q1 in how many ways can 2310 be expressed as a product of 3 factors? a 41 b 23 c 56 d 46 ans->41 q2 each family in a village has atmost two adults and the total no. of boys in this village is less than the no. of girls.similarly the no. of girls is less than the no. of adults in the village.raghubir ,the chief of this village is the only adult in his family.the min. no. of families in the village is: a 2 b 3 c 4 d none of above ans->4 q3 direction for question below:a number of 109 digits is written as follows: 1 2 3 4 5 6 7 8 9 10 11 12 13......... what is the least possible whole number which must be subtracted from the given number so that it become divisible by 3 and 11 both? ans->3.
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Q1. In how many ways can 2310 be expressed as a product of 3 factors?
a 41
b 23
c 56
d 46
Ans:
This question was asked in last year TIME mock and the answer given was ridiculous and confusing
When a number can be expressed as a product of n distinct primes,
then it can be expressed as a product of 3 numbers in (3(n - 1) + 1)/2 ways
Here 2310 = 2 x 3 x 5 x 7 x 11 i. e n =5 and answer = (34 + 1)/2 = 41
Answer to Q3.
First let’s complete the series: a number of 109 digits are written as follows: 1 2 3 4 5 6 7 8 9 10 11 12 13.........
Number of 1 digit number =9
Number of 2 digit number = 50 so the complete series: 12 3 4 . . . . . . . . .59
First we 'll check divisibility by 3 : Sum of the numbers = (59 x 60)/2 = 59 x 30 . . so its divisible by 3
Now divisibility by 11 : Difference between each number is 1 i.e 3-2 =1 ; 5-4-=1 ; . . . 59-58=1 (29 such numbers) and the first number we have excluded from the calculation is 1
So Difference between numbers at odd place and even place = 29 + 1 =30
For number to be divisible by 11 the above difference should Zero or multiple of 11
Thus if we add 3 to the above number it’ll be divisible by both 11 and 3
Hello Anita,
In your comment you have mentioned that:
Now divisibility by 11 : Difference between each number is 1 i.e 3-1 =1 ; 5-4-=1 ; . . . 59-58=1 (29 such numbers) and the first number we have excluded from the calculation is 1
So Difference between numbers at odd place and even place = 29 + 1 =30
Are you sure that the difference everytime will be 1?
I did not get that part. May be its correct. I will have to write down and solve it. If possible, please show the steps using a few numbers. Sorry for the trouble.
Thank You.
Ravi Raja
n/a
Here it is because the whole number is just a sequence of first 59 positive integers
1 2 3 4 5 . . . . . . .59
so the diff . is always 1 like 3-2=1 ; 5-4=1 . . . . . /
I think you must not have seen the Qs/Ans properly ;
Any how you can add the sum of numbers at even place 2 + 4 + . . . .58 and odd place 1 + 3 + . . . . 59 and substract them to get the same result.
Hello Anita,
But the test of 11 is to add and subtract the "DIGITS" in the odd and even places and "Not" the "NUMBERS" in the odd and even places and secondly, you need to start from the right hand side of the number and not from the left hand side otherwise the result will differ.
Take this example:
What is the remainder when 92 is divided by 11?
What you are doing is: Starting from the right hand side and getting: 9 - 2 = 7. So the remainder is 7.
But if you start from the right hand side, then according to the test, you will get: 2 - 9 = - 7. So the remainder is: - 7 + 11 = 4 and now you can divide 92 by 11 and check that the remainder is 4 and and not 7.
I hope this example is clear to you or may be your explanation is still not clear to me. But its ok. I will try solving it again and let you know what I am getting. Sorry once again for the trouble.
Thank You.
Ravi Raja
n/a
Ok Please tell me if it is correct to chk the divisibilty by Substracting S1 From S2
Where S1 = 59 + 57 + . . . . . + 1 and S2 = 58 + 56 + . . . . . + 2
Please give me some time to verify both the methods. I will check it for both the methods and then let you know.
Thank You.
Ravi Raja
n/a
Hi anita !
youR solution tells that ADD 3 to get the required number.
question is SUBSTRACT 3 to get the number.
lets first leave the single digit nos as they are:
so we are starting with 10 and going till 59.
The numbers in the even places are 1,2,3,4, 5, repeated 10 times , sum = 15 * 10 = 150
the numbers in the odd places is 1 to 9 repeated 5 times. , sum = 45 * 5 = 225
the difference bettween the consecutive digits for the single digits nos = 1.
so net after subtracting the odd places nad even places = 225 - 150 + 5 = 80.
thus we have to SUBTRACT 3 to get the required answer.
hi
secondharsa,z it a formula ? if d question had been "in howmany ways can it b xpressed as a product of 2 factors" den what would hav been d answer?
I apologize for making a mess of such an easy question.
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