Question

What is the least positive integer 'n' such that the product of 2016 and 'n' ends in
the digits 00?

Answer 1

2016 = 4*504
4*2*252
4*4*126
4*4*

much work..
2^5*3^2*7

Answer 2

so we gotta get a 100 in there

Answer 3

2^5*3^2*7
9*7,4*4*2

so... its probably 25?

Answer 4

Hmmm....

5 * 6 = 30
.........2.....You are right :o

Answer 5

ya well

Answer 6

100 = 10^2=2^2*5^2 sooo
and we got
2016 = 2^5*3^2*7
so we are missing the 5^2 in there to get to a 100 multiple

Answer 7

moree ask moree

Answer 8

to justify it more clearly incase they want like proper solution

2016 = 2^2*(2^3*3^2*7)
the part in the bracket has no factor 10 so it doesnt end in 0
now if we multiple any number not ending in 0 by 100 we will get xxxx00
so we gotta multiply by 25

2^2*(2^3*3^2*7) --> (25*2^2)*(2^3*3^2*7)
now its in form 100*some number not ending in 0

Answer 9

Yeah that makes sense. Factorizationmunizationishunnnn and getting products 2*5 = 10 * 10 blah blah blah.

Answer 10

2*5 = 10 not 10*10

Answer 11

:)