Question

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.

Answer 1

Answer: 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

Answer 2

There are 50 even numbers - divisible by 2
There are 25 multiples of 4
There are 12 multiples of 8
There are 6 multiples of 16
There are 3 multiples of 32
There is one multiple of 64
Each of these is also counted in the line above

So the unique multiples are (1 x 64) x (2 x 32) x (3 x 16) x (6 x 8) x (13 x 4) x (25 x 2)
= (1 x 2^6) x (2 x 2^5) x (3 x 2^4) x (6 x 2^3) x (13 x 2^2) x (25 x 2^1)
So n = 6 + 10 + 12 + 18 + 26 + 25 = 97

The slow (but easier way) to see this is to write down all the even numbers in factors of 2 and count the number of 2S
Eg 2, 2x2, 2x3, 2x2x2, 2x5, 2x2x3, 2x7, 2x2x2x2, 2x9, 2x2x5, 2x11, 2x2x2x3, 2x13, 2x2x7, 2x15 etc
(This only got to 30 but you should get the idea - there are 26 twos up to there)

This is a tough question to explain simply