The registration number of Lisa's car is a 10 digit number. Though she

Question

The registration number of Lisa's car is a 10 digit number. Though she forgot the last 4 digits, but she knew that product of those digits is an odd number which has more than two factors. How many trials would guarantee access to the online details of the vehicle?

1. 608
2. 610
3. 612
4. 624
5. 625

A. 608
B. 610
C. 612
D. 624
E. 625

chetan2u · Accepted Answer

This means all the 4 digits are odd..
So any of 1,3,5,7,9..
If there were no restrictions, we would get possible ways 5*5*5*5=625..

But there is a restriction of more than 2 factors, this means it cannot be a prime number  ..
Possibilities of product to be a prime.
1) All of them are 1.....1*1*1*1.....ONE way
2) 3 are 1, and the fourth is a prime so 4!/3!=4..3! for three 1s.
Since there are 3 primes, 3,5,7, the ways become 4*3=12..
Total ways not possible 1+12=13..

Answer with restrictions =625-13=612

C

sach24x7 · Answer

Product of digits is odd means all numbers should be odd

First place - 1,3,5,7,9
Second place- 1,3,5,7,9
Third place- 1,3,5,7,9
Fourth place- 1,3,5,7,9

So 5*5*5*5= 625

Only product of 1 1 1 1 will have one factor.

So answer is 624.

Hope I'm right.

Posted from my mobile device

chetan2u · Answer

You have missed out on 1113, 1131, 1115, 5111, 1117, and so on

sach24x7 · Answer

Yes. Im wrong. Missed those. Thanks!

I think 4*4 =16 such cases will add.

So total 15 and answer would be 610.

Is it ok?

Posted from my mobile device

chetan2u · Answer

No, 9 itself is non prime so the number can be 1119..
You have to consider only primes 3,5,7

sach24x7 · Answer

True! Such blind i am!

MuddyFort · Answer

The question states that the number has  more than 2 factors  which means 3 or more!
For a product of n numbers to be odd, all the n numbers must be odd. Hence, possible digits are 1, 3, 5, 7, 9.

The main points are:
- 1 is considered a factor of every number.
- It is only mentioned that the number has more than 2 factors, and not "prime" factors. Hence, having a 9 would technically account for 3 factors that is (9, 3, 1)

Using reverse logic:

num_trials = Total - E1 - E2

- Total = 5^4
- E1. Remove all the cases where only 1 factor exists. The only case would be 1111 (1*1*1*1 = 1 which has only 1 as the factor)
E1 = 1
- E2. Remove all the cases where only 2 factors exist. The cases would be 111x and all its arrangements (1*1*1*7 = 7 which has 1, 7 as the factors). Also, x!=9 (1*1*1*9 would have 1, 3, 9 as the factors which will satisfy the condition).
E2 = (4!/3!) * 3 = 12

num_trials = 5^4 - 13 = 625-13 =  612

The registration number of Lisa's car is a 10 digit number. Though she

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The registration number of Lisa's car is a 10 digit number. Though she

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