Number Play | CBSE Class 8 Maths Notes

This note covers consecutive numbers, parity, factors and multiples, algebraic reasoning, remainders, divisibility tests, digital roots, cryptarithms and the number strategy game Navakankari.

How can consecutive numbers be represented and added?

Consecutive integers follow one another with a difference of 1. Integers include negative whole numbers, zero and positive whole numbers. Whole numbers are 0, 1, 2 and so on. A difference is the result of subtraction. The symbols +, −, ×, ÷ and = mean addition, subtraction, multiplication, division and equality, respectively.

A sum is the result of addition; a product is the result of multiplication. The same number can sometimes be expressed as a sum of consecutive numbers in different ways. For example, 15 can be written as 7 + 8, as 4 + 5 + 6, and as 1 + 2 + 3 + 4 + 5.

Other examples are 7 = 3 + 4, 10 = 1 + 2 + 3 + 4 and 12 = 3 + 4 + 5. These examples invite exploration, but checking examples alone does not establish a rule for every number.

How does algebra describe a whole group?

Algebra uses letters to represent numbers. Let n represent the first integer in a group of four consecutive integers. The group is n, n + 1, n + 2 and n + 3. Here, 4n means 4 × n.

Worked example 1. Four consecutive numbers have a sum of 34. Find the numbers.

Answer: Let n be the smallest number. Then n + (n + 1) + (n + 2) + (n + 3) = 34. Combining terms gives 4n + 6 = 34, so 4n = 28 and n = 7. The numbers are 7, 8, 9 and 10.

The letters must match the information given. If p denotes the greatest of five consecutive numbers, the others are p − 1, p − 2, p − 3 and p − 4. Starting from the greatest number requires subtraction, rather than addition.

Why does changing addition to subtraction preserve parity?

Definition: Parity tells whether an integer is even or odd. An even integer has 2 as a factor, meaning it can be written as 2 times an integer. An odd integer is one more than an even integer.

Negative integers can also be even. For example, −2, −4 and −6 have a factor of 2. Being negative does not stop an integer from having even parity.

Take the consecutive numbers 3, 4, 5 and 6. Each of the three spaces between them can contain either an addition sign or a subtraction sign. This gives eight expressions. Two examples are 3 + 4 − 5 + 6 = 8 and 3 − 4 − 5 − 6 = −12.

A branching diagram begins at 3 and branches towards 4, then 5, then 6. Each stage offers an addition or subtraction sign. Its eight final branches organise the possible expressions.

Reference: NCERT Class 8, page 113, unnumbered

Property: Switching a sign does not change parity

Let a, b, c and d represent any four integers. Compare a + b − c − d with a − b − c − d. Subtracting the second expression from the first gives 2b. Their difference is even, so the two values have the same parity.

Switching any other sign also changes the value by twice an integer. Therefore, all eight expressions formed from the same four integers have the same parity. This does not mean that their numerical values are equal.

For four consecutive integers, all eight results are even. More generally, the parity rules explain the pattern: two even integers or two odd integers give an even sum or difference; one even and one odd integer give an odd sum or difference.

How can we tell whether an algebraic expression is always even?

An algebraic expression combines numbers, letters and operations. To establish that it is always even for integer inputs, show that it has a factor of 2. Checking a few inputs is useful for exploration, but does not prove a claim about every input.

How can an even factor be exposed?

Let m and q be any integers. A term is a part of an expression separated by addition or subtraction signs. In 4m + 2q, both terms are even, so their sum is even. Alternatively, write 4m + 2q = 2(2m + q). The brackets group the expression multiplied by 2.

Worked example 2. Evaluate 4m + 2q for m = 4 and q = −9, and explain its parity.

Answer: Substitution, meaning replacement of letters by their given values, gives 4 × 4 + 2 × (−9) = −2. This is even. The general form 2(2m + q) explains why every integer choice gives an even result.

Care is needed when the parity depends on the input. Let x be an integer. The notation x² means x × x, called the square of x. The square has the same parity as x, and adding 2 does not change that parity.

Worked example 3. Does x² + 2 give an even number for every integer x?

Answer: No. For x = 6, it gives 38, which is even. For x = 3, it gives 11, which is odd. The second calculation is a counterexample, a case that disproves an “always” claim.

Use examples and counterexamples together when a claim is sometimes true. Use a general argument when claiming that it is always true.

When do two even numbers add to a multiple of 4?

A multiple of a number is obtained by multiplying it by an integer. The divisor is the number used to divide another number. A number is divisible by a non-zero divisor when division leaves no remainder. The remainder is the amount left after complete groups of the divisor have been removed.

Even numbers belong to two groups when divided by 4. Some leave remainder 0; the others leave remainder 2. Let p and q represent integers used independently to describe the two numbers.

Property: Matching remainder types give a multiple of 4

1. Two multiples of 4 have the forms 4p and 4q. Their sum is 4(p + q), a multiple of 4.
2. Two even non-multiples of 4 have the forms 4p + 2 and 4q + 2. Their remainders combine to make another group of 4.
3. Their sum is 4p + 4q + 4 = 4(p + q + 1), again a multiple of 4.
4. A mixed pair gives 4p + (4q + 2) = 4(p + q) + 2, which leaves remainder 2.

Worked example 4. Explain why 12 + 16 and 22 + 6 are both divisible by 4.

Answer: Both 12 and 16 are multiples of 4, and 12 + 16 = 4(3 + 4) = 28. Neither 22 nor 6 is a multiple of 4, but each leaves remainder 2. Those remainders combine into 4, and 22 + 6 = 28.

Blue and yellow dots form rows of four. In the lower illustration, two extra blue dots and two extra yellow dots join to make one complete additional row.

Reference: NCERT Class 8, page 117, unnumbered

The useful distinction is therefore not simply “even or odd”. Both numbers are already even. The decision depends on which remainder type each number has when divided by 4.

Which properties of factors and multiples can we prove?

A factor of a number divides it exactly. Statements about factors need careful reasoning because reversing a true statement can produce a false one. A claim is always true if it holds in every allowed case, sometimes true if it holds in some cases but fails in others, and never true if it cannot hold.

Property: A common divisor divides the sum and difference

Let a be a non-zero integer, and let M and N be multiples of a. The capital letters M and N name the two numbers. Since each consists of complete groups of a, both M + N and M − N are also multiples of a.

For example, 8 divides both 16 and 56, and 16 + 56 = 72 is divisible by 8. However, a sum divisible by 8 need not have two parts separately divisible by 8: 72 = 50 + 22 is a counterexample.

Property: Divisibility passes to multiples and to factors of the divisor

If A is divisible by k, where A is an integer and k is a non-zero integer, every integer multiple of A is divisible by k. Multiplying A keeps its factor k. For instance, the multiples 28, 70 and 154 of 14 are all divisible by 7.

If A is divisible by k, it is also divisible by every non-zero factor of k. A multiple of 12 therefore has the factors of 12 among its own factors. Divisibility by a number gives divisibility by its factors, but not by all its multiples.

Why must combined divisibility tests use the LCM?

The least common multiple, abbreviated LCM, is the smallest positive number that is a multiple of each of the numbers being considered. If a number is divisible by each of two positive integers, it is divisible by their LCM.

Property: Two divisors guarantee their least common multiple

Divisibility by both 9 and 4 guarantees divisibility by 36. However, divisibility by both 6 and 4 does not guarantee divisibility by 24. The number 12 is divisible by 6 and by 4, yet is not divisible by 24.

Prime factorisation expresses a positive integer as a product of primes. A prime is a whole number greater than 1 with exactly two positive factors, 1 and itself. Combined tests must account for all the prime factors needed by the intended divisor.

For divisibility by 6, check both 2 and 3. For divisibility by 24, check both 3 and 8. Checking 4 and 6 misses part of the required factor structure, so it cannot establish divisibility by 24.

How can sets of multiples be pictured?

A Venn diagram uses regions to show relationships between groups of objects or numbers. A region inside another indicates that every member of the inner group belongs to the outer group.

The region labelled “Multiples of 32” lies inside “Multiples of 8”, which lies inside “Multiples of 4”. The nesting shows that every multiple of 32 is also a multiple of 8 and of 4.

Reference: NCERT Class 8, page 134, option iv

The nesting follows the factors of the numbers. It should not be reversed: knowing that a number is a multiple of 4 does not establish that it belongs to the smaller group of multiples of 32.

How can algebra describe and combine remainders?

Numbers that leave remainder 3 when divided by 5 are three more than multiples of 5. Let k be a non-negative integer, meaning zero or a positive integer. The expression 5k + 3 describes these non-negative numbers.

The same numbers are two less than the next multiple of 5. They can therefore also be described by 5k − 2, now taking k as a positive integer. The starting value of k matters when describing the same sequence.

How do remainders behave under addition and subtraction?

Separate each number into complete groups and a remainder. Complete groups contribute no remainder, but the combined leftover amount may itself contain another complete group. The final remainder must be non-negative and smaller than the positive divisor.

Worked example 5. When divided by 7, 4779 leaves remainder 5 and 661 leaves remainder 3. Find the remainders of their sum and difference without evaluating the full sum and difference.

Answer: Let p and q be their integer quotients, the numbers of complete groups of 7. Write 4779 = 7p + 5 and 661 = 7q + 3. Their sum is 7(p + q + 1) + 1, giving remainder 1. Their difference is 7(p − q) + 2, giving remainder 2.

Equal remainders can also be combined with the LCM. A number leaving remainder 2 on division by both 3 and 4 is two more than a common multiple of 3 and 4. Its form is 12n + 2, where n is a non-negative integer.

The examples 14, 26 and 38 correspond to positive choices of n. Allowing n = 0 also includes 2. This follows because a divisor larger than the number gives quotient zero and leaves the number itself as the remainder.

How does place value explain familiar divisibility tests?

A digit is one of the symbols 0 to 9 used to write numbers. Place value is the value contributed by a digit because of its position. In a four-digit numeral written dcba, the letters d, c, b and a represent its thousands, hundreds, tens and units digits. Here dcba names the numeral, not a product of letters.

Its expanded form is 1000d + 100c + 10b + a. For 4075, the digits are d = 4, c = 0, b = 7 and a = 5. Digits can be 0 to 9, but the first digit of a multi-digit number cannot be 0.

Why do the ending digits matter?

Every place value except the units place is a multiple of 10. Consequently, all those parts are divisible by 10, and the units digit determines whether the complete number is divisible by 10.

The same reasoning explains tests using the final two or three digits. Hundreds and higher place values are divisible by 4. Thousands and higher place values are divisible by 8. Therefore, the remaining ending portion decides the result.

For a number with fewer digits than the stated ending portion, use the whole number. These rules work because the omitted higher place values already form complete groups of the divisor.

Why do digit sums test divisibility by 9 and 3?

A digit sum is the sum of all digits of a number. Each decimal place value is one more than a multiple of 9: 10 = 9 + 1, 100 = 99 + 1 and 1000 = 999 + 1.

On expanding a number, the large multiples of 9 can be separated from the extra units. Those units add to the digit sum. Thus, the original number and its digit sum leave the same remainder when divided by 9.

Property: Divisibility by 9 is equivalent to divisibility of the digit sum

A number is divisible by 9 if and only if its digit sum is divisible by 9. “If and only if” means the implication works in both directions. The units digit alone is insufficient: 99 is divisible by 9, whereas 109 is not.

Worked example 6. Find the remainder when 7309 is divided by 9, using its digits.

Answer: Add 7 + 3 + 0 + 9 = 19. Add again to obtain 1 + 9 = 10, then 1 + 0 = 1. Therefore, 7309 leaves remainder 1 when divided by 9.

The picture splits 400 into four pieces labelled 99 with extra units, and 20 into two pieces labelled 9 with extra units. The loose units regroup into 9 and a remainder of 4.

Reference: NCERT Class 8, page 124, unnumbered

What changes for divisibility by 3?

A number is divisible by 3 when its digit sum is divisible by 3. Every multiple of 9 is a multiple of 3. The reverse does not hold: 15, 33 and 87 are multiples of 3 without being multiples of 9.

Reversing or rearranging the digits preserves the digit sum. A rearrangement of a multiple of 9 therefore remains divisible by 9. The reason concerns the sum of the digits, rather than their original order.

Worked example 7. The numeral 31z5 has an unknown digit z. Find z if the number is a multiple of 9.

Answer: Its digit sum is 3 + 1 + z + 5 = 9 + z. Since z is a digit, this is a multiple of 9 for z = 0 or z = 9. Both values are allowed.

How does the divisibility test for 11 work?

For divisibility by 11, decimal place values alternate between being one more and one less than a multiple of 11. This creates an alternating sum, in which addition and subtraction signs alternate between successive digits.

How should the test be applied?

1. Start at the units digit with an addition sign.
2. Move left, assigning subtraction and addition signs alternately.
3. Evaluate the resulting expression. A value of zero or a multiple of 11 establishes divisibility by 11.
4. If a remainder is required, express the result as a non-negative remainder smaller than 11.

Worked example 8. Test 328105 for divisibility by 11 and find its remainder.

Answer: Starting with an addition sign at the units digit gives −3 + 2 − 8 + 1 − 0 + 5 = −3. Thus, 328105 is three less than a multiple of 11, or eight more than the previous multiple. Its remainder is 8, so it is not divisible by 11.

The equivalent grouping method adds the digits in alternate places, then subtracts one group sum from the other. For 328105, the excess group totals 2 + 1 + 5 = 8, while the shortfall group totals 3 + 8 + 0 = 11.

Note: A negative alternating sum describes a shortfall from a multiple of 11. It is not the final division remainder. Keep track of which group is subtracted when using the test to find a remainder.

What is a digital root and what does it reveal?

Definition: The digital root of a non-negative whole number is the single digit obtained by repeatedly adding its digits until only one digit remains.

Worked example 9. Find the digital root of 489710.

Answer: First add 4 + 8 + 9 + 7 + 1 + 0 = 29. Next, 2 + 9 = 11. Finally, 1 + 1 = 2. The digital root is 2.

For a positive number not divisible by 9, its digital root is its remainder on division by 9. A positive multiple of 9 has digital root 9, although its remainder on division by 9 is zero. Keep these two quantities distinct.

What patterns occur in multiples?

Digital roots also help with division by 3. Roots 1, 4 and 7 correspond to remainder 1; roots 2, 5 and 8 correspond to remainder 2; roots 3, 6 and 9 correspond to remainder zero.

There is no consistent rule matching the parity of a number to the parity of its digital root. The digital root records information connected with division by 9; it does not replace every other divisibility test.

Aryabhata II’s work Mahāsiddhānta, from about 950 CE, mentions repeated digit addition. CE means Common Era. This method is known to have been used to check calculations involving arithmetic operations.

How can cryptarithms be solved by reasoning about digits?

A cryptarithm is a numerical puzzle in which letters stand for digits. The same letter keeps the same digit, different letters represent different digits, and the first digit of a multi-digit number is never zero.

Joined letters name a numeral. For example, PQ is a two-digit number whose tens digit is P and whose units digit is Q. It does not mean P multiplied by Q. The operation between complete numerals is written separately.

How can the size of a product restrict the digits?

Worked example 10. Solve PQ × 8 = RS, where P, Q, R and S are distinct digits and PQ and RS are two-digit numbers.

Answer: The product 10 × 8 = 80 repeats digit 0 for different letters. The choice 11 repeats the digit assigned to P and Q. However, 12 × 8 = 96 fits every condition. Since 13 × 8 = 104, any larger starting number gives too many digits. Thus P = 1, Q = 2, R = 9 and S = 6.

In the puzzle GH × H = 9K, G, H and K represent different digits and GH is a two-digit number. The units digit H of the first number must also be the multiplier. A numerical multiplication that ignores this repeated letter is not a solution.

How do individual columns help?

In addition puzzles, start with the units column and consider any carry, the amount transferred into the next place. Then use the tens column and the requirement that distinct letters have distinct digits.

For A1 + 1B = B0, A and B represent different non-zero digits. The units column requires 1 + B to end in 0, so B = 9 and there is a carry of 1. The tens column then gives A + 1 + 1 = 9, so A = 7.

Finally, substitute every digit back into the entire puzzle. This checks the arithmetic, repeated letters, different-letter restriction and number of digits together.

How is the strategy game Navakankari played?

Navakankari is a traditional Indian board game also known as Sālu Mane Āṭa, Chār-Pār or Navkakri. It is the same game as Nine Men’s Morris or Mills in the West. Two players try to form lines of three pawns.

A pawn is a playing piece. An intersection is a marked meeting point on the board where a pawn may be placed. Forming lines allows a player to remove an opponent’s pawn and can help block the opponent’s movement.

What are the rules and winning conditions?

1. Each player begins with 9 pawns. Players take turns placing them on marked intersections, with at most one pawn at each intersection.
2. After all pawns have been placed, players take turns moving one pawn to an adjacent empty intersection, meaning a neighbouring connected position.
3. A horizontal or vertical line of three of a player’s pawns allows that player to remove an opponent’s pawn, provided it is not part of an opponent’s line.
4. A player wins when the opponent has fewer than 3 pawns or cannot make a move.

The placement stage and the movement stage have different actions. During movement, the destination must be both adjacent and empty. Remembering these restrictions is part of analysing which moves are possible.

Glossary

• Consecutive integers — Integers arranged in order so that neighbouring numbers differ by exactly one.
• Parity — The classification of an integer as either even or odd.
• Factor — A number that divides another number exactly, leaving no remainder.
• Multiple — A number obtained by multiplying a given number by an integer.
• Divisibility — The property of leaving no remainder when divided by a specified non-zero number.
• Remainder — The non-negative amount left after division, smaller than the positive divisor.
• Least common multiple — The smallest positive number that is a multiple of every specified positive integer.
• Prime factorisation — The expression of a positive integer as a product of prime numbers.
• Counterexample — A particular case showing that a proposed general statement is not always true.
• Place value — The contribution a digit makes because of its position within a numeral.
• Digit sum — The total obtained by adding all the digits of a number.
• Digital root — The single digit obtained by repeatedly adding the digits of a number.
• Alternating sum — An expression evaluated with addition and subtraction signs alternating between successive terms.
• Cryptarithm — A numerical puzzle in which letters consistently replace digits under stated restrictions.

Common errors and misconceptions

• Misconception: Every sum of two even numbers is divisible by 4. Correct: Their remainder types on division by 4 must match. A mixed pair leaves remainder 2.
• Misconception: A sum divisible by 8 must have both parts divisible by 8. Correct: The example 72 = 50 + 22 disproves this claim.
• Misconception: Divisibility by 4 and 6 proves divisibility by 24. Correct: The number 12 passes both tests and fails the test for 24. Use 3 and 8 instead.
• Misconception: A few successful examples prove that an expression is always even. Correct: A general argument is needed. For x² + 2, x = 3 gives the odd result 11.
• Misconception: A positive multiple of 9 has digital root zero. Correct: Its remainder is zero, but its digital root is 9.
• Misconception: An alternating digit sum of −3 is a remainder of −3. Correct: For division by 11, this shortfall corresponds to remainder 8.
• Misconception: Any correct multiplication solves a cryptarithm. Correct: It must also preserve repeated letters, use distinct digits for different letters and avoid a leading zero.

Exam-style questions with model answers

Q1. Four consecutive integers have a sum of 34. Find them, showing your method. [2 marks]

1. Let n be the smallest integer. Then n + (n + 1) + (n + 2) + (n + 3) = 34, giving 4n + 6 = 34.
2. Thus 4n = 28 and n = 7. The four required integers are 7, 8, 9 and 10.

Q2. Explain all the cases in which the sum of two even integers is, or is not, divisible by 4. [3 marks]

1. Let p and q be integers. If both numbers are multiples of 4, their sum is 4p + 4q = 4(p + q), so it is divisible by 4.
2. If both leave remainder 2, their sum is (4p + 2) + (4q + 2) = 4(p + q + 1), again divisible by 4.
3. If their remainder types differ, the sum is 4p + (4q + 2) = 4(p + q) + 2. It leaves remainder 2 and is not divisible by 4.

Q3. On division by 7, 4779 leaves remainder 5 and 661 leaves remainder 3. Find the remainders of 4779 + 661 and 4779 − 661 using algebra. [4 marks]

1. Let p and q be the integer quotients of the two numbers on division by 7. Then 4779 = 7p + 5 and 661 = 7q + 3.
2. Adding gives 7(p + q) + 8. The remainder contributions add to 8, which contains one more complete group of 7.
3. Rewrite the sum as 7(p + q + 1) + 1. Therefore, its remainder on division by 7 is 1.
4. Subtracting gives 7(p − q) + 2. Since 2 is already smaller than 7, the remainder of the difference is 2.

Q4. Explain why adding digits can test divisibility by 9. Use the method to find the remainder when 7309 is divided by 9, and distinguish digital root from remainder for a positive multiple of 9. [5 marks]

1. Each decimal place value is one more than a multiple of 9. For example, 10 = 9 + 1, 100 = 99 + 1 and 1000 = 999 + 1.
2. In a number’s expanded form, separate these complete multiples of 9. The remaining units add to the digit sum, so the number and its digit sum have the same remainder.
3. For 7309, the digit sum is 7 + 3 + 0 + 9 = 19. The number is therefore 19 more than a multiple of 9.
4. Continue adding digits: 1 + 9 = 10 and 1 + 0 = 1. Thus, the remainder is 1 and 7309 is not divisible by 9.
5. A positive multiple of 9 has digital root 9, but its remainder on division by 9 is zero. Divisibility requires a zero remainder.

Q5. Use the divisibility test for 11 to decide whether 328105 is divisible by 11 and find its remainder. [3 marks]

1. Start at the units digit with an addition sign and alternate signs moving left. This gives −3 + 2 − 8 + 1 − 0 + 5.
2. The alternating sum is −3. This means the original number is three less than a multiple of 11; it does not mean the division remainder is negative.
3. Three less than the next multiple is eight more than the previous multiple. Therefore, the remainder is 8, and 328105 is not divisible by 11.

Q6. In the six-digit numeral 48a23b, a and b are digits from 0 to 9. List all ordered pairs (a, b) for which the number is divisible by 18, and justify completeness. [5 marks]

1. Divisibility by 18 requires divisibility by both 2 and 9. The notation (a, b) gives the value of a first and the value of b second.
2. For divisibility by 2, the units digit b must be 0, 2, 4, 6 or 8. These are all the possible even digits.
3. The digit sum is 4 + 8 + a + 2 + 3 + b = 17 + a + b. This must be a multiple of 9.
4. Checking each possible b gives the complete list: (1, 0), (8, 2), (6, 4), (4, 6) and (2, 8).
5. For each permitted b, these are the values of a between 0 and 9 that make the digit sum divisible by 9. Every pair also keeps the units digit even.

Q7. Solve PQ × 8 = RS. The letters P, Q, R and S represent four distinct digits, and PQ and RS are two-digit numerals with no leading zero. Explain why there is only one solution. [4 marks]

1. The smallest two-digit starting number is 10. However, 10 × 8 = 80 assigns the same digit 0 to Q and S, which violates the distinct-digit condition.
2. The next candidate, 11, assigns the same digit to P and Q. It is therefore not permitted either.
3. The calculation 12 × 8 = 96 uses four distinct digits and two-digit numerals. Hence P = 1, Q = 2, R = 9 and S = 6.
4. Since 13 × 8 = 104, every larger two-digit starting number gives a three-digit product. No further candidate can fit RS.

Q8. Explain why testing divisibility by 4 and 6 is insufficient for divisibility by 24. Give a counterexample and state a correct pair of tests. [3 marks]

1. A number divisible by two numbers is guaranteed to be divisible by their least common multiple. It need not be divisible by their product.
2. The number 12 is divisible by both 4 and 6, but it is not divisible by 24. This is a counterexample to the proposed test.
3. Instead, check divisibility by 3 and by 8. Their least common multiple is 24, so passing both tests guarantees divisibility by 24.

Key takeaways

• Represent consecutive integers with a starting letter and successive additions, adjusting the direction when the greatest integer is given.
• Changing addition to subtraction changes an expression by an even amount, preserving the parity of its value.
• Two even numbers sum to a multiple of 4 when their remainders on division by 4 match.
• A common divisor divides both the sum and difference of the numbers it divides separately.
• When combining divisibility tests, use the least common multiple of the divisors rather than assuming their product works.
• Digit sums explain divisibility by 3 and 9; alternating digit sums explain divisibility by 11.
• A positive multiple of 9 has digital root 9 and remainder zero, so these quantities must be distinguished.
• Cryptarithm solutions must satisfy the arithmetic, repeated-letter consistency, distinct-digit condition and restriction against leading zeroes.

Test yourself

If p is the greatest of five consecutive integers, what are the other four?

They are p − 1, p − 2, p − 3 and p − 4, each one step below the previous integer.

Why is 4m + 2q even for any integers m and q?

It equals 2(2m + q), so it has a factor of 2 for every permitted input.

Why does 72 = 50 + 22 matter when reasoning about divisibility by 8?

It shows that a sum can be divisible by 8 even though neither of its two parts is divisible by 8.

Which expression describes non-negative numbers leaving remainder 3 when divided by 5?

Use 5k + 3 with k a non-negative integer. The expression adds three to each multiple of 5.

Why does reversing the digits of a multiple of 9 preserve divisibility by 9?

Reversal leaves the digit sum unchanged. A digit sum divisible by 9 still establishes divisibility by 9 after reversal.

What does an alternating digit sum of −3 mean for division by 11?

The number is three short of a multiple of 11, so its non-negative division remainder is 8.

What is the digital root of 489710?

Its digits sum to 29, then to 11, and finally to 2. Its digital root is therefore 2.

In a cryptarithm, may two different letters stand for the same digit?

No. Different letters stand for different digits, while repeated occurrences of the same letter must keep the same digit.