Revision and important questions: Number Play

Number Play22 min readLesson 6 of 7

6.1 Divisibility rules at a glance Divisor Test Reason 2 units digit is even every higher place is a multiple of 10 5 units digit is 0 or 5 multiples of 10 end in 0, add 5 10 units digit is 0 same place-value argument 4 last two digits form a multiple of 4 100 is a multiple of 4 8 last three digits form a multiple of 8 1000 is a multiple of 8 3 digit sum divisible by 3 each power of 10 is 1 mod 3 9 digit sum divisible by 9 each power of 10 is 1 mod 9 6 divisible by 2 and by 3 6 = 2 \times 3 11 alternating digit sum is 0 or a multiple of 11 powers of 10 alternate between 1 and -1 mod 11 18 divisible by 2 and by 9 18 = 2 \times 9 For a composite divisor, check its coprime factors: divisibility by 24 follows from divisibility by 3 and 8, not from 4 and 6, because 12 is divisible by 4 and by 6 yet 24 does not divide it. The LCM rule covers the general case: if a number is divisible by b and c, it is divisible by LCM(b, c).

Divisor	Test	Reason
2	units digit is even	every higher place is a multiple of 10
5	units digit is 0 or 5	multiples of 10 end in 0, add 5
10	units digit is 0	same place-value argument
4	last two digits form a multiple of 4	100 is a multiple of 4
8	last three digits form a multiple of 8	1000 is a multiple of 8
3	digit sum divisible by 3	each power of 10 is 1 mod 3
9	digit sum divisible by 9	each power of 10 is 1 mod 9
6	divisible by 2 and by 3	6 = 2 × 3
11	alternating digit sum is 0 or a multiple of 11	powers of 10 alternate between 1 and -1 mod 11
18	divisible by 2 and by 9	18 = 2 × 9

6.2 Digital roots and remainders The digital root of N is the single digit obtained by repeatedly summing digits. It equals N mod 9, with 0 written as 9. Consequences worth remembering: every multiple of 9 has digital root 9; the digital root of N + 10 is one more than the digital root of N, unless the digital root is 9, when it wraps to 1; adding a multiple of 9 does not change the digital root; digital roots give a quick check on addition and multiplication, but they cannot prove a calculation right, because different numbers can have the same digital root. The book also records that Aryabhata II's Mahasiddhanta, from around 950 CE, describes this repeated digit-sum method and its use in checking calculations.

dr (N) \equiv N (mod 9)

Digital root is the remainder on division by 9, with 0 written as 9

N \equiv a_{0} - a_{1} + a_{2} - a_{3} + \dots (mod 11)

Divisibility by 11 from alternating digits starting at the units place

6.3 Cryptarithms: method Read the units column and list the possible digits and carries. Use the carry in the tens column, then the hundreds column and so on. Remember that different letters are different digits and a leading letter is never 0. Check the complete solution by substituting the digits back into the puzzle. For A1 + 1B = B0, the units column forces B = 9 with carry 1, and the tens column gives A + 2 = 9, so A = 7. For AB + 37 = 6A, the tens column gives A + 3 + carry = 6; with carry 1, A = 2, and then B = 5. For multiplication cryptarithms, bound the first number from the size of the product: in PQ \times 8 = RS the product is two digits, so PQ \leq 12, and the distinct-digit condition leaves 12 \times 8 = 96.

6.4 Ten important questions Question 1. Find the sum of four consecutive numbers whose least number is 12. Answer. 12 + 13 + 14 + 15 = 54. Algebraically 4n + 6 = 48 + 6 = 54. Question 2. Four consecutive numbers add to 58. Find them. Answer. 4n + 6 = 58, so n = 13. The numbers are 13, 14, 15 and 16. Question 3. Is 47,832 divisible by 9? Answer. Digit sum 4 + 7 + 8 + 3 + 2 = 24, and 24 is not divisible by 9, so 47,832 is not divisible by 9. It leaves remainder 6. Question 4. Find the smallest number greater than 12 that leaves remainder 2 when divided by 3 and remainder 2 when divided by 4. Answer. x - 2 is a multiple of LCM(3, 4) = 12. Taking x - 2 = 12 gives x = 14, so the numbers are 14, 26, 38, ... (the expression 12n + 2 also gives x = 2 for n = 0, but the question asks for a number greater than 12). Question 5. If 31z5 is a multiple of 9, find z. Answer. 3 + 1 + z + 5 = 9 + z must be a multiple of 9. So z = 0 (sum 9) or z = 9 (sum 18). Question 6. Find the digital root of 489710. Answer. 4 + 8 + 9 + 7 + 1 + 0 = 29, then 2 + 9 = 11, then 1 + 1 = 2. The digital root is 2. Question 7. What is the digital root of 9a + 36b + 13? Answer. The multiples of 9 do not affect the remainder mod 9; the digital root is that of 13, which is 4. Question 8. How many multiples of 9 lie between 4300 and 4400? Answer. 11, from 4302 to 4392. Question 9. If 48a23b is a multiple of 18, find all pairs (a, b). Answer. 17 + a + b must be 18 or 27, so a + b = 1 or 10; b must be even. The pairs are (1, 0), (2, 8), (4, 6), (6, 4) and (8, 2). Question 10. Solve the cryptarithm AB + 37 = 6A. Answer. At the units place B + 7 gives A with carry 1. At the tens place A + 3 + 1 = 6, so A = 2, and then B + 7 ends in 2, giving B = 5. Check: 25 + 37 = 62.

6.5 Common mistakes to avoid Checking divisibility by 9 from the units digit instead of the digit sum. Using the units digit for divisibility by 4 or 8 instead of the last two or three digits. Applying the divisibility test for 11 with the wrong starting digit; always start the alternating sum from the units digit. Treating 0 as a digital root; the convention is to write 9 for a remainder of 0. Forgetting that a cryptarithm uses different digits for different letters and no leading zero. Proving "sometimes true" with one example only; a sometimes-true claim needs both an example and a counterexample. Using 4 and 6 to test divisibility by 24; use 3 and 8 instead.

Exam

Show the test and the reason in divisibility questions: write the digit sum, the last digits or the alternating sum, then state the conclusion. In remainder questions, write both numbers in divisor form and combine the remainders. In "always/sometimes/never" questions, give one example and one non-example before the algebra.

in hinglish

Revision ke liye divisibility rules yaad rakho: 2, 5, 10 units digit se; 4 last do digits se; 8 last teen digits se; 3 aur 9 digit sum se; 11 alternating sum se; 6 ke liye 2 aur 3 dono; 18 ke liye 2 aur 9 dono. Composite numbers ke liye coprime factors use karo; 24 ke liye 3 aur 8 sahi hain, 4 aur 6 nahi. Digital root digits ko baar-baar jodkar single digit deta hai aur remainder mod 9 batata hai; 0 ki jagah 9 likhte hain. Multiples of 9 ka root 9 hota hai, aur multiple of 9 jodne se root nahi badalta. Cryptarithm mein letters digits hote hain, har letter alag aur leading digit 0 nahi. Units place se carry nikaalo, phir tens place solve karo, aur answer substitute karke check karo. Always/sometimes/never questions mein example aur counterexample dono do, phir algebra se generalise karo.