Number Play

Everything in this chapter rests on one small idea: the digits you write are not the number, they are a short code for it. When you write 573, the 5 does not mean five. It means five hundreds, purely because of the place it is sitting in. So 573 is really 500 + 70 + 3, or 100 × 5 + 10 × 7 + 1 × 3.

To reason about every number at once instead of one number at a time, replace the digits by letters. A two-digit number whose tens digit is a and whose ones digit is b equals 10a + b. A three-digit number with digits a, b and c is 100a + 10b + c. Almost every trick in this chapter falls out of these two small expressions, so it is worth writing them down and staring at them for a minute.

Reverse a two-digit number and add. Take 73 and 37: the total is 110. Take 52 and 25: the total is 77. Take 41 and 14: the total is 55. Every answer is a multiple of 11, and that is not a coincidence. In letters, (10a + b) + (10b + a) = 11a + 11b = 11(a + b). The 11 is built into the answer. The other factor is a bonus, because it hands you the digit sum: 110 ÷ 11 = 10, and sure enough 7 + 3 = 10.

Now reverse and subtract instead. 73 - 37 = 36, 52 - 25 = 27, and 41 - 14 = 27. Every answer is now a multiple of 9, because (10a + b) - (10b + a) = 9a - 9b = 9(a - b). For 73 the digit difference is 7 - 3 = 4, and 9 × 4 = 36. The rule predicts the answer before you subtract anything.

Three digits. Reverse 842 to get 248 and subtract: 842 - 248 = 594. In letters, (100a + 10b + c) - (100c + 10b + a) = 99a - 99c = 99(a - c). Notice that the middle digit b cancels out completely, so it can be anything at all and the answer will not change. Here a - c = 8 - 2 = 6, and 99 × 6 = 594, exactly as predicted.

The 1089 surprise. Pick a three-digit number whose first and last digits differ by at least 2, say 732. Reverse it and take the smaller away from the larger: 732 - 237 = 495. Now reverse that answer and add it on: 495 + 594 = 1089. Try 921 instead: 921 - 129 = 792, and 792 + 297 = 1089 again. The reason is the rule you have just proved. The first difference has to be 99 × (a - c), and with a - c running from 2 to 9 that leaves only 198, 297, 396, 495, 594, 693, 792 and 891. Every one of them has 9 in the middle and outer digits adding to 9, so adding the reverse always lands on 1089.

Cycling the digits. Take 254 and slide the digits round twice more to get 542 and then 425. Add all three: 254 + 542 + 425 = 1221. In letters the total is (100a + 10b + c) + (100b + 10c + a) + (100c + 10a + b) = 111a + 111b + 111c = 111(a + b + c). Since 111 = 3 × 37, this sum is always divisible by 3 and by 37, whichever digits you started with. Here a + b + c = 2 + 5 + 4 = 11, and 111 × 11 = 1221, with 1221 ÷ 37 = 33.

Every one of these results was found by trying examples first and proved by place value afterwards. That order, notice then explain, is the habit this chapter is really teaching.

A divisibility test is a shortcut that tells you whether one number divides another exactly, without your doing the division. What makes this chapter different from a list of rules to memorise is that you are expected to say why each test works. The reasoning is always the same: split the number into a part that is obviously a multiple of the divisor, plus a small leftover, and then only the leftover matters.

By 2, 5 and 10. Every place from the tens upwards is a whole number of tens, and 10 is a multiple of 2, of 5 and of 10 itself. So those places can never upset the division, and only the ones digit decides. A number is even when it ends in 0, 2, 4, 6 or 8; it is a multiple of 5 when it ends in 0 or 5; and a multiple of 10 only when it ends in 0.

By 4 and by 8. Since 100 = 4 × 25, every hundred is already a multiple of 4, so only the last two digits can matter. Write 6,45,732 as 6457 × 100 + 32. The first part is a multiple of 4 whatever 6457 happens to be, and 32 = 4 × 8, so the whole number is a multiple of 4. In exactly the same way 1000 = 8 × 125, so for 8 you look at the last three digits. In 5,34,672 the last three digits give 672 = 8 × 84, so that number is divisible by 8.

By 3 and by 9. This is the test worth understanding properly. Notice that 10 = 9 + 1, 100 = 99 + 1 and 1000 = 999 + 1. So a three-digit number splits like this: 100a + 10b + c = (99 + 1)a + (9 + 1)b + c = (99a + 9b) + (a + b + c). The bracket 99a + 9b is a multiple of 9 no matter what a and b are, so the number is a multiple of 9 exactly when the leftover a + b + c, the digit sum, is a multiple of 9. Because 9 and 99 are also multiples of 3, the very same split proves the test for 3. Longer numbers behave identically, since 9999, 99999 and so on are all multiples of 9.

By 6, and one trap. A number is divisible by 6 when it is divisible by 2 and by 3, because 2 and 3 are co-prime, meaning they share no factor except 1. The same reasoning lets you combine 3 and 4 to get 12. But be careful: divisible by 4 and by 6 does not guarantee divisible by 24, because 4 and 6 share the factor 2. The number 12 is divisible by 4 and by 6 and is certainly not divisible by 24.

By 11. Here 10 is 1 less than 11, 100 is 1 more than 99, and 1000 is 1 less than 1001, which is 11 × 91. The pattern keeps flipping, so starting from the ones digit you add, subtract, add, subtract and so on. For 4752 that gives 2 - 5 + 7 - 4 = 0, and 0 counts as a multiple of 11, so 4752 is divisible by 11. Checking by division, 4752 ÷ 11 = 432. For 5,34,672 the alternating total is 2 - 7 + 6 - 4 + 3 - 5 = -5, which is not a multiple of 11, so that number is not.

A full worked check. Test 5,34,672 against everything. It ends in 2, so it is divisible by 2. Its digit sum is 5 + 3 + 4 + 6 + 7 + 2 = 27, a multiple of 9, so it is divisible by both 3 and 9. Its last two digits give 72 = 4 × 18, so it passes for 4, and 672 = 8 × 84 gives 8 as well. Divisible by 2 and by 3 means divisible by 6. It fails three of the tests: it does not end in 0 or 5, so it is not a multiple of 5 or of 10, and its alternating total of -5 rules out 11.

Filling in a missing digit. These tests are strong enough to recover a digit you cannot see. For which digit y is the number 45y2 a multiple of 9? Its digit sum is 4 + 5 + y + 2 = 11 + y, and because y is a single digit this total lies somewhere between 11 and 20. The only multiple of 9 in that range is 18, so y = 7 and the number is 4572. Checking, 4572 ÷ 9 = 508. Sometimes there are two answers instead of one, so always check the whole range before writing down a single value.

One more useful fact: 1001 = 7 × 11 × 13. That is why writing a three-digit number twice over, as in 123123, always produces a multiple of 7, of 11 and of 13, since 123123 = 123 × 1001. Here 123123 ÷ 7 = 17589.

A sequence is a list of numbers written in a fixed order, and each number in the list is called a term. Describing a sequence properly means finding a rule that takes you straight to any term you want, without writing out all the ones before it. That rule is called the general term, and it is written using n, where n stands for the position of the term.

Sequences with a constant jump. Look at 7, 12, 17, 22, 27, and so on. Each term is 5 more than the one before, so the jump, usually called the common difference d, is 5. The first term a is 7. To reach the nth term you start at 7 and take that jump (n - 1) times, so the general term is 7 + (n - 1) × 5 = 5n + 2. Test it: n = 1 gives 5 + 2 = 7, n = 2 gives 10 + 2 = 12, and n = 3 gives 15 + 2 = 17. The 40th term is 5 × 40 + 2 = 202, and you never had to write out the 39 terms in front of it.

The general rule is nth term = a + (n - 1)d, which tidies up to dn + (a - d). The number multiplying n is always the jump, which gives you a quick way to read off a rule. For 4, 7, 10, 13, the jump is 3 and the first term is 4, so the rule is 3n + 1. Check the far end too: n = 4 gives 12 + 1 = 13, which matches.

Going downwards. Nothing breaks if the jump is negative. For 40, 37, 34, 31, the jump is -3 and the first term is 40, so the general term is 40 + (n - 1)(-3) = 43 - 3n. The 14th term is 43 - 42 = 1 and the 15th is 43 - 45 = -2, so this sequence slips into negative numbers between those two positions.

Is a given number in the sequence? This is where a general term really earns its keep. Is 101 a term of 5n + 2? Set 5n + 2 = 101, so 5n = 99 and n = 19.8. A position has to be a counting number, so 101 is not in the list. Is 102? Then 5n = 100 and n = 20, so yes, 102 is the 20th term. Whenever n comes out as a fraction, the answer is simply no.

When the jump is not constant. Look at 2, 5, 10, 17, 26. The jumps are 3, 5, 7 and 9, which are not equal, so the rule cannot be dn + something. But those jumps form a tidy run of odd numbers, and that is a strong hint that squares are involved. Compare the terms with 1, 4, 9, 16, 25. Each term is exactly 1 more, so the general term is n × n + 1. The 12th term is 12 × 12 + 1 = 144 + 1 = 145. Looking at the differences, and then at the differences of the differences, is the standard first move when a pattern will not sit still.

Sequences worth recognising instantly. Square numbers 1, 4, 9, 16, 25 have general term n × n. Odd numbers 1, 3, 5, 7 have general term 2n - 1, and even numbers 2, 4, 6, 8 have 2n. Triangular numbers 1, 3, 6, 10, 15, which count dots stacked in a triangle, have general term n(n + 1) ÷ 2, so the 20th triangular number is 20 × 21 ÷ 2 = 210. Doubling sequences such as 1, 2, 4, 8, 16 are different in kind, because you multiply instead of adding; the general term is 2^{n - 1}, so the 10th term is 2⁹ = 512. These grow far faster than constant-jump sequences, which is worth remembering.

A sequence you might actually use. Suppose you save ₹150 in the first week and then ₹40 more each week than the week before. The amounts are 150, 190, 230, 270, and so on, a constant-jump sequence with a = 150 and d = 40, so in week n you save 40n + 110 rupees. In week 10 that is 400 + 110 = ₹510. For the ten-week total, pair the first amount with the last: 150 + 510 = 660, and ten terms make five such pairs, so the total saved is 5 × 660 = ₹3,300.

Patterns in addition and multiplication are the part of this chapter where a picture often explains more than a calculation does. The aim is not to add faster, but to see why the answer is forced to come out the way it does.

Odd numbers build squares. Add the odd numbers from the beginning: 1 = 1, then 1 + 3 = 4, then 1 + 3 + 5 = 9, then 1 + 3 + 5 + 7 = 16, then 1 + 3 + 5 + 7 + 9 = 25. The answers are exactly the square numbers. Picture a square array of dots to see why. To grow a 4 by 4 square into a 5 by 5 square you add an L-shaped border: 4 dots down one side, 4 along the top and 1 in the corner, which is 9 dots in all, the next odd number. Each new odd number is precisely the border needed for the next square. So the sum of the first n odd numbers is n × n. That instantly gives 1 + 3 + 5 + ... + 39, because 39 is the 20th odd number (2 × 20 - 1 = 39), so the total is 20 × 20 = 400 with nothing added at all.

Counting numbers. To add 1 + 2 + 3 + ... + 100, pair the first with the last: 1 + 100 = 101, 2 + 99 = 101, 3 + 98 = 101, and so on. There are 50 such pairs, so the total is 50 × 101 = 5050. The same pairing gives the general rule 1 + 2 + ... + n = n(n + 1) ÷ 2. For n = 60 that is 60 × 61 ÷ 2 = 3660 ÷ 2 = 1830.

Even numbers. The sum 2 + 4 + 6 + ... + 2n is just twice 1 + 2 + ... + n, so it equals n(n + 1). Check it with four terms: 2 + 4 + 6 + 8 = 20, and 4 × 5 = 20.

Patterns that look like magic. Try 1 × 9 + 2 = 11, then 12 × 9 + 3 = 111, then 123 × 9 + 4 = 1111. The next line should be 1234 × 9 + 5, and since 1234 × 9 = 11106, adding 5 gives 11111 exactly as promised. A close cousin runs 1 × 8 + 1 = 9, then 12 × 8 + 2 = 98, then 123 × 8 + 3 = 987, and the next line is 1234 × 8 + 4 = 9872 + 4 = 9876. Repeated ones do something pretty too: 11 × 11 = 121, 111 × 111 = 12321, and 1111 × 1111 = 1234321, the digits climbing up and then coming back down.

Squares that talk to each other. Two neighbouring squares differ by an odd number: 25 × 25 - 24 × 24 = 625 - 576 = 49, which is 2 × 25 - 1. That is the same L-shaped border as before, which is why odd numbers and squares keep turning up together. More generally a × a - b × b = (a + b)(a - b), which converts an awkward subtraction into an easy multiplication: 43 × 43 - 37 × 37 = (43 + 37) × (43 - 37) = 80 × 6 = 480.

Squaring a number that ends in 5. Drop the 5, multiply what is left by the next counting number, and write 25 after the result. For 35, take 3 × 4 = 12 and write 1225. For 65, take 6 × 7 = 42 and write 4225. For 85, take 8 × 9 = 72 and write 7225. Place value explains it: (10a + 5) × (10a + 5) = 100 × a × a + 100a + 25 = 100 × a × (a + 1) + 25, and the final 25 is why the answer always ends in those two digits.

Numbers sitting either side of a round number. Since 19 × 21 = (20 - 1)(20 + 1) = 400 - 1 = 399, a product of two numbers spaced equally around a round number is just that square minus a small bit. The same idea handles numbers close to 100: 98 × 97 = (100 - 2)(100 - 3) = 10000 - 300 - 200 + 6 = 9506. Notice how much of this chapter comes back to one habit, which is rewriting a number as something convenient plus or minus a little.

Puzzles are where all the reasoning so far gets used. In a cryptarithm, the digits are hidden behind letters. Two rules apply: each letter stands for just one digit and each digit is shown by just one letter, so two different letters can never be the same digit; and no number may begin with 0. You solve one column at a time, starting from the ones place, exactly the way you do ordinary addition, and you keep track of carries.

Puzzle one. Find the digits A and B if 3A + 25 = B2, where 3A means the two-digit number with tens digit 3 and ones digit A. Start with the ones column: A + 5 must end in 2. Since A is a single digit, A + 5 is at most 14, so the only possibility is A + 5 = 12, giving A = 7 and a carry of 1 into the tens column. The tens column then reads 3 + 2 + 1 = 6, so B = 6. Checking the whole sum, 37 + 25 = 62, which does have 6 in the tens place and 2 in the ones place.

Puzzle two. Find A and B if AB + 37 = 6A. The tens column says A + 3 plus any carry equals 6. Suppose first that there is no carry. Then A = 3, and the ones column would need B + 7 to end in 3, which forces B + 7 = 13 and B = 6, but that produces a carry, contradicting the assumption. So there must be a carry of 1. Then A + 3 + 1 = 6 gives A = 2, and the ones column needs B + 7 = 12, so B = 5. Checking, 25 + 37 = 62, and 62 does have 6 in the tens place and A = 2 in the ones place. Ruling out one case to force the other is the main skill in these puzzles.

Puzzle three, with multiplication. Find A if 1A × A = 9A, meaning a two-digit number starting with 1, multiplied by the single digit A, gives a two-digit answer starting with 9 and ending in A. Work through the digits, noting that the product (10 + A) × A grows as A grows. Any A of 3 or less gives at most 13 × 3 = 39, nowhere near 90, so the search can start at A = 4. A = 4 gives 14 × 4 = 56, no. A = 5 gives 15 × 5 = 75, no. A = 6 gives 16 × 6 = 96, which starts with 9 and ends in 6, so it works. A = 7 gives 17 × 7 = 119, already a three-digit number, and larger values only get worse. So A = 6 is the only answer, and a short organised search beat any clever formula.

Think-of-a-number tricks. Ask a friend to think of any number, add 5, double the result, subtract 8, halve it, and finally take away the number they first thought of. Their answer is 1, every single time. Algebra shows why. Call their number x. Adding 5 gives x + 5; doubling gives 2x + 10; subtracting 8 gives 2x + 2; halving gives x + 1; and taking away x leaves 1. The x disappears, which is exactly why you can announce the answer without knowing their number. Test it with 7: 12, then 24, then 16, then 8, then 1. A shorter version is to multiply by 3, add 6, divide by 3 and subtract the original number, which always leaves 2, because (3x + 6) ÷ 3 - x = x + 2 - x = 2.

A chain that always ends in the same place. Take any four-digit number whose digits are not all the same, say 3524. Arrange its digits in descending order and in ascending order, then subtract the smaller from the larger: 5432 - 2345 = 3087. Repeat with the answer: 8730 - 0378 = 8352. Repeat once more: 8532 - 2358 = 6174. Now you are stuck, because 7641 - 1467 = 6174 all over again. This number 6174 is known as Kaprekar's constant, and every allowed starting number reaches it within seven steps.

Making palindromes. A palindrome reads the same forwards and backwards, like 121 or 4554. Take a number, add its reverse, and repeat until the answer is a palindrome. Starting at 68: 68 + 86 = 154, then 154 + 451 = 605, then 605 + 506 = 1111, which is a palindrome. Most small starting numbers give in within a few steps, though a handful are famously stubborn. What these puzzles share is that a rule which looks like magic turns ordinary the moment you write it out in place value or in letters.

Behind all the patterns sits the basic building material of number play: factors and primes.

Primes and composites. A prime number has exactly two factors, 1 and itself. A composite number has more than two. The number 1 is neither, because it has only one factor, and calling it prime would spoil the fact that every number breaks into primes in exactly one way. Primes thin out as numbers grow larger but they never run out. Pairs of primes that are two apart, such as 11 and 13, 17 and 19, or 29 and 31, are called twin primes. Two numbers whose only common factor is 1, such as 8 and 15, are called co-prime, and they need not be prime themselves.

Testing whether a number is prime. You do not have to try every possible divisor. If a number had a factor larger than its square root, the matching factor would have to be smaller than the square root, and you would have found it already. So it is enough to test the primes up to the square root. Take 197. Its square root is a little over 14, so test 2, 3, 5, 7, 11 and 13. It is odd, so 2 fails; its digit sum 1 + 9 + 7 = 17 is not a multiple of 3; it does not end in 0 or 5; 7 × 28 = 196 leaves a remainder of 1; 11 × 17 = 187 leaves 10; and 13 × 15 = 195 leaves 2. Nothing divides it, so 197 is prime. Compare 221, which looks similar: the same search reaches 13 × 17 = 221, so it is composite.

Prime factorisation and counting factors. Break 360 down step by step: 360 = 2 × 180 = 2 × 2 × 90 = 2 × 2 × 2 × 45 = 2 × 2 × 2 × 3 × 3 × 5. So 360 is made of three 2s, two 3s and one 5. Now for a neat consequence. Any factor of 360 is built by choosing how many 2s to take (0, 1, 2 or 3, which is 4 choices), how many 3s to take (0, 1 or 2, which is 3 choices) and how many 5s to take (0 or 1, which is 2 choices). Every different set of choices gives a different factor, so 360 has 4 × 3 × 2 = 24 factors, and you never had to list a single one.

Why squares are the odd ones out. Factors normally come in pairs, one on each side of the square root, so most numbers have an even number of factors. A perfect square is the exception, because its square root pairs with itself. Take 36 = 2 × 2 × 3 × 3, where the count is 3 × 3 = 9, an odd number, and listing them confirms it: 1, 2, 3, 4, 6, 9, 12, 18 and 36. So a number with an odd number of factors must be a perfect square.

Last digits of squares. Square the digits 0 to 9 and record only the last digit of each answer: 0, 1, 4, 9, 6, 5, 6, 9, 4, 1. So a perfect square can end only in 0, 1, 4, 5, 6 or 9. That gives a one-second test. The number 2,05,382 ends in 2, so it cannot possibly be a perfect square, and you needed no square root to say so. Remember that this test only rules numbers out: ending in 6 does not prove that a number is a square.

Two more curiosities. A perfect number equals the sum of all its factors apart from itself: 6 = 1 + 2 + 3, and 28 = 1 + 2 + 4 + 7 + 14. They are rare, and searching for them has kept people busy for centuries. Finally, for any two numbers, HCF × LCM equals the product of the numbers. For 12 and 18 the HCF is 6 and the LCM is 36, and 6 × 36 = 216, which is exactly 12 × 18. That relation lets you find one of the four quantities as soon as you know the other three, and it is a fast check on any HCF and LCM answer.