5.5 Going Deeper: Enrichment & Exam Preparation

Where Number Play Shows Up in Real Life

Math in the Real World

  • Barcodes (EAN-13). Every retail barcode carries a check digit computed from the other twelve. The thirteenth digit is chosen so that (1st + 3rd + 5th + … digits) +\,3\times(2nd + 4th + … digits) lands on a multiple of 10. When a scanner misreads one digit, this sum no longer works out and the till rejects the scan, the same “make it divide cleanly” idea you used for divisibility.
  • Book ISBNs. An old ISBN-10 weights its digits by 10, 9, 8, \dots, 2 and picks a final check digit so the total is a multiple of 11 (an “X” stands for the check value 10). This is exactly the kind of weighted divisibility test the chapter is about, built to catch a single mistyped digit or two swapped digits.
  • Aadhaar and account numbers. Identity and bank account numbers use check digits of the same family (the Verhoeff and Luhn schemes) so that a slip of one digit is caught before money moves.
  • Casting out nines. Because the digital root equals the remainder on division by 9, you can sanity-check any long multiplication in seconds: \text{droot}(478)\times\text{droot}(23) = 1\times 5 = 5, and indeed 478\times 23 = 10994 has digital root 5. If the two digital roots disagree, the answer is certainly wrong (though agreement alone does not prove it right).

Quick Reference: Divisibility Tests

Knowing these by heart turns slow trial-division into instant judgement, a real advantage under exam time pressure.

Divisor Quick test
2 Units digit is 0,2,4,6 or 8 (even).
3 Sum of the digits is divisible by 3.
4 Number formed by the last two digits is divisible by 4.
5 Units digit is 0 or 5.
6 Divisible by both 2 and 3.
7 Drop the units digit, double it, subtract from the rest; repeat, result divisible by 7.
8 Number formed by the last three digits is divisible by 8.
9 Sum of the digits is divisible by 9.
10 Units digit is 0.
11 Alternating sum of digits (from the units, + - + -) is 0 or a multiple of 11.

Combine tests only when the factors share no common prime: 6 = 2\times 3 works, and 24 = 8\times 3 works, but 24 as “4 and 6” does not (both carry a 2, so they only guarantee 12).

Use the multi-checker below to enter any number and see which of 2 to 11 divide it, each with the reason:

Divisibility multi-checker

Exam Tip When a question asks for divisibility by a composite like 6, 12, 18 or 36, split it into coprime factors first. For 36, test 4 and 9 (since \gcd(4,9)=1 and 4\times 9 = 36), never “test 6 and 6”. Picking factors that share a prime is the single most common slip, and it loses easy marks.

Memory Tricks & One-Page Revision

Quick Revision Card

  • Parity laws: odd \pm odd = even, even \pm even = even, odd \pm even = odd. So a\pm b always share a parity.
  • Digit-sum tests: divisible by 3 or 9 \iff the digit sum is. (Powers of 10 are all one more than a multiple of 9.)
  • Alternating-sum test (11): add and subtract digits from the units; result 0 or \pm 11, \pm 22,\dots means divisible by 11.
  • Double-and-subtract (7): chop the units digit, double it, subtract from the rest; repeat.
  • Digital root = remainder mod 9 (a digital root of 9 means an exact multiple of 9). This powers casting out nines.
  • Combining tests: only multiply coprime factors, 6=2\times3, 24 = 8\times3. Avoid overlapping primes.
  • Powers of two (1,2,4,8,16,\dots) are the only numbers that cannot be written as a sum of two or more consecutive numbers.
  • Cryptarithms: distinct letters = distinct digits, no leading zero; chase the units digit and carries first.

Spot the Mistake

Common Exam Mistakes

  • Testing 24 with “4 and 6”. They overlap in a factor of 2, so they only certify divisibility by 12. Use 8 and 3 instead.
  • Confusing the digit sum with the alternating sum: the digit sum is for 3 and 9; the alternating sum is for 11.
  • Reading “digital root 9” as “remainder 9”. A digital root of 9 means remainder 0 (an exact multiple of 9).
  • Thinking casting out nines proves an answer correct. It only catches errors, agreement of digital roots is necessary, not sufficient.
  • In a cryptarithm, letting two different letters share a digit, or allowing a leading zero, both are forbidden.
  • Claiming “odd + even can be a multiple of 6”. Odd + even is odd, and every multiple of 6 is even.

Crack the Cryptarithm

A quick puzzle drill: a fresh cryptarithm appears, you read off the digits, type the hidden value, and build a streak.

Cryptarithm cracker
Find the digit the highlighted letter stands for.
Score 0 · Streak 0

Exam Corner: CBSE-Style Practice

Mixed Practice (objective, short, long, HOTS)

Objective type (1 mark each)

  1. The digital root of 85764 is ______.
  2. Is 90871 divisible by 11? (Yes / No)
  3. In the cryptarithm SO + SO = TOO (distinct digits, T\neq 0), the digit T is ______.

Short answer (2 marks each)

  1. For how many single digits p is the number 7p2 divisible by 3? List them.
  2. Use the double-and-subtract test for 7 to decide whether 3439 is divisible by 7, and state the remainder if it is not.

Long answer (3 marks each)

  1. How many four-digit numbers are multiples of 9? State the smallest and the largest, and the count.
  2. Solve the cryptarithm AB\times 3 = CAB, where the three-digit product shows the original two digits A,B after a new leading digit C (distinct digits, A\neq 0). Show it has only one solution.

HOTS (Higher Order Thinking)

  1. Take any three-digit number whose first and last digits differ, and subtract its reverse (for example 631 - 136). Prove that the result is always a multiple of 99, and explain why its digital root is always 9.
  2. The ISBN-10 code 0\text{-}13\text{-}468599\text{-}? is missing its final check digit, chosen so that 10a_1 + 9a_2 + \dots + 2a_9 + a_{10} is a multiple of 11. Find the missing check digit.

Assertion–Reason (Choose: (a) both true, R explains A; (b) both true, R does not explain A; (c) A true, R false; (d) A false, R true.)

  1. Assertion (A): 90871 is divisible by 11.   Reason (R): A number is divisible by 11 exactly when the alternating sum of its digits (from the units) is 0 or a multiple of 11.
  1. Digit sum 8+5+7+6+4 = 30, then 3+0 = \mathbf{3}. (This is also 85764’s remainder on division by 9.)
  2. Alternating sum from the units: 1-7+8-0+9 = 11, a multiple of 11, so Yes, 90871 is divisible by 11 (90871 = 11\times 8261).
  3. T = \mathbf{1}. Doubling a two-digit number to give a three-digit number forces the hundreds digit to be 1, and 50+50 = 100 gives S=5,\,O=0,\,T=1.
  4. 7+p+2 = 9+p must be a multiple of 3, so p \in \{\mathbf{0, 3, 6, 9}\}, four values.
  5. Strip the units digit: 343 - 2\times 9 = 325; again 32 - 2\times 5 = 22. Since 22 is not a multiple of 7, 3439 is not divisible by 7. Direct division gives remainder \mathbf{2}.
  6. The smallest four-digit multiple of 9 is 1008 = 9\times 112 and the largest is 9999 = 9\times 1111. The count is 1111 - 112 + 1 = \mathbf{1000}.
  7. Tripling AB must leave its last two digits unchanged, so 3B ends in B, forcing B=0 (then B=5 would need a carry that breaks the tens digit). With B=0, 3A ends in A only for A=5, giving 50\times 3 = 150: A=5,\,B=0,\,C=1. A full check of all two-digit AB confirms this is the only solution.
  8. A three-digit number is 100a + 10b + c and its reverse is 100c + 10b + a; their difference is 99a - 99c = 99(a-c), always a multiple of 99. Since 99(a-c) is a multiple of 9, its digital root is 9 (whenever a\neq c, so the difference is non-zero). For instance 631 - 136 = 495 = 99\times 5, with digit sum 4+9+5 = 18 \to 9.
  9. The weighted sum of the known digits is 10(0)+9(1)+8(3)+7(4)+6(6)+5(8)+4(5)+3(9)+2(9) = 192. The check digit a_{10} must make 192 + a_{10} a multiple of 11. Since 192 = 11\times 17 + 5 leaves remainder 5, we need a_{10} = 11 - 5 = 6, giving 198 = 11\times 18. The missing check digit is \mathbf{6}.
  10. (a), Both statements are true and R is exactly the rule that makes A true: the alternating sum of 90871 is 1-7+8-0+9 = 11, a multiple of 11.

Connections to Other Chapters

How This Chapter Links Forward

  • Chapter 6 (We Distribute, Yet Things Multiply): the algebra of parity and divisibility (6x+9y = 3(2x+3y)) is the distributive law at work, factoring out a common multiple.
  • Chapter 13 (Algebra Play): writing numbers as 5k+3 or 12n+1 to describe whole families at once is the leap from arithmetic to algebraic generalisation.
  • Chapters 3 & 8 (A Story of Numbers, Fractions in Disguise): factors, multiples and the LCM/HCF reasoning here underpin simplifying fractions and finding common denominators.

Glossary

Key Terms

  • Parity: whether a number is even or odd; preserved and combined by fixed rules under +,-,\times.
  • Divisibility test: a digit-based shortcut for deciding whether one number divides another without doing the full division.
  • Digit sum: the total of all the digits of a number; the basis of the 3- and 9-tests.
  • Alternating sum: digits added and subtracted in turn from the units; the basis of the 11-test.
  • Digital root: the single digit reached by repeatedly summing digits; equals the remainder on division by 9.
  • Casting out nines: an error-check that compares the digital roots of the inputs and the output of a calculation.
  • Check digit: an extra digit appended to a code (barcode, ISBN, account number) so that a single mistyped digit breaks a divisibility rule and is caught.
  • Cryptarithm: a puzzle where letters stand for distinct digits, with no leading zero.