Math Magic Explorer

1

For what values of natural number n, 4ⁿ can end with the digit 6?

Step 1: Let's observe the pattern of 4ⁿ for different values of n.

Step 2: 4¹ = 4 (ends with 4)

Step 3: 4² = 16 (ends with 6)

Step 4: 4³ = 64 (ends with 4)

Step 5: 4⁴ = 256 (ends with 6)

Conclusion: We see that when n is even, 4ⁿ ends with 6.

2

If m, n are natural numbers, for what values of m, does 2ⁿ × 5^m ends in 5?

Step 1: For a number to end with 5, it must be divisible by 5 but not by 2.

Step 2: The expression is 2ⁿ × 5^m.

Step 3: If n ≥ 1, the number will be even (divisible by 2) and thus cannot end with 5.

Step 4: The only possibility is when n = 0, but since n is a natural number (n ≥ 1), this isn't possible.

Wait! Actually, if n = 0 (if we consider it), then the number becomes 5^m which ends with 5 for any m ≥ 1.

But: Typically natural numbers start from 1, so there may be no solution under standard definition.

Alternative: If we allow n=0, then for all m ≥ 1, the number ends with 5.

3

Find the HCF of 252525 and 363636.

Step 1: Let's find prime factors of both numbers.

Step 2: 252525 = 25 × 10101 = 5² × 3 × 7 × 13 × 37

Step 3: 363636 = 36 × 10101 = 2² × 3² × 7 × 13 × 37

Step 4: Common factors are 3, 7, 13, and 37.

Step 5: HCF = 3 × 7 × 13 × 37 = 10101

Fun Fact: Notice how 252525 = 25 × 10101 and 363636 = 36 × 10101!

4

If 13824 = 2^a × 3^b, then find a and b.

Step 1: Let's factorize 13824 completely.

Step 2: 13824 ÷ 2 = 6912

Step 3: 6912 ÷ 2 = 3456

Step 4: Continue dividing by 2 until odd: 3456 ÷ 2 = 1728 → 864 → 432 → 216 → 108 → 54 → 27

Step 5: We've divided by 2 9 times, so a = 9.

Step 6: Now factorize 27: 27 = 3³, so b = 3.

Verification: 2⁹ × 3³ = 512 × 27 = 13824 ✓

5

If p₁^x₁ × p₂^x₂ × p₃^x₃ × p₄^x₄ = 113400 where p's are primes in ascending order and x's are integers, find the values.

Step 1: First, factorize 113400 completely.

Step 2: 113400 ÷ 2 = 56700 → 28350 → 14175 (divided by 2 three times)

Step 3: 14175 ÷ 3 = 4725 → 1575 → 525 → 175 (divided by 3 three times)

Step 4: 175 ÷ 5 = 35 → 7 (divided by 5 two times)

Step 5: 7 ÷ 7 = 1 (divided by 7 one time)

Step 6: So, 113400 = 2³ × 3⁴ × 5² × 7¹

Solution: p₁ = 2, x₁ = 3
p₂ = 3, x₂ = 4
p₃ = 5, x₃ = 2
p₄ = 7, x₄ = 1

6

Find the LCM and HCF of 408 and 170 by applying the fundamental theorem of arithmetic.

Step 1: Find prime factorizations.

Step 2: 408 = 2³ × 3¹ × 17¹

Step 3: 170 = 2¹ × 5¹ × 17¹

Step 4: HCF = product of lowest powers of common primes = 2¹ × 17¹ = 34

Step 5: LCM = product of highest powers of all primes = 2³ × 3¹ × 5¹ × 17¹ = 2040

Check: HCF × LCM = 34 × 2040 = 69360 = 408 × 170 ✓

7

Find the greatest 6-digit number exactly divisible by 24, 15, and 36.

Step 1: First find LCM of 24, 15, and 36.

Step 2: 24 = 2³ × 3, 15 = 3 × 5, 36 = 2² × 3²

Step 3: LCM = 2³ × 3² × 5 = 8 × 9 × 5 = 360

Step 4: Largest 6-digit number is 999999.

Step 5: Divide 999999 by 360 → 999999 ÷ 360 ≈ 2777.775

Step 6: Take integer part: 2777, multiply by 360 → 2777 × 360 = 999720

Solution: The greatest 6-digit number divisible by 24,15,36 is 999720.

8

What is the smallest number that when divided by 35, 56 and 91 leaves remainder 7 in each case?

Step 1: We need a number N such that N ≡ 7 mod 35, N ≡ 7 mod 56, N ≡ 7 mod 91.

Step 2: This means N-7 is divisible by 35, 56, and 91.

Step 3: Find LCM of 35, 56, 91.

Step 4: 35 = 5 × 7, 56 = 2³ × 7, 91 = 7 × 13

Step 5: LCM = 2³ × 5 × 7 × 13 = 8 × 5 × 7 × 13 = 3640

Step 6: So N-7 = 3640 ⇒ N = 3640 + 7 = 3647

Verification: 3647 ÷ 35 = 104 remainder 7 ✓
3647 ÷ 56 = 65 remainder 7 ✓
3647 ÷ 91 = 40 remainder 7 ✓

9

Find the least number that is divisible by the first ten natural numbers.

Step 1: We need LCM of numbers 1 through 10.

Step 2: Prime factorizations:
1 = 1
2 = 2
3 = 3
4 = 2²
5 = 5
6 = 2 × 3
7 = 7
8 = 2³
9 = 3²
10 = 2 × 5

Step 3: Take highest powers of all primes: 2³, 3², 5, 7

Step 4: LCM = 8 × 9 × 5 × 7 = 2520

Fun Fact: 2520 is the smallest number divisible by all numbers 1 through 10!