Question 10: If the product of all the numbers from 1 to 100 is divisible by 2^n, then what is the maximum possible value for n: A. 128; B. 87; C. 64; D. 97; E. 70.

D

Step-by-step explanation:

100!=2*3*4*...*100

Counting which numbers are divisible by 2:

2,4,6,8,10,12,14,...,96,98,100 : 50 numbers (100/2=50)

Counting which numbers are divisible by 4=2²:

4,8,12,16,20,...96,100: 25 numbers (100/4=25)

Counting which numbers are divisible by 8=2³:

8,16,24,32,40,...,96: 12 numbers (100/8=12.5 ==> 12)

Counting which numbers are divisible by 16=2⁴:

16,32,48,...,96: 6 numbers (100/16=6.25 ==> 6)

Counting which numbers are divisible by 32=2⁵:

32,64,96 : 3 numbers(100/32=3.125 ==>3)

Counting which numbers are divisible by 64=2⁶:

64 : 1 number (100/64=1.5625 ==>1)

Sum of those numbers= 50+25+12+6+3+1=97

100! is divisible by 2⁹⁷.

Answer D

6 (3) - 4 + 32

hope this

0

step-by-step explanation:

you can not divide by 0 so 4 ÷ 0 would be 0

0 × 9999 =0

mark brainliest   : )