Problem Solving

How many POSITIVE INTEGERS less than 1000 are divisible by 5 which can be formed using the digits 0 to 9 inclusive (no repetition) ?

(A) 144
(B) 154
(C) 155
(D) 182
(E) 214

OA: B

Hi ,

Is there any another method to solve this?

Please advise.

Best regards ,

Shreyans

@shreyans

You can take the following approach to solve this problem:-

Step1 - You can find the total no. of positive integers less than 1000 which are divisible by 5 using the Arthimetic Progression

995= 5+(n-1)*5
=> n = 199

Step 2 - Now determine the number of 1-digit, 2-digit and 3-digit integers which are divisible by 5 BUT have repetitive digits in them and subtract the total number of them from 'n' in Step1.

1-digit numbers which are divisible by 5

Since 1-digit number cannot have repetitive digits.

Hence, no. of 1 digit no.'s = 0



2-digit numbers which are divisible by 5

2-digit numbers divisible by 5 are 10,15,20,25,...,50,55,60,...95.

Only 55 has repetitive digits.

Hence, no. of 2-digit no.'s = 1



3-digit numbers which are divisible by 5

Between 100-195, numbers which are divisible by 5 = 100,105,110,115,120,125,130,135,140,145,150,155,160,165,170,175,180,185,190,195

You can see that there are exactly 4 numbers between 100-195 which have repetitive digits.

Similarly, there will be 4 numbers with repetitive digits between 200-299, 300-399 and so on..except for numbers between 500-599 where there will be 12 numbers with repetitive digits - 500,505,515, 525, 535, 545, 550,555, 565, 575, 585,595.

Hence, no. of 3-digit no.'s = 4*8+12=44

Step3 -

Hence, Total number of positive numbers less than 1000 which are divisible by 5 and with no repeated digit

=199-0-1-44

=154

j_shreyans wrote:Hi ,

Is there any another method to solve this?

Please advise.

Best regards ,

Shreyans

I would solve using Brent's approach.
That said, here's an alternate solution:

Multiples of 5 with no repeated digits = (all multiples of 5) - (multiples of 5 with repeated digits).

All multiples of 5 between 1 and 1000:
Here, any multiple of 5 between 001 and 999, inclusive, is valid.

Number of options for the rightmost digit = 2. (0 or 5.)
Number of options for the leftmost digit = 10.
Number of options for the middle digit = 10.
To combine these options, we multiply:
2*10*10 = 200.

Of these 200 options, one is invalid: 000.
Subtracting this one invalid option, we get:
200-1 = 199.

Bad Case 1: 2-digit multiples of 5 with repeated digits
55.
Total options = 1.

Bad Case 2: 3-digit multiples of 5 with exactly 2 digits the same
Bad Case 2A: X00 or X55
Number of options for the repeated digit = 2. (X00 or X55.)
Number of options for the hundreds digit = 9. (Any digit 1 through 9.)
To combine these options, we multiply:
Total options = 2*9 = 18.
Of these 18 options, one is invalid: 555
Subtracting this one invalid option, we get:
18-1 = 17.

Bad Case 2B: 5X5
Number of options for the tens digit = 9. (Any digit but 5.)

Bad Case 2C: XX5 or XX0
Number of options for the units digit = 2. (0 or 5.)
Number of options for the repeated digit = 9. (Any digit 1 through 9.)
To combine these options, we multiply:
2*9 = 18.
Of these 18 options, one is invalid: 555.
Subtracting this one invalid option, we get:
18-1 = 17.

Bad Case 3: 3-digit multiples of 5 with 3 repeated digits
555.
Total options = 1.

Subtracting the bad cases from the total, we get:
199 - (1 + 17 + 9 + 17 + 1) = 154.

The correct answer is B.

Another way to count the multiples of 5 between 1 and 1000.

For any set of EVENLY SPACED INTEGERS:
Total number of integers = (biggest - smallest)/(interval) + 1.
The interval is the distance between successive terms.

In this case:
Biggest = 995.
Smallest = 5.
Interval = 5.
Thus:
Total number of multiples of 5 = (995-5)/5 + 1 = 199.

Another way to count the multiples of 5 between 1 and 1000.

For any set of EVENLY SPACED INTEGERS:
Total number of integers = (biggest - smallest)/(interval) + 1.
The interval is the distance between successive terms.

In this case:
Biggest = 995.
Smallest = 5.
Interval = 5.
Thus:
Total number of multiples of 5 = (995-5)/5 + 1 = 199.


Hi Guru ,

I was also solving like above method.

What should be next step because we have to find no repeated digits.

Please advise.

Thanks ,

Shreyans [/quote]

Here's a simpler way, I think.

Let our number be _ _ _. We'll start with 0 0 5 (that is, 5 itself) as a special case, so that gives us one number.

Now suppose we have _ _ 5. The first digit can be anything other than 5, so there are nine options. (0, 1, 2, 3, 4, 6, 7, 8, 9). The second digit can be anything other than 5 or the first digit, so there are eight options. That gives us 9 * 8 = 72 numbers.

Now suppose we have 0 _ 0. There are nine options here (010, 020, 030, 040, ..., 090).

Now suppose we have _ _ 0, but the first digit is NOT 0. We again have 9 * 8 = 72 options, much like in our _ _ 5 case.

So in all we have 1 + 72 + 9 + 72 = 154 options.

There are total 1000/5 -1 = 199 numbers divisible by 5.
Now we ll eliminate the repetitions.
Repetitions:
1 digit: None.
2 digit: only 55. just 1.
3 digit: _ _ 0 or _ _ 5.
_ _ 0 : 110, 220,...., 990. -> 9.
100, 200,...., 900. -> 9. Total 18.
_ _ 5 : 115, 225,...., 995. -> 9.
155, 255,...., 955. -> 8.(except 555, which is repeated)
505, 515,...., 595. -> 9.(except 555, which is repeated) So total 9+8+9=26.
So, total no of repetitions = 1+18+26 = 45.
Answer = 199 - 45 = 154.
Thanks Brent for showing the easiest way.

~Binit.