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Number Theory
If both 11^2 and 3^3 are factors of the number a * 4^3 * 6^2 * 13^11, then what is the smallest possible value of a?
A. 121
B. 3267
C. 363
D. 33
E. None of the above
- by Maze
Sun Feb 21, 2010 10:40 am- Forum: Problem Solving
- Topic: Number Theory
- Replies: 3
- Views: 1281
P = 200C40 * 160C40 * 120C40 * 80C40 * 40C40 = 200!/(40!)^5
Powers of 5 in 200! = 200/5 + 40/5 + 8/5(only take the quotient) = 49
Powers of 5 in 40! = 40/5 + 8/5 = 9
Powers of 5 in (40!)^5 = 9*5 = 45
Hence Maximum Power of 5 in P = 49-45 = 4
- by Maze
Sat Feb 06, 2010 12:51 am- Forum: Problem Solving
- Topic: Probability + Power
- Replies: 3
- Views: 1187
shashank.ism wrote:P = 200!/((25!)^8)
The highest power of 5 in 200! = 40 +8+ 1 = 49
The highest power of 5 in 25! = 5 + 1 = 6.
The highest power of 5 in 25!^8 = 6 x 8 =48
Therefore, the highest power of 5 in P = 49 – 48 = 1
can you explain how u obtained the value of P??
- by Maze
Sat Feb 06, 2010 12:33 am- Forum: Problem Solving
- Topic: Probability + Power
- Replies: 3
- Views: 1187