## Few Quant Doubts

Hello Every one
Please help me in solving the following questions
1. How many of the numbers under 100 that are divisible by 10 can you make by adding four consecutive integers?
2. Determine whether or not there exists a positive integer n such that n id divisible by exactly 2000 different prime numbers and 2^{n} is divisible by n.
3 If n= 1 – 1/2 + 1/3 – 1/4 + . . . . . – 1/1318 + 1/1319
Then
(a) n is divisible by1979
(b) n is divisible by1989
(c) n is divisible by1879
(d) n is divisible by1889
(e) None of the above

## ques 1.
Let the integers be

ques 1.
Let the integers be n-1, n , n+1 , n+2 adding we will have 4n+2 , n v [0, 98]
so 4n+2 = 10 k , k = 1 , 2 ... 9
thus we have to find k such that 10k -2 is divisible by 4
clearly k = 1, 3 , 5 , 7 , 9 .
Tus the answer is 5 i guess.

## Ques 2 : The answer is no a

Ques 2 : The answer is no and i cannot give the solution because the question is plain silly.

## Ques 3 : are you sure the

Ques 3 : are you sure the question is correct ? if yes i think the ans should be none of the above . anyone else thinks different !

## Thanks for the first two

Thanks for the first two answers and i had missed something in third question the correct question is
If n/m= 1 – 1/2 + 1/3 – 1/4 + . . . . . – 1/1318 + 1/1319
Then
(a) n is divisible by1979
(b) n is divisible by1989
(c) n is divisible by1879
(d) n is divisible by1889
(e) None of the above
And the answer is a

## Solution to Q 1.

There are 9 numbers under 100 divisible by 10 viz 10,20,30,40,50,60,70,80,90.
Dividing each of these by 4, we get 2.5,5,7.5,10,12.5,15,17.5,20,22.5.
For 5 of these numbers we get a fractional quantity . Now 4 numbers can be written equidistantly from this quantity , e.g. for 2.5, we can write 1,2,3,4 and these sum up to 10.
Hence, the answer is 5 (10,30,50,70,90).

## Solution to Q 2.

The required number must satisfy the following conditions :
1. There should be exactly 2000 prime numbers in its factorisation.
2. 2 should be one of the 2000 numbers and the power of 2 should be greater than or equal to n.
Since infinite numbers satisfy these conditions, the number of required numbers is infinite.

## g_nik ! n divides 2^n . that

g_nik ! n divides 2^n . that means if n had any other factor other than a power of 2 IT WILL NOT BE ABLE TO DIVIDE 2^n . so there is no question of having 2000 or whatever prime numbers.

## What is the answer of

What is the answer of question 2
None of the above is the answer ? ? ?

## what is the answer or are

what is the answer or are you sure you wrote the question correctly !

## The question cant be wrong

The question cant be wrong as it was asked in mathematic olympiad

## what i meant is did you

what i meant is did you write it correctly ! and which olymapiad . there are lots of sites which post answers to olympiad questions

## Question 1

1. How many of the numbers under 100 that are divisible by 10 can you make by adding four consecutive integers?
Solution:
Let the four consecutive integers be x, x + 1, x + 2 and x + 3
Now, their sum = x + (x + 1) + (x + 2) + (x + 3) = 4x + 6
This should be divisible by 10.
That is, 4x + 6 = 10k for k = 1, 2, 3, 4, 5, 6, 7, 8, 9.
4x + 6 = 10k
or, 4x + 6 – 10 = 10k – 10
or, 4x – 4 = 10k – 10
or, 4(x – 1) = 10(k – 1)
or, 2(x – 1) = 5(k – 1)
Now, it is clear that k – 1 has to be divisible by 2.
For that the possible values of k are 1, 3, 5, 7 and 9.
Hence there are altogether 5 such groups of numbers consisting of four consecutive integers such that their sum is divisible by 10.
Thank You
Ravi Raja