, then 868 + 2 × 3 = 874, then 87 + 2 × 4 = 95 = 19 × 5. Divisible.

How to remember the multipliers. Think of a multiple of the divisor that sits next to a round number. 21 = 20 + 1, so for 7 multiply by 2 and subtract. 39 = 40 − 1, so for 13 multiply by 4 and add. 51 = 50 + 1, so for 17 multiply by 5 and subtract. 19 = 20 − 1, so for 19 multiply by 2 and add. One more than a round number means subtract; one less means add. The same idea works for 23 (69 = 70 − 1, so multiply by 7 and add) and for any divisor ending in 1, 3, 7 or 9.

The 1001 method for 7, 11 and 13 together. 1001 = 7 × 11 × 13. Split the number into groups of three digits from the right, then add and subtract the groups alternately, starting with a plus on the rightmost group. If the result is divisible by 7, 11 or 13, so is the number. For 4516694 the groups are 4, 516 and 694, and 694 − 516 + 4 = 182 = 2 × 7 × 13. So 4516694 is divisible by 7 and by 13, but not by 11.

In an options question, run the easy tests (2, 3, 5, 11) first. The chain method is for the one or two options left over, and with practice it takes about fifteen seconds per number.

Composite Divisors: Split into Co-prime Factors

To test a number for a composite divisor, break the divisor into factors that are co-prime (their only common factor is 1) and check each factor separately. The number must pass every part.

A quick run-through. Is 5,24,392 divisible by 88? For 8, the last three digits are 392 = 8 × 49, so yes. For 11, the digits from the right are 2, 9, 3, 4, 2, 5. Odd places: 2 + 3 + 2 = 7. Even places: 9 + 4 + 5 = 18. The difference is 11, so yes. It passes both, so 5,24,392 is divisible by 88.

The 24 trap. 24 = 4 × 6 is true, but 4 and 6 share the factor 2, so checking 4 and 6 is not enough. 36 is divisible by 4 and by 6, yet 36 ÷ 24 = 1.5. A number that passes 4 and 6 is only sure to be divisible by their LCM, which is 12. For 24 always use 3 × 8.

The same mistake turns up with other splits:

Why co-prime works. If two co-prime numbers both divide a number, their product divides it too. If they share a factor, only their LCM is guaranteed. (LCM is covered in Section 5.)

For 99 there is an even faster route. Split the number into pairs of digits from the right and add the pairs. If the total is divisible by 99, so is the number. For 5,13,216 the pairs are 16, 32 and 51, which add up to 99, so 5,13,216 is divisible by 99 (it is 99 × 5184). This works because 100 leaves remainder 1 when divided by 99.

Missing-Digit Questions

A missing-digit question gives a number with one or two unknown digits and a divisibility condition. The method is the same every time.

1. Turn each rule into a condition on the unknowns. Digit sums for 3 and 9, alternate sums for 11, the tail for 2, 4, 5, 8 and 25.
2. Start with the rule that fixes a digit fastest. End-digit rules (5, 10, 25, 8) usually pin down the last digit straight away. Then use the digit sum or the 11 rule for the other unknown.
3. Remember the limits. Each unknown is a single digit from 0 to 9, and the first digit of a number cannot be 0.
4. Apply the extra condition last: smallest, largest, $Y - X = 6$, both digits prime, and so on.

One unknown. If $3x5217$ is divisible by 9, the digit sum 3 + 5 + 2 + 1 + 7 + $x$ = $18 + x$ must be a multiple of 9. So $x = 0$ or $x = 9$, and the question has to tell you which one it wants (the smallest or the largest). For 3 there are usually three or four answers that differ by 3 (like 1, 4 and 7), so you will almost always be asked for the smallest or the largest.

With the 11 rule, set the difference equal to 0 first. If that gives something outside 0 to 9, try 11 or −11.

Two unknowns. Use one rule for each unknown, or two rules that together give two equations.

• Divisible by 8 and 5 together means divisible by 40. The last digit must be 0, and then the last three digits must be divisible by 8.
• Divisible by 9 with a condition like $Y - X = 6$. The digit sum gives the possible values of $X + Y$ (often two choices, such as 3 or 12). Pair each one with $Y - X = 6$. Since $X + Y$ and $Y - X$ are always both even or both odd, you can throw out a choice with the wrong parity at once.
• Divisible by 99: use 9 to get $X + Y$ and 11 to get $X - Y$, then solve.
• Divisible by 72: 8 first (last three digits), then 9.
• Divisible by 75: 25 first (the last two digits must be 00, 25, 50 or 75), then 3.
• Divisible by 36: 4 first (last two digits), then 9.
• Digits must be prime: only 2, 3, 5 and 7 are allowed.
• Maximise or minimise something like $2A + 3B$: list every valid pair, then pick. Give the digit with the bigger coefficient the bigger value wherever you can, and check the next pair down to be sure.

[SSC Tip] When the options give values, test them instead of solving. If the question asks for the largest $x$ and the options are 2, 5, 8 and 9, try 9 first and work down. This often saves half a minute.

Checking Options Fast, and the Least Number to Add or Subtract

"Which of these is divisible by …?" Split the divisor into co-prime parts and run the cheapest test first. Most options fail an easy test, and you never need the hard one. For 72 (8 × 9) the digit sum is quicker than the 8 test, so check 9 first and then 8 on whatever is left. For 88 (8 × 11) the 8 test on the last three digits usually knocks out two or three options before you need the 11 rule.

"Which of these is NOT divisible by …?" One failed test is enough. Setters like a near miss here: a number that passes 9 and 2 but fails 4, so it is divisible by 18 and not by 36. Check every part of the split, not just the first.

Least number to subtract or add. Divide and find the remainder $r$.

• To make $N$ divisible by $d$ by subtracting, subtract $r$.
• To do it by adding, add $d - r$ (or nothing if $r = 0$).

For 5000 and 17: 5000 = 17 × 294 + 2. So subtract 2 (to get 4998) or add 15 (to get 5015).

Nearest multiple. Compare $r$ with $d - r$ and take the smaller. The multiple of 17 nearest to 5000 is 4998, only 2 away.

Remainders by 9 and 3 from the digit sum. A number leaves the same remainder on division by 9 as its digit sum does. 7,84,326 has digit sum 30, and 30 ÷ 9 leaves 3, so 7,84,326 leaves remainder 3 when divided by 9. To make it divisible by 9, subtract 3 or add 6. The same trick works for 3. Remainders are covered fully in Section 4.

Largest and smallest n-digit multiples.

• Largest: divide the largest n-digit number (9999 for four digits) by $d$ and subtract the remainder. 9999 = 37 × 270 + 9, so the largest four-digit multiple of 37 is 9999 − 9 = 9990.
• Smallest: divide the smallest n-digit number (1000 for four digits) by $d$ and add $d - r$. 1000 = 37 × 27 + 1, so the smallest four-digit multiple of 37 is 1000 + 36 = 1036.

When a question names several divisors ("the largest four-digit number divisible by 12, 15 and 18"), find their LCM first and then use the same steps. That is covered in Section 5.

[SSC Tip] "Least number to add" and "least number to subtract" give different answers, and both usually sit among the options. Read which one is asked before you mark.

Counting Multiples in a Range

From 1 to N. The number of multiples of $k$ from 1 to $N$ is the whole-number part of $\frac{N}{k}$. From 1 to 500 there are 71 multiples of 7, because 500 ÷ 7 = 71.4…

From A to B. Find the first and last multiples inside the range. Then

$$\text{count} = \frac{\text{last} - \text{first}}{k} + 1$$

Multiples of 7 from 100 to 500: the first is 105 and the last is 497, so the count is (497 − 105) ÷ 7 + 1 = 56 + 1 = 57. You can also take the 71 multiples up to 500 and remove the 14 up to 99, which again gives 57.

"Between" or "from … to". "From 140 to 280" includes both ends. "Between 140 and 280" usually leaves them out. This only matters when an end point is itself a multiple, so look at the end points. Multiples of 7 from 140 to 280 inclusive: (280 − 140) ÷ 7 + 1 = 21. Strictly between them: 19.

Divisible by both. A number divisible by both $a$ and $b$ is divisible by their LCM. From 1 to 1000, the numbers divisible by both 6 and 9 are the multiples of 18, and 1000 ÷ 18 gives 55.

Divisible by either. Add the two counts and take away the overlap, so nothing is counted twice:

$$n(a \text{ or } b) = n(a) + n(b) - n(\text{both})$$

From 1 to 1000, divisible by 6 or 9: 166 + 111 − 55 = 222.

Divisible by one but not the other. Take the overlap away from that one count. Divisible by 9 but not by 6, from 1 to 1000: 111 − 55 = 56.

Divisible by neither. Total minus the "either" count: 1000 − 222 = 778.

[Tier 2] Three divisors. Numbers from 1 to 300 divisible by 2, 3 or 5: add the three single counts, take away the three pair counts, and add back the count for all three.

$$150 + 100 + 60 - 50 - 30 - 20 + 10 = 220$$

So 300 − 220 = 80 numbers from 1 to 300 are divisible by none of 2, 3 and 5.

[SSC Tip] "Two-digit numbers" means 10 to 99. For two-digit multiples of 7: there are 14 multiples up to 99 and 1 up to 9, so 13 in all. Forgetting to remove the one-digit multiples is a common slip.

Sums and Averages of Number Patterns

Multiples of a number in a range are evenly spaced, so their average is (first + last) ÷ 2, and the sum is count × average. This works for any evenly spaced list: consecutive numbers, odd numbers, even numbers, multiples of 7.

Take the sum of all multiples of 8 between 100 and 300. The first is 104 and the last is 296. The count is (296 − 104) ÷ 8 + 1 = 25. The average is (104 + 296) ÷ 2 = 200. So the sum is 25 × 200 = 5000.

Quick checks with $n = 10$: 1 + 2 + … + 10 = 55; the first ten odd numbers 1 + 3 + … + 19 add to 100; the first ten even numbers 2 + 4 + … + 20 add to 110; the squares up to $10^2$ add to 385; and the cubes up to $10^3$ add to 3025, which is $55^2$. The sum of the first $n$ cubes is always the square of the sum of the first $n$ numbers.

Part of a range. Subtract two full sums. 21 + 22 + … + 50 = (1 + … + 50) − (1 + … + 20) = 1275 − 210 = 1065.

Counting the terms. There are 50 odd numbers from 1 to 99, not 49, and their sum is $50^2 = 2500$. To count the terms of any list with a fixed gap, use (last − first) ÷ gap + 1. Don't leave out the "+ 1".

Divisibility of Power Sums and $a^n \pm b^n$

Power sums: take out the smallest power. When a sum has several powers of the same base, take the smallest power out as a common factor and look at what is left in the bracket.

$$4^{12} + 4^{13} + 4^{14} + 4^{15} = 4^{12}(1 + 4 + 16 + 64) = 4^{12} \times 85$$

85 = 5 × 17, so the sum is divisible by 5, 17 and 85, and also by 10 and 20, since $4^{12}$ supplies plenty of 2s.

$$6^{25} + 6^{26} + 6^{27} + 6^{28} = 6^{25}(1 + 6 + 36 + 216) = 6^{25} \times 259$$

259 = 7 × 37, so this sum is divisible by 7 and by 37 (and by every factor of $6^{25}$).

The rules for $a^n - b^n$ and $a^n + b^n$.

• $a^n - b^n$ is divisible by $a - b$ for every $n$.
• $a^n - b^n$ is divisible by $a + b$ when $n$ is even.
• $a^n + b^n$ is divisible by $a + b$ when $n$ is odd.

Here is how the rules work with $b = 1$. For $8^m - 1$: it is always divisible by 8 − 1 = 7, and when $m$ is even it is also divisible by 8 + 1 = 9, so by 63. $5^7 + 1$ is divisible by 5 + 1 = 6 because 7 is odd.

One more rule that settles many option lists: if $m$ divides $n$, then $a^m - 1$ divides $a^n - 1$. So $2^{12} - 1 = 4095$ is divisible by $2^3 - 1 = 7$, by $2^4 - 1 = 15$ and by $2^6 - 1 = 63$.

Products of consecutive integers.

• $n^3 - n = (n - 1)n(n + 1)$ is a product of three consecutive integers, so it is always divisible by 6. When $n$ is odd it is divisible by 24, because $n - 1$ and $n + 1$ are then two consecutive even numbers and one of them is a multiple of 4.
• In general, the product of $r$ consecutive integers is divisible by $r!$ (that is, 1 × 2 × … × r). Two in a row give a multiple of 2, three give a multiple of 6, four give a multiple of 24, five give a multiple of 120.
• So $n(n + 1)(n + 2)(n + 3)$ is always divisible by 24. For example, 5 × 6 × 7 × 8 = 1680 = 24 × 70.

[SSC Tip] When one option is "all of these", test every listed number with these rules before you pick a single one. An expression like $a^{24} - 1$ often passes two or three of them at once.

Repeated-Digit and Pattern Numbers

Numbers made by repeating digits have fixed factors. Write the number as a product and the answer is usually staring at you.

For example, 5,85,585 = 585 × 1001 and 4,44,444 = 4 × 1,11,111. Any number made of the same digit repeated 6, 12, 18 … times is divisible by 7, 11 and 13, because 1,11,111 is.

A number and its reverse. For a two-digit number $\overline{ab}$:

$$\overline{ab} + \overline{ba} = 11(a + b) \qquad \overline{ab} - \overline{ba} = 9(a - b)$$

For three digits, $\overline{abc} - \overline{cba} = 99(a - c)$, so the difference is always a multiple of 99. The sum of a three-digit number and its reverse has no fixed factor like this, so do not assume one.

Cyclic sums. Rotating the digits of $\overline{abc}$ gives $\overline{bca}$ and $\overline{cab}$. Each digit then appears once in every place, so

$$\overline{abc} + \overline{bca} + \overline{cab} = 111(a + b + c) = 3 \times 37 \times (a + b + c)$$

The sum is always divisible by 3, 37, 111 and by $a + b + c$. For 254: 254 + 542 + 425 = 1221 = 111 × 11.

A number minus its digit sum is always a multiple of 9. For 5372 the digit sum is 17, and 5372 − 17 = 5355 = 9 × 595.

Solved Examples

Question 1

Which of these numbers is divisible by 8? (a) 7,35,428 (b) 4,82,316 (c) 6,19,544 (d) 3,50,762

Answer. Only the last three digits matter, and I use the hundreds-digit trick. In 428 the hundreds digit 4 is even, so 28 itself must be a multiple of 8, and it isn't. In 316 the 3 is odd, so I check 16 + 4 = 20, not a multiple of 8. In 762 the 7 is odd, and 62 + 4 = 66, no. In 544 the 5 is odd, and 44 + 4 = 48 = 8 × 6, yes. The answer is (c) 6,19,544.

Question 2

Is 2,73,581 divisible by 7?

Answer. I keep cutting off the last digit, doubling it and subtracting.

• 27358 − 2 = 27356
• 2735 − 12 = 2723
• 272 − 6 = 266
• 26 − 12 = 14

14 is 7 × 2, so yes, 2,73,581 is divisible by 7. (It is 7 × 39,083.)

Question 3

Is 52,853 divisible by 17? Is it divisible by 19?

Answer. For 17 I multiply the last digit by 5 and subtract: 5285 − 15 = 5270, then 527 − 0 = 527, then 52 − 35 = 17. That is 17 itself, so 52,853 is divisible by 17.

For 19 I double the last digit and add: 5285 + 6 = 5291, then 529 + 2 = 531, then 53 + 2 = 55. 55 is not a multiple of 19, so 52,853 is not divisible by 19. (In fact 52,853 = 17 × 3109.)

Question 4

A student finds that 3612 is divisible by both 4 and 6, and says it must be divisible by 24. Is he right?

Answer. No. 4 and 6 share the factor 2, so passing both only proves that 3612 is divisible by their LCM, 12. For 24 I need the co-prime split 3 × 8. The digit sum is 12, so 3 is fine. For 8 I look at 612: the hundreds digit 6 is even, so 12 would have to be a multiple of 8, and it isn't. So 3612 is not divisible by 24 (3612 ÷ 24 = 150.5).

Question 5

Which of these is divisible by 88? (a) 5,27,304 (b) 6,43,148 (c) 7,18,256 (d) 3,92,502

Answer. 88 = 8 × 11, and 8 and 11 are co-prime. I do the 8 test first on the last three digits.

• 304 = 8 × 38, passes.
• 148: the hundreds digit 1 is odd, and 48 + 4 = 52, which is not a multiple of 8. Out.
• 256 = 8 × 32, passes.
• 502 is not even a multiple of 4 (02 is not). Out.

Now 11 on the two that are left. For 5,27,304 the odd places from the right hold 4, 3 and 2 (sum 9) and the even places hold 0, 7 and 5 (sum 12). The difference is 3, so no. For 7,18,256 the odd places hold 6, 2 and 1 (sum 9) and the even places hold 5, 8 and 7 (sum 20). The difference is 11, so yes. The answer is (c) 7,18,256.

Question 6

Find the digit $x$ if $73x452$ is divisible by 9.

Answer. The known digits add up to 7 + 3 + 4 + 5 + 2 = 21. So $21 + x$ has to be a multiple of 9. The next multiple of 9 after 21 is 27, so $x = 6$. The one after that, 36, would need $x = 15$, which is not a digit. So $x = 6$.

Question 7

Find the digit $x$ if $58x642$ is divisible by 11.

Answer. Reading from the right, the digits are 2, 4, 6, $x$, 8, 5. Odd places: 2 + 6 + 8 = 16. Even places: 4 + $x$ + 5 = $9 + x$. The difference is $16 - (9 + x) = 7 - x$, and it must be 0 or a multiple of 11. With $x$ between 0 and 9, only $7 - x = 0$ works. So $x = 7$, and the number is 5,87,642.

Question 8

The number $46x2y$ is divisible by both 8 and 5. How many values can $x$ take?

Answer. Divisible by 8 and 5 means divisible by 40, so the last digit must be 0. That gives $y = 0$, and the last three digits $x20$ must now be divisible by 8. Using the hundreds-digit trick: 20 is not a multiple of 8, but 20 + 4 = 24 is, so the hundreds digit $x$ has to be odd. That means $x$ = 1, 3, 5, 7 or 9, so $x$ can take 5 values.

Question 9

The number $7X3Y5$ is divisible by 9, and $Y - X = 6$. Find $X \times Y$.

Answer. The digit sum is 7 + 3 + 5 + $X + Y$ = $15 + X + Y$. For this to be a multiple of 9, $X + Y$ must be 3 or 12. It can't be 21, because two digits add up to at most 18.

$X + Y$ and $Y - X$ are always both even or both odd. $Y - X = 6$ is even, so $X + Y = 3$ is ruled out. With $X + Y = 12$ and $Y - X = 6$, adding gives $2Y = 18$, so $Y = 9$ and $X = 3$. The number is 73,395, and $X \times Y$ = 27.

Question 10

If $3x52y$ is divisible by 99, find $x + y$.

Answer. 99 = 9 × 11, so I use both rules.

For 9: 3 + 5 + 2 + $x + y$ = $10 + x + y$, so $x + y$ is 8 or 17.

For 11: from the right the digits are $y$, 2, 5, $x$, 3. Odd places: $y$ + 5 + 3 = $8 + y$. Even places: 2 + $x$. The difference is $6 + y - x$, which must be 0 or 11. So $x - y = 6$ or $y - x = 5$.

Now I pair them up. $x + y = 8$ with $x - y = 6$ gives $x = 7$ and $y = 1$. The other three pairings give halves or a digit bigger than 9. So the number is 37,521, and $x + y$ = 8.

Question 11

If $4a63b$ is divisible by 72, find $a + b$.

Answer. 72 = 8 × 9. I start with 8 because it involves only $b$. The last three digits $63b$ must be divisible by 8. Between 630 and 639 the only multiple of 8 is 632 = 8 × 79, so $b = 2$.

For 9: 4 + $a$ + 6 + 3 + 2 = $15 + a$, so $a = 3$. The number is 43,632 = 72 × 606, and $a + b$ = 5.

Question 12

The number $34x2y$ is divisible by 75. How many such numbers are there?

Answer. 75 = 3 × 25. For 25, the last two digits $2y$ must be 00, 25, 50 or 75, and only 25 fits. So $y = 5$.

For 3: 3 + 4 + $x$ + 2 + 5 = $14 + x$ must be a multiple of 3, so $x$ is 1, 4 or 7. That gives 34,125, 34,425 and 34,725. There are 3 such numbers.

Question 13

The number $x567y$ is divisible by 36. Find the largest such number.

Answer. 36 = 4 × 9. For 4, the last two digits $7y$ must be a multiple of 4, so they are 72 or 76. That makes $y$ = 2 or 6.

For 9, the digit sum is $18 + x + y$.

• If $y = 2$, then $x + 2$ must be a multiple of 9, so $x = 7$. The number is 75,672.
• If $y = 6$, then $x + 6$ must be a multiple of 9, so $x = 3$. The number is 35,676.

The largest is 75,672.

Question 14

The number $A35B2$ is divisible by 12. Find the largest possible value of $2A + 3B$.

Answer. 12 = 3 × 4.

For 4: the last two digits $B2$ must be a multiple of 4. 12, 32, 52, 72 and 92 all work, so $B$ is odd: 1, 3, 5, 7 or 9.

For 3: $A$ + 3 + 5 + $B$ + 2 = $10 + A + B$ must be a multiple of 3.

$B$ has the bigger coefficient, so I try $B = 9$ first. Then $19 + A$ must be a multiple of 3, so $A$ is 2, 5 or 8. Taking $A = 8$ gives 2 × 8 + 3 × 9 = 16 + 27 = 43. With $B = 7$ the best I can do is $A = 7$, which gives only 14 + 21 = 35. So the largest value is 43, from the number 83,592.

Question 15

What is the least number that must be subtracted from 8,46,375 to make it divisible by 13? What is the least number that must be added?

Answer. I divide first: 846375 = 13 × 65105 + 10, so the remainder is 10.

To subtract, I just take away the remainder: 10, which gives 8,46,365.

To add, I need 13 − 10 = 3, which gives 8,46,378.

Question 16

Find the largest five-digit number divisible by 47 and the smallest six-digit number divisible by 83.

Answer. The largest five-digit number is 99,999. 99999 = 47 × 2127 + 30, so I subtract the 30 and get 99,969.

The smallest six-digit number is 1,00,000. 100000 = 83 × 1204 + 68. I need 83 − 68 = 15 more, so the answer is 1,00,015.

Question 17

How many numbers between 300 and 800 are divisible by 9?

Answer. The first multiple of 9 after 300 is 306 (3 + 0 + 6 = 9). The last one before 800 is 792 (7 + 9 + 2 = 18). The count is (792 − 306) ÷ 9 + 1 = 54 + 1 = 55.

Neither 300 nor 800 is a multiple of 9, so "between" and "from … to" give the same answer here.

Question 18

How many numbers from 1 to 600 are divisible by 4 or 6? How many are divisible by 4 but not by 6?

Answer. Multiples of 4: 600 ÷ 4 = 150. Multiples of 6: 600 ÷ 6 = 100. Divisible by both means divisible by the LCM, 12: 600 ÷ 12 = 50.

Divisible by 4 or 6: 150 + 100 − 50 = 200.

Divisible by 4 but not 6: 150 − 50 = 100.

Question 19

Find the sum and the average of all the multiples of 6 between 100 and 400.

Answer. The first multiple is 102 and the last is 396. The count is (396 − 102) ÷ 6 + 1 = 49 + 1 = 50. The average is (102 + 396) ÷ 2 = 249, so the sum is 50 × 249 = 12,450.

Question 20

$7^{20} + 7^{21} + 7^{22}$ is divisible by which of these? (a) 17 (b) 23 (c) 57 (d) 61

Answer. I take out the smallest power: $7^{20}(1 + 7 + 49) = 7^{20} \times 57$. So 57 divides it straight away. The answer is (c) 57.

Question 21

$17^{24} - 1$ is divisible by: (a) 16 (b) 18 (c) 288 (d) all of these

Answer. $a^n - 1$ is always divisible by $a - 1$, so 16 works. The power 24 is even, so it is also divisible by $a + 1 = 18$. And since $17^{24} - 1 = (17^2)^{12} - 1$, it is divisible by $17^2 - 1 = 288$. All three work, so the answer is (d) all of these.

Question 22

A six-digit number is formed by writing a three-digit number twice, like 3,62,362. Which of these always divides such a number? (a) 17 (b) 77 (c) 111 (d) 101

Answer. Writing $\overline{abc}$ twice gives $\overline{abc} \times 1000 + \overline{abc} = \overline{abc} \times 1001$, and 1001 = 7 × 11 × 13. So 7, 11 and 13 always divide it, and so does 77 = 7 × 11. The other options depend on which three-digit number was chosen. The answer is (b) 77.