filippocarta01 wrote:
Let s be the set of all 6-digit numbers of the form \(54x,y12\), where each of \(x\) and \(y\) can be any of the digits 3, 5, 8, and x = y is allowed. What is the probability that a randomly chosen number from set S is divisible by 8?
(A) \(0\)
(B) \(\frac{1}{9}\)
(C) \(\frac{2}{9}\)
(D) \(\frac{1}{3}\)
(E) \(\frac{2}{3}\)
Let me discuss on certain observations/analysis, which could help you in arriving at answer faster.
Divisibility rule of 8If last 3 digits are divisible then the entire number is divisible by 8.
Why?Because any number with last three digits zero is always divisible by 8. For example, 123000 or 876000 or 111000. (1000 is 8*125 and any of the numbers can be written as 111*1000 or 111*8*125).
We utilise the above fact for our divisibility rule.
Any number abcdef can be written as abc000+d00+ef. Here abc000 is divisible by 8.
So, we check def for divisibility by 8.
Some more factsWhen you divide 100, 300, 500, 700 or 900 with 8, it will leave a remainder 4, so the last two digits should also give 4 as remainder.
For example 128 = 100+28 = 100+24+4
When you divide 200, 400, 600, 800 or 000 with 8, it will leave a remainder 0, so the last two digits should also give 0 as remainder.
For example 224 = 200+24
Back to questionLast 2 digits are 12, so the remainder is 4, implying that the number should have odd hundreds digit=> 3 and 5 are the possibility for y
As the thousands digit does not matter, y could be any of the three digits.
Total ways number is divisible by 8 is 3*2 or 6
Total ways to select x and y is 3*3 or 9
Hence the probability is \(\frac{6}{9}\) or \(\frac{2}{3}\)
E