Figure it Out
Q1. The sum of four consecutive numbers is 34. What are these numbers?
Answer:
Let the four consecutive numbers be x, x+1, x+2, x+3. Their sum is x + (x+1) + (x+2) + (x+3) = 4x + 6. Set equal to 34: 4x + 6 = 34 → 4x = 28 → x = 7. The numbers are 7, 8, 9, 10.
Q2. Suppose p is the greatest of five consecutive numbers. Describe the other four numbers in terms of p.
Answer:
If p is the greatest, the five consecutive numbers are p-4, p-3, p-2, p-1, and p. The other four numbers are p-1, p-2, p-3, p-4.
Q3. For each statement below, determine whether it is always true, sometimes true, or never true. Explain your answer. Mention examples and non-examples as appropriate. Justify your claim using algebra. (i) The sum of two even numbers is a multiple of 3. (ii) If a number is not divisible by 18, then it is also not divisible by 9. (iii) If two numbers are not divisible by 6, then their sum is not divisible by 6. (iv) The sum of a multiple of 6 and
Answer:
(i) Sometimes true. Example: 2+4=6 (multiple of 3), non-example: 2+6=8 (not multiple of 3). Algebra: even numbers = 2a, 2b, sum = 2(a+b). This is even, but not necessarily multiple of 3.
(ii) Never true. Counterexample: 9 is not divisible by 18 but is divisible by 9. If a number is not divisible by 18, it could still be divisible by 9 (e.g., 9, 27). So the statement is false.
(iii) Sometimes true. Example: 5+7=12 (divisible by 6), non-example: 5+5=10 (not divisible by 6). Algebra: numbers not divisible by 6 can be of forms 6k+1,6k+2,...,6k+5; sum may or may not be divisible by 6.
(iv) The statement is incomplete in the text; hence no solution can be provided.
Q1. Find, without dividing, whether the following numbers are divisible by 9. (i) 123 (ii) 405 (iii) 8888 (iv) 93547 (v) 358095
Answer:
Using the divisibility rule for 9: A number is divisible by 9 if the sum of its digits is divisible by 9.
(i) 123: Sum = 1+2+3 = 6, not divisible by 9 → Not divisible by 9.
(ii) 405: Sum = 4+0+5 = 9, divisible by 9 → Divisible by 9.
(iii) 8888: Sum = 8+8+8+8 = 32, not divisible by 9 → Not divisible by 9.
(iv) 93547: Sum = 9+3+5+4+7 = 28, not divisible by 9 → Not divisible by 9.
(v) 358095: Sum = 3+5+8+0+9+5 = 30, not divisible by 9 → Not divisible by 9.
Q2. Find the smallest multiple of 9 with no odd digits.
Answer:
We need a multiple of 9 where all digits are even (0,2,4,6,8). The smallest such number is 0, but typically we consider positive numbers. Check multiples: 9 (odd digit 9), 18 (odd digit 1), 27 (odd), 36 (odd digit 3), 45 (odd), 54 (odd), 63 (odd), 72 (odd), 81 (odd), 90 (odd digit 9), 99 (odd), 108 (odd digit 1), 117 (odd), 126 (odd), 135 (odd), 144 (odd), 153 (odd), 162 (odd), 171 (odd), 180 (odd), 189 (odd), 198 (odd), 207 (odd), 216 (odd), 225 (odd), 234 (odd), 243 (odd), 252 (odd), 261 (odd), 270 (odd), 279 (odd), 288 (all digits even? 2,8,8 are even) → 288. So the smallest is 288.
Q3. Find the multiple of 9 that is closest to the number 6000.
Answer:
Divide 6000 by 9: 9 × 666 = 5994, remainder 6. So 5994 is 6 less than 6000. The next multiple is 6003 (9 × 667), which is 3 more than 6000. Therefore, the closest multiple is 6003 (since 3 < 6). Answer: 6003.
Q4. How many multiples of 9 are there between the numbers 4300 and 4400?
Answer:
Find the smallest multiple ≥ 4300: 4300 ÷ 9 = 477 remainder 7, so next multiple = 4300 + (9-7) = 4302 (9×478 = 4302). Largest multiple ≤ 4400: 4400 ÷ 9 = 488 remainder 8, so largest = 4400 - 8 = 4392 (9×488 = 4392). The multiples are 9×478 to 9×488. Count = 488 - 478 + 1 = 11. Thus, there are 11 multiples.
Q1. The digital root of an 8-digit number is 5. What will be the digital root of 10 more than that number?
Answer:
If the digital root of a number is 5, then the number ≡ 5 (mod 9). Adding 10 gives: (number + 10) ≡ 5 + 10 ≡ 15 ≡ 6 (mod 9). So the digital root of 10 more is 6 (since digital root is the remainder mod 9, except that 0 becomes 9). But 6 ≠ 9, so digital root = 6.
Q2. Write any number. Generate a sequence of numbers by repeatedly adding 11. What would be the digital roots of this sequence of numbers? Share your observations.
Answer:
Choose any number, say 1. Sequence: 1, 12, 23, 34, 45, 56, 67, 78, 89, 100, 111, ... Digital roots: 1, 3, 5, 7, 9, 2, 4, 6, 8, 1, 3, ... Observation: The digital roots cycle through the odd numbers 1,3,5,7,9 then even numbers 2,4,6,8 and repeat every 9 terms. This is because adding 11 adds 1+1=2 to the digital root mod 9.
Q3. What will be the digital root of the number 9a + 36b + 13?
Answer:
9a is divisible by 9, 36b = 9×4b is divisible by 9. So 9a + 36b + 13 ≡ 13 (mod 9). 13 ≡ 4 (mod 9). Therefore digital root is 4.
Q4. Make conjectures by examining if there are any patterns or relations between (i) the parity of a number and its digital root. (ii) the digital root of a number and the remainder obtained when the number is divided by 3 or 9.
Answer:
(i) Parity (even/odd) and digital root are not directly related. For example, even numbers can have any digital root, and odd numbers can also have any digital root. However, we can observe that digital root gives remainder mod 9, which doesn't determine parity. (ii) The digital root is exactly the remainder when the number is divided by 9, except that a remainder of 0 gives digital root 9. For division by 3, digital root mod 3 gives the remainder when divided by 3 (with 0 replaced by 3 if digital root is 3,6,9). So there is a direct relation.
Exercises
Q7. When divided by 7, the number 661 leaves a remainder of 3, and 4779 leaves a remainder of 5. Without calculating, can you say what remainders the following expressions will leave when divided by 7? Show the solution both algebraically and visually. (i) 4779 + 661 (ii) 4779 – 661
Answer:
Algebraically: Let 661 = 7a + 3 and 4779 = 7b + 5, where a,b are integers.
(i) 4779 + 661 = (7b+5) + (7a+3) = 7(a+b) + 8 = 7(a+b+1) + 1. Remainder = 1.
(ii) 4779 – 661 = (7b+5) – (7a+3) = 7(b–a) + 2. Remainder = 2.
Visually: Draw two groups: 661 as 7 groups of some size plus 3 leftover dots; 4779 as 7 groups plus 5 leftover dots. For addition, combine leftovers: 3+5=8 dots; 8 can form one more group of 7, leaving 1 dot. For subtraction, remove 3 dots from 5; remaining 2 dots.
Q8. Find a number that leaves a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5. What is the smallest such number? Can you give a simple explanation of why it is the smallest?
Answer:
Let the number be N. N ≡ 2 (mod 3), N ≡ 3 (mod 4), N ≡ 4 (mod 5). Observe that each condition means N+1 is divisible by 3, 4, and 5 respectively (since 2+1=3, 3+1=4, 4+1=5). So N+1 is a multiple of LCM(3,4,5)=60. Thus N+1 = 60k, so N = 60k - 1. Smallest positive k=1 gives N=59. Check: 59÷3=19 rem 2, 59÷4=14 rem 3, 59÷5=11 rem 4. Yes. Why smallest? Because if k=0, N=-1 (negative), so the smallest positive is when k=1.
Q1. Find, without dividing, whether the following numbers are divisible by 9. (i) 123 (ii) 405 (iii) 8888 (iv) 93547 (v) 358095
Answer:
(i) 1+2+3=6, not divisible by 9 → No. (ii) 4+0+5=9, divisible → Yes. (iii) 8+8+8+8=32, not → No. (iv) 9+3+5+4+7=28, not → No. (v) 3+5+8+0+9+5=30, not → No.
Q2. Find the smallest multiple of 9 with no odd digits.
Answer:
Multiples of 9: 9,18,27,36,45,54,63,72,81,90,99,108,... Check digits: 9 (odd), 18 (1 odd), 27 (odd), 36 (3 odd), 45 (5 odd), 54 (5 odd), 63 (6 even,3 odd), 72 (7 odd), 81 (1 odd), 90 (9 odd), 99 (odd), 108 (1 odd), 117 (odd), 126 (1 odd), 135 (1,3,5 odd), 144 (1 odd), 153 (1,3,5 odd), 162 (1 odd), 171 (1,7 odd), 180 (1,8 even but 1 odd), ... Continue: 288 (2,8,8 all even). So smallest is 288.
Q3. Find the multiple of 9 that is closest to the number 6000.
Answer:
6000 ÷ 9 = 666 remainder 6. 9×666 = 5994, 9×667 = 6003. Differences: |6000-5994|=6, |6000-6003|=3. So closest is 6003.
Q4. How many multiples of 9 are there between the numbers 4300 and 4400?
Answer:
First multiple ≥4300: 4300÷9=477.77..., 9×478=4302. Last multiple ≤4400: 4400÷9=488.88..., 9×488=4392. Number of terms = 488-478+1 = 11.
Q1. The digital root of an 8-digit number is 5. What will be the digital root of 10 more than that number?
Answer:
Digital root 5 means number ≡ 5 mod 9. Adding 10: ≡ 5+10=15≡6 mod 9. So digital root = 6.
Q2. Write any number. Generate a sequence of numbers by repeatedly adding 11. What would be the digital roots of this sequence of numbers? Share your observations.
Answer:
Start with 1: 1,12,23,34,45,56,67,78,89,100,111,122,... Digital roots: 1,3,5,7,9,2,4,6,8,1,3,5,... The pattern cycles every 9 steps through all digits 1-9. Adding 11 adds 2 to digital root mod 9.
Q3. What will be the digital root of the number 9a + 36b + 13?
Answer:
9a and 36b are divisible by 9, so digital root = digital root of 13 = 1+3=4.
Q4. Make conjectures by examining if there are any patterns or relations between (i) the parity of a number and its digital root. (ii) the digital root of a number and the remainder obtained when the number is divided by 3 or 9.
Answer:
(i) No direct relation; both even and odd numbers can have any digital root. (ii) Digital root equals remainder mod 9 (0 becomes 9). Remainder mod 3 = digital root mod 3 (or digital root itself if ≤3, else reduce mod 3). So digital root determines both remainders.