✨ Math Magic Portal ✨

Where numbers come alive and problems become puzzles!

Problem Solving Adventure

1
Prove that n² - n is divisible by 2 for every positive integer n.
Step 1: Let's consider the expression n² - n
Step 2: Factorize it: n(n - 1)
Step 3: Observe that for any positive integer n, either n or (n-1) must be even (divisible by 2)
Step 4: Therefore, the product n(n-1) is always divisible by 2
Conclusion: n² - n is divisible by 2 for every positive integer n Q.E.D.
2
A milkman has 175 litres of cow's milk and 105 litres of buffalo's milk. He wishes to sell the milk by filling the two types of milk in cans of equal capacity. Calculate:
  1. Capacity of a can
  2. Number of cans of cow's milk
  3. Number of cans of buffalo's milk
Step 1: Find the HCF of 175 and 105 to determine the largest possible equal capacity
Step 2: Prime factors of 175 = 5 × 5 × 7
Step 3: Prime factors of 105 = 3 × 5 × 7
Step 4: Common factors are 5 and 7, so HCF = 5 × 7 = 35
Solution i: Capacity of each can = 35 litres
Solution ii: Number of cow's milk cans = 175 ÷ 35 = 5 cans
Solution iii: Number of buffalo's milk cans = 105 ÷ 35 = 3 cans
3
When the positive integers a, b and c are divided by 13 the respective remainders are 9, 7 and 10. Find the remainder when a + 2b + 3c is divided by 13.
Step 1: Express a, b, c in terms of multiples of 13 plus remainders:
a = 13k + 9
b = 13m + 7
c = 13n + 10
Step 2: Compute a + 2b + 3c:
= (13k + 9) + 2(13m + 7) + 3(13n + 10)
= 13k + 9 + 26m + 14 + 39n + 30
Step 3: Separate terms divisible by 13 and remainder terms:
= 13(k + 2m + 3n) + (9 + 14 + 30)
= 13(k + 2m + 3n) + 53
Step 4: Find remainder when 53 is divided by 13:
53 ÷ 13 = 4 with remainder 1 (since 13 × 4 = 52 and 53 - 52 = 1)
Final Answer: The remainder is 1
4
Show that 107 is of the form 4q + 3 for any integer q.
Step 1: We need to express 107 in the form 4q + 3
Step 2: Subtract 3 from 107: 107 - 3 = 104
Step 3: Divide 104 by 4: 104 ÷ 4 = 26
Step 4: Therefore, we can write:
107 = 4 × 26 + 3
Conclusion: 107 is indeed of the form 4q + 3 where q = 26
5
If (m+1)th term of an A.P. is twice the (n+1)th term, then prove that (3m+1)th term is twice the (m+n+1)th term.
Step 1: Recall the general term of an A.P.: aₖ = a + (k-1)d
Step 2: Given: aₘ₊₁ = 2 × aₙ₊₁
⇒ a + md = 2(a + nd)
⇒ a + md = 2a + 2nd
⇒ -a + md - 2nd = 0
⇒ a = (m - 2n)d ...(1)
Step 3: Now consider (3m+1)th term:
a₃ₘ₊₁ = a + (3m)d = [(m - 2n)d] + 3md = (4m - 2n)d ...from (1)
Step 4: Consider (m+n+1)th term:
aₘ₊ₙ₊₁ = a + (m+n)d = [(m - 2n)d] + (m+n)d = (2m - n)d
Step 5: Now observe:
2 × aₘ₊ₙ₊₁ = 2 × (2m - n)d = (4m - 2n)d = a₃ₘ₊₁
Conclusion: Hence proved that (3m+1)th term is twice the (m+n+1)th term
6
Find the 12th term from the last term of the A.P. -2, -4, -6, ..., -100.
Step 1: First, find the total number of terms in the A.P.
Given A.P.: -2, -4, -6, ..., -100
First term (a) = -2, common difference (d) = -2
aₙ = a + (n-1)d ⇒ -100 = -2 + (n-1)(-2)
⇒ -100 = -2 - 2n + 2 ⇒ -100 = -2n ⇒ n = 50
Step 2: The 12th term from the last is the (50 - 12 + 1) = 39th term from the start
Step 3: Calculate the 39th term:
a₃₉ = a + 38d = -2 + 38(-2) = -2 - 76 = -78
Final Answer: The 12th term from the last is -78
7
Two A.P.'s have the same common difference. The first term of one A.P. is 2 and that of the other is 7. Show that the difference between their 10th terms is the same as the difference between their 21st terms, which is the same as the difference between any two corresponding terms.
Step 1: Let both A.P.s have common difference d
Step 2: First A.P.: a₁ = 2, so nth term = 2 + (n-1)d
Step 3: Second A.P.: a₁ = 7, so nth term = 7 + (n-1)d
Step 4: Difference between nth terms:
= [7 + (n-1)d] - [2 + (n-1)d] = 7 - 2 = 5
Step 5: For 10th terms: difference = 5
Step 6: For 21st terms: difference = 5
Conclusion: The difference between any two corresponding terms is always 5, which is simply the difference between their first terms.
8
A man saved ₹16500 in ten years. In each year after the first he saved ₹100 more than he did in the preceding year. How much did he save in the first year?
Step 1: This forms an A.P. where:
Sum S₁₀ = 16500, n = 10, d = 100
Step 2: Sum of A.P. formula: Sₙ = n/2 [2a + (n-1)d]
16500 = 10/2 [2a + 9×100]
16500 = 5 [2a + 900]
3300 = 2a + 900
2a = 2400 ⇒ a = 1200
Final Answer: He saved ₹1200 in the first year.
9
Find the G.P. in which the 2nd term is 6 and the 6th term is 96.
Step 1: In a G.P., nth term = arⁿ⁻¹
Step 2: Given: a₂ = ar = 6 ...(1)
Step 3: Given: a₆ = ar⁵ = 96 ...(2)
Step 4: Divide equation (2) by (1):
ar⁵ / ar = 96/6 ⇒ r⁴ = 16 ⇒ r = ±2
Step 5: Case 1: r = 2
From (1): a × 2 = 6 ⇒ a = 3
G.P.: 3, 6, 12, 24, 48, 96, ...
Step 6: Case 2: r = -2
From (1): a × (-2) = 6 ⇒ a = -3
G.P.: -3, 6, -12, 24, -48, 96, ...
Final Answer: There are two possible G.P.s: 3, 6, 12, 24, ... or -3, 6, -12, 24, ...
10
The value of a motorcycle depreciates at the rate of 15% per year. What will be the value of the motorcycle 3 years hence, which is now purchased for ₹45,000?
Step 1: Initial value (P) = ₹45,000
Step 2: Depreciation rate (r) = 15% per year
Step 3: Time (n) = 3 years
Step 4: Value after depreciation formula: A = P(1 - r/100)ⁿ
A = 45000 × (1 - 0.15)³ = 45000 × (0.85)³
Step 5: Calculate (0.85)³ = 0.85 × 0.85 × 0.85 ≈ 0.614125
Step 6: Final value = 45000 × 0.614125 ≈ 27635.625
Final Answer: The value after 3 years will be approximately ₹27,635.63