Final Answer: All numbers of the form 3k + 2 where k = 0, 1, 2, 3, ...
2
Flower Pot Arrangement
A man has 532 flower pots. He wants to arrange them in rows such that each row contains 21 flower pots. Find the number of completed rows and how many flower pots are left over.
1
We need to divide 532 flower pots into rows of 21 pots each.
2
This is a division problem: 532 ÷ 21
3
Let's perform the division:
21 × 25 = 525
25 complete rows (525 pots) with remaining:
4
Subtract from total:
532 - 525 = 7
7 leftover flower pots
5
Final Answer: 25 complete rows with 7 flower pots left over.
3
Consecutive Integers
Prove that the product of two consecutive positive integers is divisible by 2.
1
Let the two consecutive integers be n and n+1.
n
n+1
2
In any two consecutive integers, one must be even (divisible by 2).
3
Therefore, the product n(n+1) must include this even number.
4
Since the product contains a factor of 2, it is divisible by 2.
5
Conclusion: The product of any two consecutive positive integers is always divisible by 2.
Example: 3 × 4 = 12 (divisible by 2)
Example: 6 × 7 = 42 (divisible by 2)
4
Remainder Sum
When the positive integers a, b and c are divided by 13, the respective remainders are 9, 7 and 10. Show that a+b+c is divisible by 13.
1
Express each number using the division algorithm:
a = 13k + 9
b = 13m + 7
c = 13n + 10
2
Add them together:
a + b + c = 13(k + m + n) + (9 + 7 + 10)
3
Calculate the remainder sum:
9 + 7 + 10 = 26
9
+
7
+
10
=
26
4
Since 26 is divisible by 13 (26 = 13 × 2), the entire sum is divisible by 13.
5
Conclusion: a + b + c is divisible by 13.
5
Square Remainders
Prove that square of any integer leaves the remainder either 0 or 1 when divided by 4.
1
Any integer n can be either even or odd.
2
Case 1: n is even (n = 2k)
n² = (2k)² = 4k² → divisible by 4 (remainder 0)
Even number squared:
(2k)² = 4k² ≡ 0 mod 4
3
Case 2: n is odd (n = 2k + 1)
n² = (2k + 1)² = 4k² + 4k + 1 = 4(k² + k) + 1 → remainder 1
Odd number squared:
(2k+1)² = 4(k² + k) + 1 ≡ 1 mod 4
4
These are the only two possibilities for any integer.
5
Conclusion: The square of any integer leaves remainder 0 or 1 when divided by 4.
Examples:
2² = 4 ≡ 0 mod 4
3² = 9 ≡ 1 mod 4
4² = 16 ≡ 0 mod 4
5² = 25 ≡ 1 mod 4
6
Euclid's Algorithm
Use Euclid's Division Algorithm to find the Highest Common Factor (HCF) of:
340 and 412
867 and 255
10224 and 9648
84, 90 and 120
i
340 and 412:
412 = 340 × 1 + 72
340 = 72 × 4 + 52
72 = 52 × 1 + 20
52 = 20 × 2 + 12
20 = 12 × 1 + 8
12 = 8 × 1 + 4
8 = 4 × 2 + 0
HCF = 4
ii
867 and 255:
867 = 255 × 3 + 102
255 = 102 × 2 + 51
102 = 51 × 2 + 0
HCF = 51
iii
10224 and 9648:
10224 = 9648 × 1 + 576
9648 = 576 × 16 + 432
576 = 432 × 1 + 144
432 = 144 × 3 + 0
HCF = 144
iv
84, 90 and 120:
First find HCF of 84 and 90:
90 = 84 × 1 + 6
84 = 6 × 14 + 0 → HCF = 6
Now find HCF of 6 and 120:
120 = 6 × 20 + 0 → HCF = 6
Final HCF = 6
7
Common Divisor with Remainder
Find the largest number which divides 1230 and 1926 leaving remainder 12 in each case.
1
If a number divides 1230 and leaves remainder 12, then it divides 1230 - 12 = 1218 exactly.
2
Similarly, it divides 1926 - 12 = 1914 exactly.
3
So we need to find HCF of 1218 and 1914.
4
Using Euclid's algorithm:
1914 = 1218 × 1 + 696
1218 = 696 × 1 + 522
696 = 522 × 1 + 174
522 = 174 × 3 + 0
5
Final Answer: The largest number is 174.
8
Linear Combination
If d is the Highest Common Factor of 32 and 60, find x and y satisfying d = 32x + 60y.
1
First find HCF of 32 and 60 using Euclid's algorithm:
60 = 32 × 1 + 28
32 = 28 × 1 + 4
28 = 4 × 7 + 0
HCF d = 4
2
Now express 4 as a linear combination of 32 and 60:
4 = 32 - 28 × 1
3
From step 1: 28 = 60 - 32 × 1
4
Substitute into previous equation:
4 = 32 - (60 - 32 × 1) × 1
4 = 32 - 60 + 32
4 = 2 × 32 - 1 × 60
Comparing with d = 32x + 60y:
x = 2
y = -1
5
Final Answer: x = 2, y = -1
9
Changing Divisors
A positive integer when divided by 88 gives the remainder 61. What will be the remainder when the same number is divided by 11?
1
Let the number be N. Then N = 88k + 61 for some integer k.