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The integers 1,2,3, ... 100 are written on the board. What is the smallest number of these integers that can be wiper off so that the product of the remaining integers ends in 2?

a) 20

b) 21

c) 22

d) 23

e) 24

 Jul 17, 2020
edited by Guest  Jul 17, 2020
 #1
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If you divided 100! [which is 1 x 2 x 3 x 4 x 5..........x 98 x 99 x 100] by 5, 23 times and the 24th time divide it by 10, you will end up with 2 at the end of the still very large number. In fact, this is the number that you will end up with after dividing 100! by 5 some 23 times and divide it by 10 on the 24th time:

 

7 8287703748 2793470738 9236886635 8969387002 3897110337 7849044106 8567513352 1044008415 9164912362 4209837854 8214060408 2346158362 5595826977 8892685312 - which is 141 digits long after bringing it down from 158 digits (which is 100!).

 Jul 17, 2020
 #2
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You can actually do it in removing  20 numbers: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95 and the last and 20th number is 10. I think!

 Jul 17, 2020

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