The digits $1,2,3,4,5,6$ are arranged to form a 6-digit number. The probability that the number formed is divisible by 4 can be expressed as $\frac{a}{b}$ where $a$ and $b$ are coprime positive integers. What is the value of $a + b$ ?

**
Details and assumptions
**

Each digit is only used once in forming the 6-digit number.

The answer is 19.

**
This section requires Javascript.
**

You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.

Thanks!

Kou$htav Chakrabarty
- 7 years, 6 months ago

I was so close! (7/30) or 37

Rhoy Omega
- 7 years, 6 months ago

nice

Pamela Hsu
- 7 years, 6 months ago

the same as mine.:)

Kim Chan
- 7 years, 6 months ago

i was close. i mistakenly wrote 8/30. i wrote the answer as 38

Pranav Kirsur
- 7 years, 6 months ago

- Divisibilty by 4: Number must be finished (in this case): $(12, 16, 24, 32, 36, 52, 56, 64) \rightarrow 8$ cases..
- When we fill 2 last numbers, remain 4 numbers to permutation.
- $P_{4}$ : $4!$
- All numbers wanted: $8 \cdot 4!$

3 Helpful
0 Interesting
0 Brilliant
0 Confused

If the last two digits are divisible by 4, the whole number will be divisible by 4. The only possible endings for the six digit numbers will thus be the numbers 12, 16, 24, 32, 36, 52, 56 and 64.

With these eight endings, there will only by four numbers left to place, giving in total

$(4 \times 3 \times 2) \times 8 = 192$

numbers divisable by 4.

There are $6! =720$ possible numbers in all, giving the probability as

$\frac{192}{720}=\frac{4}{15}$

$4+15= \boxed{19}$

3 Helpful
0 Interesting
0 Brilliant
0 Confused

The total no. of 6-digit nos. using these 6 digits=6!=720;

For a no. to be divisible by 4,last 2 digits should be divisible by 4.Thus the last two digits can be 12,16,24,32,36,52,56 or 64.The no. of required nos. with two digits fixed=4!=24;

Thus,total no. of nos. divisible by 4=8*24=192;

Thus probability=192/720=4/15;

4+15=19.

2 Helpful
0 Interesting
0 Brilliant
0 Confused

The number would be divisible by $4$ if and only if last $2$ digits of it are divisible by $4$ . Now consider possible two-digit numbers which can be formed using digits from the set $(1,2,3,4,5,6)$ using one element at a time. These are: $(12,16,24,32,36,52,56,64)$ . Thus there are $8$ such digits. Now if we choose any of these as last two digits, remaining four digits can be permuted in $4!$ ways. Also, without any restriction, we can form $6!$ total numbers. Hence the required probability is:

$\frac{4!*8}{6!}=\frac{4}{15}$ .

Hence the required answer is: $4+15=\boxed{19}$

1 Helpful
0 Interesting
0 Brilliant
0 Confused

1 Helpful
0 Interesting
0 Brilliant
0 Confused

Numbers that are divisible by 4 are any number has the last two digits divisible by 4. for these six digits numbers are 12,16,24,32,36,52,56,64 (8 numbers)

so we have 8*(4P4)=192 numbers are divisible by 4 and we have 6P6=720 numbers could be formed from 6 digits

so the probability is 192/720=4/15 so a+b= 19

0 Helpful
0 Interesting
0 Brilliant
0 Confused

0 Helpful
0 Interesting
0 Brilliant
0 Confused

×

Problem Loading...

Note Loading...

Set Loading...

First we have to know that numbers divisible by $4$ have to have two last algarisms divisible by $4$ . Then the possibilities to this is ${12,16,24,32,36,52,56,64 = 8}$ possibilities. However are still missing the four firsts algarisms that be organized by $4!$ differents ways, then we have $8 * 4! = 192$ . But our amostral space is $6!$ that are the all forms to creat differents six algarisms numbers using $1,2,3,4,5,6$ . In this way, we have the probability is $P = \frac{192}{720} = \frac{4}{15}$ . The answer is $4+15 = 19$ .