Question

A box contains 100 tickets, numbered from 1 to 100. A person picks out three tickets from the box, such that the product of the numbers on two of the tickets yields the number on the third ticket. Which of the following tickets can never be picked as third ticket?

• A. 10
• B. 12
• C. 25
• D. 26

Answer 1

The correct option is C 25

$\left(a\right)100=5\times 2possible$
$\left(b\right)12=6\times 2possible$
(c) 25 = 5 x 5 not possible as only one 5 Numbers ticket
$\left(d\right)30=15\times 2possible$