Problem Solving

Came across this good one-

A binary number is said to be 'Tri-one', if it has exactly three 1s

1)If the 'Tri-one' numbers are arranged in ascending order, then find the rank of the least eight digit 'Tri-one' number

a) 35 (b) 32 (c) 34 (d) 40 (e) 36

2) How many 'Tri-one' numbers less than 110, when converted to decimals are divisible by 5? [ note- 110 and 5 are to base 10]

a) 5 (b) 6 (c) 7 (d) 8 (e) 10.

OA to follow soon..

Before I or somebody answers, I would like to say a big HELLO.

Welcome back Jyothi

Hey Anil.. Thanks a lot for the warm welcome I will be active on the forum for a few more months again. Feels great to be back!

Is the answer for first question [spoiler]e)[/spoiler] ?

I am very very bad at counting numbers so used my common sense ( this could be very time consuming in the actual GMAT).

The least 8 digit binary number will be 10000011 ( viz. 131). Since this number is the least 8 digit number,all other numbers smaller than this number will be 3,4,5,6 or 7 digit.

You can have three 1's in 15 different ways for a 7 digit number, 10 different ways for a 6 digit number, 6 different ways for a 5 digit number, 3 different ways for a 4 digit number and 1 way for a 3 digit number.

15 + 10 + 6+ 3 + 1 = 35.

Therefore, 131 is the 36th number.

Or, three ones can be placed in a seven digit number in the following number of ways :

7 ! / (4!)(3!) = 35. This covers all number lesser than the least 8 digit tri - number. So this number is 36th in the list.

Hey perfect answer...

I was about to post the second method of solution, which is quite quick and easy.

The first one, I tried to compute for every digit:

abcd is a 4 digit number and we need to arrange 3 ones's in it.

so in essence 'a' has to be 1 for the digit to be four digit and the other b, c and d have 2 options either 1 or 0.

so 1*2*2*2/3! (3! for the 3 ones) -> it seems to result in something wrong.

Answer A for second question?

I had to convert all multiples of 5 less than 110 to binary to check if they are tri-ones. I would appreciate for a better solution.

Hello, for the first method, considering a four digit number we can solve as following ( not using formulae) :

A B C D

1 1 _ _

Now the third "1" can be placed in 2 ways ( C and D). Now we keep the first "1" at A, move the second "1" to C. The third "1" can now be placed only at D ( You cannot place it at B as that option has already been accounted for). So you have 3 ways in which three "1" can be placed in a four digit number.

For the second question, I thought of the following solution :

110 is seven digits in the binary system. So lets start with seven digit numbers

Lets write down the binary system :

64 32 16 8 4 2 1

1 _ _ _ _ _ _

The above number represents 64. To be divisible by 5 the number must end with "5" or "0". So we have to place the two remaining "1" in a way that the units place of sum of these two "1" equals to 1 or 6. 1 is not possible. We can get 6 if we place "1" below 2 and 4 , and also below 32 and 4

64 32 16 8 4 2 1

1 _ _ _ 1 1 _


64 32 16 8 4 2 1

1 1 _ _ 1 _ _


So we have two number here viz. 70 and 100

Try similarly for 6 digit numbers and you will get 2 numbers there as well : 50 and 35

For 5 digit numbers you get 1 number : 25

None for 4 digit numbers and none for 3 digit numbers.

So total = 5

Hope it is correct.

Very very tedious process...took me close 5 minutes to do it this way..but I am no genius...I am sure someone will come up with a much better solution..

Sorry the binary system is not alligned with the "1" ...hope you guys can make out what I am trying to out across.