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If x = 100!, y = 100!/60!, and z = 100^40 , order them from least to greatest without using a calculator.

 Feb 6, 2024
 #1
avatar+129895 
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100 ! / 60!   < 100^40 < 100!

 

 

cool cool cool

 Feb 6, 2024
 #2
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Can you tell me how you solved it?

Curry30shivank  Feb 6, 2024
 #3
avatar+129895 
+1

100!  / 60!  =   100 * 99 * 98 * .....* 62 * 61

 

This is definitely smaller than 100^40

 

100!  = 

(100*99) * (98 *97) * (96 * 95) * (94 * 93) * (92 * 91) * ......* (10 * 11)

 

Note that we  have   ( 100 - 10) / 2 + 1 =  46 terms in this partial evaluation of 100!....each is > 100

 

So

 

100! > 100^40

 

So

 

1001/ 60! < 100^40 < 100!

 

cool cool cool

 Feb 6, 2024
 #4
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0

Thank you, that made a lot of sense now.

Curry30shivank  Feb 6, 2024

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