Find the number of zeros in last 11*22*33*...*2525
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Answer:
i hope its helpful for you
Step-by-step explanation:
A = 11 * 22 * 33 * 44 * 55 * .... 99 * 110
A = 11¹⁰ * 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10
A zero at the end of A appear as many as tens in the product PLUS the number of 2 X 5 products. So we have one 5 and one 10.
===> So two zeroes for this product.
Answered by
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Number of zeros in last 11*22*33*...*2525 = 6
Given Expression is :
- 11 * 22 * 33 * ... * 2525
To Find:
- Number of zeroes in last
Write Each number as multiple of 2 and 5 where ever possible
- 11 = 11 x 1
- 22 = 11 x 2
- 33 = 11 x 3
- 44 = 11 x 2 x 2
- 55 = 11 x 5
- 66 = 11 x 2 x 3
- 77 = 11 x 7
- 88 = 11 x 2 x 2 x 2
- 99 = 11 x 3 x 3
- 1010 = 101 x 2 x 5
- 1111 = 11 x 101
- 1212 = 101 x 3 x 2 x 2
- 1313 = 101 x 14
- 1414 = 101 x 7 x 2
- 1515 = 101 x 3 x 5
- 1616 = 101 x 2 x 2 x 2 x 2
- 1717 = 101 x 17
- 1818 = 101 x 2 x 3 x 3
- 1919 = 101 x 19
- 2020 = 101 x 5 x 2 x 2
- 2121 = 101 x 3 x 7
- 2222 = 101 x 11 x 2
- 2323 = 101 x 23
- 2424 = 101 x 2 x 2 x 2 x 3
- 2525 = 101 x 5 x 5
Number of 5 as factors = 6
Number of 2 as factors are much more than 6
Hence number of zeroes at end is 6 ( as 5 x 2 = 10)
number of zeros in last 11*22*33*...*2525 = 6
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