if the product (15^17)(28^18)(35^18) was evaluated, it would end with a string of consecutive zeros. How many zeros will be in the string?
EG: 10X10X10= 3個零
請解答(WITH STEPS AND EXPLAINATIONS PLS)
數學題一問
2007-10-22 8:42 pm
回答 (2)
2007-10-22 9:24 pm
✔ 最佳答案
15^17=9852612533569335937528^18=111903730358193158266814464
35^18=6211904899255558013916015625
(28^17)(35^17)(15^17)
=[(28*35*15)^17](35*28)
=(14700^17)*980
=(147^17)*(100^17)*98*10
=(147^17)*98*10^35
35個零
2007-10-23 1:10 am
(15^17)(28^18)(35^18)
= [(3x5)^17]x[(2x2x7)^18]x[(5x7)^18]
= (3^17 x 5^17) x(2^18 x 2^18 x 7^18) x (5^18 x 7^18)
= 3^17 x (5^17 x 2^17 x 2) x (2^18 x 5^18) x (7^18 x 7^18)
= 3^17 x (2 x (5x2)^17) x ((2x5)^18) x 7^36
= 3^17 x 2 x 10^17 x 10^18 x 7^36
= 2 x 3^17 x 7^36 x 10^35
Since the expression has 2x3^17x7^36 will not produce any trailing zeroes and 10^35 has 35 trailing zeroes, the expression has 35 consecutive trailing zeroes.
= [(3x5)^17]x[(2x2x7)^18]x[(5x7)^18]
= (3^17 x 5^17) x(2^18 x 2^18 x 7^18) x (5^18 x 7^18)
= 3^17 x (5^17 x 2^17 x 2) x (2^18 x 5^18) x (7^18 x 7^18)
= 3^17 x (2 x (5x2)^17) x ((2x5)^18) x 7^36
= 3^17 x 2 x 10^17 x 10^18 x 7^36
= 2 x 3^17 x 7^36 x 10^35
Since the expression has 2x3^17x7^36 will not produce any trailing zeroes and 10^35 has 35 trailing zeroes, the expression has 35 consecutive trailing zeroes.
收錄日期: 2021-04-13 14:05:28
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https://hk.answers.yahoo.com/question/index?qid=20071022000051KK01032