Part I, Chapter 5Number Play

(i) Not every natural number can be written as a sum of two or more consecutive natural numbers.

Checking the numbers from 1 to 70, the only ones that cannot be written are \(1, 2, 4, 8, 16, 32, 64\).

These are the powers of 2.

All the other numbers can be written, for example \(3 = 1 + 2\), \(5 = 2 + 3\), \(6 = 1 + 2 + 3\), \(7 = 3 + 4\), \(9 = 4 + 5\), \(10 = 1 + 2 + 3 + 4\).

(ii) Some numbers can be written in more than one way.

\(9 = 4 + 5 = 2 + 3 + 4\)

\(15 = 7 + 8 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5\)

Numbers that have more odd factors (like 15 or 45) can be written in more ways.

(iii) Many even numbers can be written, but not all.

\(6 = 1 + 2 + 3\), \(12 = 3 + 4 + 5\), \(14 = 2 + 3 + 4 + 5\), \(18 = 5 + 6 + 7\), \(20 = 2 + 3 + 4 + 5 + 6\)

But \(2, 4, 8, 16, \ldots\) cannot be written as a sum of consecutive natural numbers.

(iv) Yes, if we allow negative numbers.

\(0 = (-1) + 0 + 1\) and \(0 = (-2) + (-1) + 0 + 1 + 2\)

With negative numbers, every number can be written.

The numbers from \(-(n-1)\) to \(n-1\) cancel in pairs, leaving \(n\).

For example, \(4 = (-3) + (-2) + (-1) + 0 + 1 + 2 + 3 + 4\).

The first number keeps its sign.

Each of the other three numbers can take \(+\) or \(-\).

So there are \(2 \times 2 \times 2 = 8\) possibilities.

The table gives the value of all 8 expressions.

The signs are for the 2nd, 3rd and 4th numbers.

General form: let the numbers be \(n, n+1, n+2, n+3\).

\(n + (n+1) + (n+2) + (n+3) = 4n + 6\)

\(n + (n+1) + (n+2) - (n+3) = 2n\)

\(n + (n+1) - (n+2) + (n+3) = 2n + 2\)

\(n + (n+1) - (n+2) - (n+3) = -4\)

\(n - (n+1) + (n+2) + (n+3) = 2n + 4\)

\(n - (n+1) + (n+2) - (n+3) = -2\)

\(n - (n+1) - (n+2) + (n+3) = 0\)

\(n - (n+1) - (n+2) - (n+3) = -2n - 6\)

Observations: all 8 results are even numbers, for every set of 4 consecutive numbers.

Three of the results do not depend on \(n\) at all: \(-4\), \(-2\) and \(0\).

The other five are \(4n+6,\ 2n,\ 2n+2,\ 2n+4,\ -2n-6\).

Each has a factor 2, so it is even for every whole number \(n\).

The table shows the 8 results for two sets of numbers that are not consecutive.

For \(2, 5, 6, 9\) all 8 results are even.

For \(2, 5, 6, 8\) all 8 results are odd.

Observation: for any 4 numbers, all 8 results have the same parity.

Why: we know \(\text{odd} \pm \text{odd} = \text{even}\), \(\text{even} \pm \text{even} = \text{even}\) and \(\text{odd} \pm \text{even} = \text{odd}\).

So \(a + b\) and \(a - b\) have the same parity, whatever the parities of \(a\) and \(b\).

In the same way, \(a \pm b + c\) and \(a \pm b - c\) have the same parity.

Continuing, all the expressions \(a \pm b \pm c \pm d\) have the same parity.

Replace \(-c\) by \(+c\):

\((a + b + c - d) - (a + b - c - d) = 2c\)

Replace \(-d\) by \(+d\):

\((a + b - c + d) - (a + b - c - d) = 2d\)

Replace both:

\((a + b + c + d) - (a + b - c - d) = 2c + 2d = 2(c + d)\)

Each difference is even.

Conclusion: switching any sign changes the value by an even number (twice the number whose sign is switched).

Two numbers whose difference is even have the same parity.

Starting from any one expression, we can reach all the other 7 by switching one or more signs.

So all 8 expressions have the same parity.

Represent \(a, b, c, d\) by that many tokens each.

A \(+\) sign means positive tokens, and a \(-\) sign means negative tokens.

The value of the expression is found by cancelling positive tokens against negative tokens.

Each cancelling pair removes 2 tokens.

The total number of tokens at the start is \(a + b + c + d\), whatever the signs are.

Removing pairs (2 at a time) does not change whether this total is odd or even.

So the tokens left over after cancelling have the same parity as \(a + b + c + d\).

The value of the expression is that number of tokens (positive or negative).

Hence every sign choice gives an answer with the same parity as \(a + b + c + d\).

No.

It happens for any number of terms.

Switching the sign of any one term changes the value by twice that term, which is even.

So all the expressions \(a \pm b \pm c \pm \cdots\) have the same parity as \(a + b + c + \cdots\).

Example with 3 numbers: \(5 + 6 + 7 = 18\), \(5 + 6 - 7 = 4\), \(5 - 6 + 7 = 6\), \(5 - 6 - 7 = -8\).

All four results are even, as predicted.

\(43 + 37\): odd + odd = even.

Even.

\(672 - 348\): even − even = even.

Even.

\(4 \times 347 \times 3\): it has the even factor 4.

Even.

\(708 - 477\): even − odd = odd.

Not even.

\(809 + 214\): odd + even = odd.

Not even.

\(119 \times 303\): odd × odd = odd.

Not even.

\(543 - 479\): odd − odd = even.

Even.

\(513^3 = 513 \times 513 \times 513\): odd × odd × odd = odd.

Not even.

\(2a + 2b = 2(a + b)\): 2 is a factor, so it is always even.

Example: \(a = 3, b = 4\) gives 14.

\(3g + 5h\): not always even.

\(g = 1, h = 1\) gives 8 (even).

\(g = 1, h = 2\) gives 13 (odd).

\(4m + 2n = 2(2m + n)\): always even.

Example: \(m = 4, n = -9\) gives \(-2\).

\(2u - 4v = 2(u - 2v)\): always even.

Example: \(u = 5, v = 3\) gives \(-2\).

\(13k - 5k = 8k\): always even.

Example: \(k = 3\) gives 24.

\(6m - 3n = 3(2m - n)\): not always even.

\(m = 1, n = 2\) gives 0 (even).

\(m = 1, n = 1\) gives 3 (odd).

\(x^2 + 2\): not always even.

\(x = 6\) gives 38 (even).

\(x = 3\) gives 11 (odd).

\(b^2 + 1\): not always even.

\(b = 3\) gives 10 (even).

\(b = 2\) gives 5 (odd).

\(4k \times 3j = 12kj\): 12 is a factor, so it is always even.

Example: \(k = 1, j = 5\) gives 60.

An expression is always even if it has 2 as a factor.

Some examples:

\(2p\)

\(6p + 4q = 2(3p + 2q)\)

\(10m - 8n = 2(5m - 4n)\)

\(8a \times b = 2(4ab)\)

\(x^2 + x = x(x + 1)\), the product of two consecutive numbers, since one of them is even.

An even number leaves remainder 0 or 2 when divided by 4.

So an even number is either \(4p\) (a multiple of 4) or \(4q + 2\) (not a multiple of 4).

Trying some pairs:

\(4 + 8 = 12\), a multiple of 4.

\(2 + 6 = 8\) and \(6 + 10 = 16\), multiples of 4.

\(4 + 6 = 10\), \(8 + 2 = 10\) and \(12 + 14 = 26\), not multiples of 4.

Case 1: both are multiples of 4.

\(4p + 4q = 4(p + q)\)

The sum is a multiple of 4.

Case 2: both leave remainder 2.

\((4p + 2) + (4q + 2) = 4p + 4q + 4 = 4(p + q + 1)\)

The sum is a multiple of 4.

Case 3: one is a multiple of 4 and the other leaves remainder 2.

\(4p + (4q + 2) = 4(p + q) + 2\)

The sum leaves remainder 2, so it is not a multiple of 4.

A multiple of 4 is \(4p\).

An even number that is not a multiple of 4 leaves remainder 2, so it is \(4q + 2\).

\(4p + (4q + 2)\)

\(= 4p + 4q + 2\)

\(= 4(p + q) + 2\)

Visualisation: \(p\) rows of 4 and \(q\) rows of 4 (with 2 extra) join to make \((p + q)\) rows of 4, and the 2 extra counters are left over.

So the sum always leaves a remainder 2 when divided by 4.

It is never a multiple of 4.

Examples: \(4 + 6 = 10\), \(8 + 2 = 10\), \(12 + 10 = 22\), \(16 + 14 = 30\).

Each leaves remainder 2 on division by 4.

Comparison: yes, it is similar to even + odd.

An even number plus an odd number is never even.

A multiple of 4 plus an even number that is not a multiple of 4 is never a multiple of 4.

Let the two numbers be \(8a\) and \(8b\), both multiples of 8.

\(8a - 8b = 8(a - b)\)

8 is a factor of the difference, so the difference is a multiple of 8.

Visually: take \(b\) rows of 8 away from \(a\) rows of 8.

\((a - b)\) rows of 8 are left, so the difference is a multiple of 8.

Examples: \(56 - 16 = 40 = 8 \times 5\), \(120 - 80 = 40\), \(16 - 56 = -40 = 8 \times (-5)\).

Statement 6: Always true.

A number divisible by both 9 and 4 is divisible by \(\text{LCM}(9, 4)\).

9 and 4 have no common factor, so \(\text{LCM}(9, 4) = 9 \times 4 = 36\).

Examples: 36, 72, 108 are divisible by 9 and 4, and also by 36.

Statement 7: Sometimes true.

A number divisible by both 6 and 4 is divisible by \(\text{LCM}(6, 4) = 12\), not necessarily by 24.

Example (true): 24 and 48 are divisible by 6, 4 and 24.

Non-example: 12 and 36 are divisible by both 6 and 4, but not by 24.

\(3\) leaves remainder 3 on division by 5.

More such numbers are \(8, 13, 18, 23, 28, \ldots\)

These are 3 more than a multiple of 5.

Testing each expression:

(i) \(3k + 5\): \(k = 2\) gives 11, which leaves remainder 1.

No.

(ii) \(3k - 5\): \(k = 3\) gives 4, which leaves remainder 4.

No.

(iii) \(\frac{3k}{5}\): \(k = 1\) gives \(0.6\), not even a whole number.

No.

(iv) \(5k + 3\): \(k = 0, 1, 2, 3, \ldots\) gives \(3, 8, 13, 18, \ldots\) Yes.

The picture shows \(k\) full rows of 5 and 3 extra counters, which is \(5k + 3\).

(v) \(5k - 2 = 5(k - 1) + 3\): \(k = 1, 2, 3, \ldots\) gives \(3, 8, 13, \ldots\) Yes.

(vi) \(5k - 3 = 5(k - 1) + 2\): it leaves remainder 2, for example \(k = 2\) gives 7.

No.

Yes.

If we add or subtract any multiple of 5 to \(5k + 3\), the remainder is still 3.

\(5k + 8 = 5(k + 1) + 3\)

\(5k + 13 = 5(k + 2) + 3\)

\(5k - 7 = 5(k - 2) + 3\)

In general, \(5k + 3 + 5m = 5(k + m) + 3\) for any whole number \(m\).

Example: \(5k + 8\) with \(k = 0, 1, 2, 3\) gives \(8, 13, 18, 23\).

Write the number as \(\ldots + 1000d + 100c + 10b + a\), where \(a, b, c, d\) are the units, tens, hundreds and thousands digits.

Divisibility by 2 and 5:

\(10b, 100c, 1000d, \ldots\) are all multiples of 10.

So each of them is a multiple of 2 and of 5.

The number is divisible by 2 exactly when \(a\) is divisible by 2, that is, when the units digit is 0, 2, 4, 6 or 8.

The number is divisible by 5 exactly when \(a\) is divisible by 5, that is, when the units digit is 0 or 5.

Divisibility by 4:

\(100 = 4 \times 25\), so \(100c, 1000d, \ldots\) are all multiples of 4.

The number is divisible by 4 exactly when \(10b + a\) is divisible by 4.

That is, when the number formed by the last two digits is divisible by 4.

Divisibility by 8:

\(1000 = 8 \times 125\), so \(1000d, 10000e, \ldots\) are all multiples of 8.

The number is divisible by 8 exactly when \(100c + 10b + a\) is divisible by 8.

That is, when the number formed by the last three digits is divisible by 8.

Every place value is 1 more than a multiple of 9: \(10 = 9 + 1\), \(100 = 99 + 1\), \(1000 = 999 + 1\), and so on.

So for any number, \(\text{number} = (\text{a multiple of } 9) + (\text{sum of its digits})\).

For example, \(7309 = (7 \times 999 + 3 \times 99 + 0 \times 9) + (7 + 3 + 0 + 9)\).

This means a number and the sum of its digits leave the same remainder when divided by 9.

(i) Correct.

If the number is divisible by 9, its remainder is 0.

So the digit sum has remainder 0 too.

(ii) Correct.

If the digit sum is divisible by 9, the number has remainder 0.

So it is divisible by 9.

(iii) Correct.

If the number is not divisible by 9, its remainder is not 0.

So the digit sum has the same non-zero remainder.

(iv) Correct.

If the digit sum is not divisible by 9, the number leaves the same non-zero remainder.

So it is not divisible by 9.

Remainders of powers of 10 on division by 3:

\(10 = 9 + 1 = 3 \times 3 + 1\), remainder 1.

\(100 = 99 + 1 = 3 \times 33 + 1\), remainder 1.

\(1000 = 999 + 1 = 3 \times 333 + 1\), remainder 1.

In general, \(10^k = 99\ldots9 + 1\), and \(99\ldots9\) is a multiple of 9, so it is also a multiple of 3.

So every power of 10 leaves remainder 1 when divided by 3.

Why the method works: take the number \(\ldots + 1000d + 100c + 10b + a\).

Write each place value as (multiple of 3) + 1:

\(= d(999 + 1) + c(99 + 1) + b(9 + 1) + a\)

\(= (999d + 99c + 9b) + (d + c + b + a)\)

The first bracket is a multiple of 3 (it is even a multiple of 9).

So the number leaves the same remainder as the sum of its digits \(d + c + b + a\) when divided by 3.

Example: \(7309\) has digit sum 19.

\(19 = 3 \times 6 + 1\), so 7309 leaves remainder 1 when divided by 3.

The place values alternate: units is 1 more than a multiple of 11, tens is 1 less, hundreds is 1 more, thousands is 1 less, and so on.

For 462:

4 hundreds: 400 is 4 more than a multiple of 11.

6 tens: 60 is 6 less than a multiple of 11.

2 units: 2 is 2 more than a multiple of 11.

So 462 is (multiple of 11) \(+ 4 - 6 + 2\) = (multiple of 11) \(+ 0\).

Check: \(462 = 11 \times 42\).

General method:

Add the digits in the units, hundreds, ten-thousands, \(\ldots\) places (the places that are 1 more than a multiple of 11).

Add the digits in the tens, thousands, \(\ldots\) places (the places that are 1 less).

Subtract the second sum from the first.

If the difference is 0 or a multiple of 11, the number is divisible by 11.

The number is (a multiple of 11) \(+\) (the difference).

If the difference is 11, 22, 33, \(\ldots\) (or 0), it is itself a multiple of 11.

So the number is (multiple of 11) \(+\) (multiple of 11) = a multiple of 11.

Method: excess = sum of digits at units, hundreds, ten-thousands, \(\ldots\) places.

Short = sum of digits at tens, thousands, \(\ldots\) places.

Remainder = excess − short (add 11 if it is negative).

(i) 158: excess \(8+1 = 9\), short \(5 = 5\), difference \(9 - 5 = 4\).

Remainder 4.

(ii) 841: excess \(1+8 = 9\), short \(4 = 4\), difference \(9 - 4 = 5\).

Remainder 5.

(iii) 481: excess \(1+4 = 5\), short \(8 = 8\), difference \(5 - 8 = -3\).

The number is 3 short of a multiple of 11, so the remainder is \(11 - 3 = 8\).

(iv) 5529: excess \(9+5 = 14\), short \(2+5 = 7\), difference \(14 - 7 = 7\).

Remainder 7.

(v) 90904: excess \(4+9+9 = 22\), short \(0+0 = 0\), difference \(22 - 0 = 22\).

Divisible by 11.

(vi) 857076: excess \(6+0+5 = 11\), short \(7+7+8 = 22\), difference \(11 - 22 = -11\).

Divisible by 11.

The two methods are the same.

Only the order of work is different.

Earlier method for 328105:

Excess (units, hundreds, ten-thousands): \(5 + 1 + 2 = 8\).

Short (tens, thousands, lakhs): \(0 + 8 + 3 = 11\).

Difference: \(8 - 11 = -3\).

New method for 328105:

The + signs fall on 5, 1, 2 and the − signs fall on 0, 8, 3.

\(-3 + 2 - 8 + 1 - 0 + 5 = (2 + 1 + 5) - (3 + 8 + 0) = 8 - 11 = -3\)

The new method does the adding and subtracting in a single line.

Quick way: use the shortcuts instead of dividing.

By 2: the units digit is even.

By 5: the units digit is 0 or 5.

By 10: the units digit is 0.

By 4: the last two digits form a multiple of 4.

By 8: the last three digits form a multiple of 8.

By 3: the digit sum is a multiple of 3.

By 9: the digit sum is a multiple of 9.

By 6: divisible by both 2 and 3.

By 11: (sum of digits at units, hundreds, \(\ldots\)) − (sum of digits at tens, thousands, \(\ldots\)) is 0 or a multiple of 11.

Check each number for 2 (units digit even) and for 3 (digit sum a multiple of 3).

38: units digit 8 is even.

Digit sum \(3 + 8 = 11\) is not a multiple of 3.

Not divisible by 6.

Check: \(38 = 6 \times 6 + 2\).

225: units digit 5 is odd, so not divisible by 2.

Digit sum 9 is a multiple of 3.

Not divisible by 6.

Check: \(225 = 6 \times 37 + 3\).

186: units digit 6 is even.

Digit sum \(1 + 8 + 6 = 15\) is a multiple of 3.

Divisible by 6.

Check: \(186 = 6 \times 31\).

64: units digit 4 is even.

Digit sum \(6 + 4 = 10\) is not a multiple of 3.

Not divisible by 6.

Check: \(64 = 6 \times 10 + 4\).

Why it works: \(6 = 2 \times 3\), and 2 and 3 have no common factor other than 1.

A number divisible by both 2 and 3 is divisible by their LCM, which is 6.

The prime factorisation of 24 is \(24 = 2^3 \times 3\).

Why 3 and 8 work: \(8 = 2^3\).

A number divisible by 8 has \(2^3\) in its prime factorisation.

A number divisible by 3 has 3 in it.

So a number divisible by both has \(2^3 \times 3 = 24\) in its prime factorisation.

It is divisible by 24.

Also, 8 and 3 have no common prime factor, and \(\text{LCM}(8, 3) = 24\).

Why 4 and 6 are not enough: \(4 = 2^2\) and \(6 = 2 \times 3\).

A number divisible by both needs only \(2^2\) and \(3\), which is \(2^2 \times 3 = 12\).

It does not need the third factor 2.

Here \(\text{LCM}(4, 6) = 12\), not 24.

Example: 12 and 36 are divisible by both 4 and 6.

But \(12 \div 24\) and \(36 \div 24\) are not whole numbers.

A number and the sum of its digits leave the same remainder when divided by 9.

Adding the digits again and again never changes this remainder.

So the digital root leaves the same remainder as the original number, when divided by 9.

Example: \(489710 = 9 \times 54412 + 2\), and its digital root is 2.

Property: if the remainder on division by 9 is 1, 2, \(\ldots\), 8, the digital root is that remainder.

If the number is a multiple of 9, the digital root is 9 (never 0, as long as the number is not 0).

A number \(6bc\) has digit sum \(6 + b + c\).

Its digital root is the remainder of this sum on division by 9.

(i) Digital root 5: \(6 + b + c\) is 14 or 23, so \(b + c = 8\) or \(17\).

608, 617, 626, 635, 644, 653, 662, 671, 680, 689, 698

(ii) Digital root 7: \(6 + b + c\) is 7 or 16, so \(b + c = 1\) or \(10\).

601, 610, 619, 628, 637, 646, 655, 664, 673, 682, 691

(iii) Digital root 3: \(6 + b + c\) is 12 or 21, so \(b + c = 6\) or \(15\).

606, 615, 624, 633, 642, 651, 660, 669, 678, 687, 696

Take the 12 consecutive numbers 100, 101, \(\ldots\), 111.

Observations:

The digital roots go 1, 2, 3, 4, 5, 6, 7, 8, 9 and then start again from 1.

Each number has a digital root 1 more than the number before it.

After 9 comes 1.

The pattern repeats after every 9 numbers.

Why: the digital root is the remainder on division by 9 (with 9 in place of 0).

The next number has a remainder 1 more.

After remainder 8 comes 0, which is written as 9.

(i) Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, \(\ldots\)

Digital roots: 3, 6, 9, 3, 6, 9, 3, 6, 9, \(\ldots\)

(ii) Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, \(\ldots\)

Digital roots: 4, 8, 3, 7, 2, 6, 1, 5, 9, \(\ldots\)

(iii) Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, \(\ldots\)

Digital roots: 6, 3, 9, 6, 3, 9, 6, 3, 9, \(\ldots\)

Pattern: the roots repeat.

For 3 the block is 3, 6, 9.

For 4 the block is 4, 8, 3, 7, 2, 6, 1, 5, 9.

For 6 the block is 6, 3, 9.

Explanation: the digital root is the remainder on division by 9 (9 in place of 0).

Each step adds the same number to the previous multiple, so the remainder goes up by the same amount each time, going round the cycle \(1, 2, \ldots, 9\).

Multiples of 3: each step adds 3 to the root.

Starting at 3: 3, 6, 9, and then 12 gives 3 again.

Multiples of 6: each step adds 6, which is the same as going back 3 (since \(6 = 9 - 3\)).

So the roots are 6, 3, 9.

Multiples of 4: each step adds 4.

Since 4 and 9 have no common factor, the roots visit all nine values 1 to 9 before repeating.

Numbers that are 1 more than a multiple of 6: 7, 13, 19, 25, 31, 37, \(\ldots\)

Digital roots: 7, 4, 1, 7, 4, 1, 7, 4, 1, \(\ldots\)

Notice: the roots go 7, 4, 1 and repeat.

Explanation: the numbers \(6k\) have roots 6, 3, 9 in a cycle (see the previous question).

Adding 1 to each root gives 7, 4, 1 (where \(9 + 1 = 10\) is written as 1).

So the roots of \(6k + 1\) are 7, 4, 1 in a cycle.

Only these three values occur.

The largest odd single-digit number is 9.

The number of digits is 9, the sum of the digits is 9, and the digital root is 9.

Each digit is the smallest odd digit, which is 1.

So the number has nine digits, all equal to 1.

Digit sum \(= 1 \times 9 = 9\), and the digital root is 9, not 1.

(This is the “no shared ground with root 1”.)

The number is \(111111111\).

In the Indian system: 11,11,11,111, which is eleven crore eleven lakh eleven thousand one hundred eleven.

(i) Parity and digital root

Even numbers 10, 12, 20, 22, 30 have digital roots 1, 3, 2, 4, 3.

Odd numbers 11, 13, 27, 29 have digital roots 2, 4, 9, 2.

Even numbers can have an odd root (10 has root 1) or an even root (20 has root 2).

Odd numbers can have an even root (11 has root 2) or an odd root (27 has root 9).

Conjecture: there is no relation between the parity of a number and its digital root.

(ii) Digital root and remainders

Write the number as \(9m + r\), where \(r\) is its digital root (\(r = 1, 2, \ldots, 9\)).

Division by 9: if \(r < 9\), the remainder is \(r\).

If \(r = 9\), the number is \(9(m + 1)\), so the remainder is 0.

Division by 3: \(9m\) is a multiple of 3, so the remainder is the remainder of \(r\) on division by 3.

Conjectures:

Digital root 3, 6 or 9 means the number is a multiple of 3.

Digital root 1, 4 or 7 means remainder 1 on division by 3.

Digital root 2, 5 or 8 means remainder 2.

Remainder on division by 9 equals the digital root, except that digital root 9 gives remainder 0.

(i) A1 + 1B = B0

Units column: \(1 + B\) must end in 0, so \(B = 9\) (\(1 + 9 = 10\)).

Write 0 and carry 1.

Tens column: \(A + 1 + 1 = B = 9\), so \(A = 7\).

Check: \(71 + 19 = 90\).

(ii) AB + 37 = 6A

Units column: \(B + 7\) ends in \(A\).

If there were no carry, \(A + 3 = 6\) would give \(A = 3\) and \(B + 7 = 3\), which is not possible.

So there is a carry: \(B + 7 = A + 10\) and \(A + 3 + 1 = 6\), giving \(A = 2\).

Then \(B + 7 = 12\), so \(B = 5\).

Check: \(25 + 37 = 62\).

(iii) ON + ON + ON = PO, that is, \(3 \times ON = PO\).

The product has 2 digits, so \(ON \le 33\).

The units digit of \(3N\) must be \(O\).

\(N = 1\): \(3N = 3\), so \(O = 3\).

\(31 \times 3 = 93\).

So \(P = 9\).

\(N = 4\): \(3N = 12\), so \(O = 2\).

\(24 \times 3 = 72\).

So \(P = 7\).

\(N = 7\): \(3N = 21\), so \(O = 1\).

\(17 \times 3 = 51\).

So \(P = 5\).

The other values of \(N\) give \(O = N\), \(O = 0\), or \(ON \times 3\) with 3 digits.

All three solutions satisfy the conditions, because the letters stand for different digits in each.

(iv) QR + QR + QR = PRR, that is, \(3 \times QR = PRR\).

Units digit: \(3R\) ends in \(R\), so \(R = 0\) or \(R = 5\).

If \(R = 0\): \(3 \times 10Q = 100P\) gives \(3Q = 10P\), which has no digit solution (with \(Q \neq 0\)).

If \(R = 5\): \(3(10Q + 5) = 100P + 55\), so \(30Q = 100P + 40\), that is, \(3Q = 10P + 4\).

\(P = 2\) gives \(3Q = 24\), so \(Q = 8\).

Check: \(85 \times 3 = 255\).

No, PQ cannot be 13.

\(13 \times 8 = 104\), which has 3 digits.

But RS has only 2 digits.

For every 2-digit number greater than 12, the product with 8 is 104 or more, so it is a 3-digit number.

So PQ can only be 10, 11 or 12.

\(10 \times 8 = 80\): here \(Q = 0\) and \(S = 0\), so two different letters stand for the same digit.

Rejected.

\(11 \times 8 = 88\): here \(P = Q = 1\) (and \(R = S = 8\)), so different letters stand for the same digit.

Rejected.

\(12 \times 8 = 96\): \(P = 1,\ Q = 2,\ R = 9,\ S = 6\), all different.

Accepted.

In \(GH \times H\), the 1-digit multiplier is the same as the units digit of the 2-digit number.

Also the three letters \(G, H, K\) must stand for different digits.

Check each option:

\(11 \times 9\): units digit of 11 is 1, not 9.

Rejected.

\(12 \times 8\): units digit of 12 is 2, not 8.

Rejected.

\(46 \times 2\): units digit of 46 is 6, not 2.

Rejected.

\(47 \times 2\): units digit of 47 is 7, not 2.

Rejected.

\(31 \times 3\): units digit of 31 is 1, not 3.

Rejected.

\(16 \times 6\): the units digit 6 matches the multiplier.

The product is 96, so \(K = 6 = H\).

But \(H\) and \(K\) must be different digits.

Rejected.

\(24 \times 4\): the units digit 4 matches the multiplier.

The product is 96, so \(G = 2\), \(H = 4\), \(K = 6\), all different.

Accepted.

Since \(B = 1\), the number BYE is \(100 + 10Y + E\).

Y cannot be 7 or more: if \(Y = 7\), then \(170 \times 6 = 1020\), which has 4 digits.

The product must have 3 digits, so \(BYE \le 166\).

So \(Y \le 6\).

Y is even: the units digit of the product is the units digit of \(6 \times E\).

\(6 \times E\) is always even, so \(Y\) is even.

So \(Y\) can only be 0, 2, 4 or 6.

Odd digits, and 8 and 9, are not possible.

Trying each value of Y:

\(Y = 0\): \(6E\) ends in 0, so \(E = 0\) or \(5\).

\(E = 0\) is the same as \(Y\), so \(E = 5\).

\(105 \times 6 = 630\), so \(R = 6\) and \(A = 3\).

The digits 1, 0, 5, 6, 3 are all different.

Accepted.

\(Y = 2\): \(6E\) ends in 2, so \(E = 2\) or \(7\).

\(E = 2\) is the same as \(Y\).

\(127 \times 6 = 762\) gives \(R = 7 = E\).

Rejected.

\(Y = 4\): \(6E\) ends in 4, so \(E = 4\) or \(9\).

\(E = 4\) is the same as \(Y\).

\(149 \times 6 = 894\) gives \(A = 9 = E\).

Rejected.

\(Y = 6\): \(6E\) ends in 6, so \(E = 1\) or \(6\).

\(E = 1\) is the same as \(B\), and \(E = 6\) is the same as \(Y\).

Rejected.

(i) UT × 3 = PUT

Units digit: \(3T\) ends in \(T\), so \(T = 0\) or \(T = 5\).

If \(T = 5\): \(30U + 15 = 100P + 10U + 5\), so \(2U + 1 = 10P\).

Odd cannot equal even, so this is not possible.

If \(T = 0\): \(30U = 100P + 10U\), so \(20U = 100P\), that is, \(U = 5P\).

\(P = 1\) gives \(U = 5\).

Check: \(50 \times 3 = 150\).

(ii) AB × 5 = BC

The product has 2 digits, so \(AB \le 19\), and \(A = 1\).

\(5(10 + B) = 10B + C\), so \(50 = 5B + C\), that is, \(C = 50 - 5B\).

\(C\) is a digit, so \(50 - 5B \le 9\), which gives \(B \ge 9\).

So \(B = 9\) and \(C = 5\).

Check: \(19 \times 5 = 95\).

(iii) L2N × 2 = 2NP

The product starts with 2, so \(L2N \times 2\) lies between 200 and 299, so \(L2N\) lies between 100 and 149.

Hence \(L = 1\).

\(2(120 + N) = 200 + 10N + P\), so \(40 = 8N + P\).

\(N = 4\) gives \(P = 8\).

\(N = 5\) gives \(P = 0\).

(\(N \ge 6\) makes \(P\) negative, and \(N \le 3\) makes \(P\) more than 9.)

Check: \(124 \times 2 = 248\) and \(125 \times 2 = 250\).

(iv) XY × 4 = ZX

The product has 2 digits, so \(XY \le 24\).

\(4(10X + Y) = 10Z + X\), so \(39X + 4Y = 10Z\).

\(X = 1\): \(39 + 4Y\) is odd, so it cannot be a multiple of 10.

\(X = 2\): \(78 + 4Y\) must end in 0, so \(4Y\) ends in 2.

\(Y = 3\) or \(Y = 8\).

But \(XY \le 24\), so \(Y = 3\).

Then \(Z = 9\).

Check: \(23 \times 4 = 92\).

(v) PP × QQ = PRP

\(PP = 11P\) and \(QQ = 11Q\), so the product is \(121 \times P \times Q\).

This must be a 3-digit number, so \(P \times Q \le 8\).

Let \(m = P \times Q\).

Then \(121m\) is 242, 363, 484, 605, 726, 847, 968 for \(m = 2, 3, \ldots, 8\).

The first and last digits must both be \(P\).

This happens only for \(m = 2, 3, 4\), and then \(P = m\).

\(P = P \times Q\) gives \(Q = 1\).

So \(P = 2, 3\) or \(4\).

\(22 \times 11 = 242\), \(33 \times 11 = 363\), \(44 \times 11 = 484\).

(vi) JK × 6 = KKK

\(KKK = 111K\), so \(6 \times JK = 111K\), that is, \(JK = 18.5K\).

So \(K\) is even.

\(K = 2\): \(JK = 37\), which does not end in 2.

\(K = 4\): \(JK = 74\), which ends in 4.

This works.

\(K = 6\) and \(K = 8\) give \(JK = 111\) and \(148\), which are not 2-digit numbers.

Check: \(74 \times 6 = 444\).

Let the numbers be \(n, n + 1, n + 2\).

\(n\) is a multiple of 2, \(n + 1\) is a multiple of 3, \(n + 2\) is a multiple of 4.

One set: 2, 3, 4.

Finding all of them: \(n + 1\) is a multiple of 3, so \(n\) is 1 less than a multiple of 3.

This is the same as \(n\) leaving remainder 2 when divided by 3.

\(n + 2\) is a multiple of 4, so \(n\) leaves remainder 2 when divided by 4.

\(n\) is a multiple of 2 and leaves remainder 2 on division by 4, which already holds.

So \(n - 2\) is a multiple of both 3 and 4, that is, a multiple of \(\text{LCM}(3, 4) = 12\).

\(n = 12k + 2\) for \(k = 0, 1, 2, \ldots\)

The sets are 2, 3, 4;

14, 15, 16;

26, 27, 28;

38, 39, 40;

\(\ldots\)

\(36 = 4 \times 9\), and 4 and 9 have no common factor.

A number is a multiple of 36 if it is divisible by both 4 and 9.

Digit sum of 45000 is 9, and its last two digits are 00.

So 45000 is a multiple of 36: \(45000 = 36 \times 1250\).

The next multiples of 36 are found by adding 36 each time.

\(45000 + 36 = 45036\)

\(45036 + 36 = 45072\)

\(45072 + 36 = 45108\)

\(45108 + 36 = 45144\)

\(45144 + 36 = 45180\)

Check one: 45036 has last two digits 36, which is divisible by 4.

The digit sum is \(4 + 5 + 0 + 3 + 6 = 18\), a multiple of 9.