How many terminal zeroes does 40^8 * 75^18 have?
Well, 40 to the 8th will have 8 terminal zeros. However many terminal zeros in the base, there will be that many times the exponent in the answer. examples: 102=100 103=1,000 104=10,000 1002=10,000 1,0002=1,000,000
As regards the second factor, any number ending with 5 no matter to what exponent you raise it, the answer will always end with a 5. examples: 52=25 152=225 252=625 1052=11,025 etc. So the second factor adds no terminal zeros to the product.
Multiplying the two factors together, there are 8 terminal zeros in the product.
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+128474 Note that we can write 40^8 * 75^18 as
( 4 * 10)^8 * ( 3 * 5^2)^18 =
( 2^3 * 5)^8 * ( 3 * 5^2)^18 =
(2 ^3)^8 * (5)^8 * 5^36 * 3 ^18 =
2^24 * 5^44 * 3^18 =
2^24 * 5^24 * 5^20 * 3^18
(2 * 5)^24 * 5^20 * 3^18 =
5^20 * 3^18 * ( 2 * 5) ^24
[5^20 * 3^18 ] * (10)^24 which has 24 terminal zeroes
I'm the Guest who posted the first answer. I see how I made a mistake. 408 is 65,536 x 108.
The 6 at the end of 65,536 times the 5 at the end of 7518 will produce at least one additional terminal zero.
I need go no further than that to realize the approach to my answer was simplistic, and, worse, it was wrong.
I'm mortified to make such an elementary arithmetic oversight, and I apologize if my blunder led anyone astray.
edited to add: Thank you, CPhill, for posting the correct answer.
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