Concepts of Indices | Moderate
The sum of digits of the number (625)⁶⁵ × (128)³⁶ is ______
Answer & Explanation
Final answer 25
Given: (625)⁶⁵ × (128)³⁶
Express the bases as powers of 5 and 2: 625 = 5⁴ and 128 = 2⁷
(625)⁶⁵ × (128)³⁶
= 5^(4×65) × 2^(7×36)
= 5^260 × 2^252
= 5^8 × 5^252 × 2^252
= 5^8 × 10^252
Now 5⁸ = 390625
So the full number is:
390625 followed by 252 zeros.
Sum of digits = 3 + 9 + 0 + 6 + 2 + 5 = 25
Answer: 25
Concepts of Logs + Maxima Minima | Hard
If log₆₄(x²) + log₈(√y) + 3 log₅₁₂( √(y)z) = 4, where x, y and z are positive real numbers, then the minimum possible value of (x + y + z) is
A) 36
B) 48
C) 96
D) 24
Answer & Explanation
Correct Answer B (48)
Given log₆₄(x²) + log₈(√y) + 3 log₅₁₂( √(y)z) = 4 x, y, z > 0
Step 1: Convert all logs to the same base (base 8)
64 = 8²; 512 = 8³
Convert log₆₄(x²) to base 8: log₆₄(x²) = (1/2) log₈(x²) = log₈x
Convert 3 log₅₁₂(√(y)z) to base 8: 3 log₅₁₂(√(y)z) = 3 (1/3) log₈(√(y)z) = log₈(√(y)z)
Substitute the converted terms back into the original equation: log₈x + log₈(√y) + log₈(√(y)z) = 4 Combine the logarithmic terms using the property logₐA + logₐB = logₐ(AB): log₈(x ⋅ √y ⋅ √y ⋅ z) = 4 => log₈(xyz) = 4
Convert the logarithmic equation to an exponential equation: xyz = 8⁴ => xyz = 4096
Apply the Arithmetic Mean – Geometric Mean (AM-GM) inequality.
For positive real numbers x, y, and z, the minimum value of their sum occurs when x = y = z.
(x + y + z) / 3 ≥ ³√(xyz)
Substitute the value of xyz: (x + y + z) / 3 ≥ ³√(4096)
Calculate the cube root: Since 16³ = 4096, we have: (x + y + z) / 3 ≥ 16
Solve for the minimum value of (x + y + z): x + y + z ≥ 3 × 16 x + y + z ≥ 48
Concepts of AP GP + Quadratic Equations | Hard
Let an be the nth term of a decreasing infinite geometric progression.
If a₁ + a₂ + a₃ = 52 and a₁a₂ + a₂a₃ + a₃a₁ = 624, then the sum of this geometric progression is
A) 54
B) 60
C) 63
D) 57
Answer & Explanation
Correct Answer: 54
Let the GP be a, ar, ar² with common ratio r (0 < r < 1 since it is decreasing and infinite).
Given:
a + ar + ar² = 52 → a(1 + r + r²) = 52
a·ar + ar·ar² + ar²·a = 624 → a²(r + r² + r³) = 624
From these, form a relation involving only r.
Divide equation (2) by [equation (1)]²:
[a²(r + r² + r³)] / [a²(1 + r + r²)²] = 624 / 52²
So:
(r + r² + r³) / (1 + r + r²)² = 624 / 2704 = 39 / 169
Now just try simple values of r between 0 and 1.
Try r = 1/2:
r + r² + r³ = 1/2 + 1/4 + 1/8 = 7/8
1 + r + r² = 1 + 1/2 + 1/4 = 7/4 → squared = 49/16
Ratio = (7/8) / (49/16) = 2/7 (not 39/169).
Try r = 1/3:
r + r² + r³ = 1/3 + 1/9 + 1/27 = 13/27
1 + r + r² = 1 + 1/3 + 1/9 = 13/9 → squared = 169/81
Ratio = (13/27) / (169/81) = 39/169 (matches).
So r = 1/3.
From a(1 + r + r²) = 52 → a(1 + 1/3 + 1/9) = 52 → a(13/9) = 52 → a = 36.
Sum of infinite GP = a / (1 − r) = 36 / (1 − 1/3) = 36 / (2/3) = 54.
