## Help me in solving the following Quant questions

Help me in solving the following Quant questions :
Q1. How many sets containing one or more consecutive natural numbers are there, such that the sum of the elements in the set is 500?
a. 1
b. 2
c. 3
d. 4
Q2. How many three-digit numbers have the product of their digits as a perfect square as well as a perfect cube greater than 0?
a. 11
b. 12
c. 15
d. 17

## Let a be any number

Let a be any number
Then a^{2} ll be a perfect square and a^{3} ll be a perfect cube.
Now a number which is both perfect square and cube 'll be of the form a^{6} (say n)

For a = 1 , n =1
a = 2, n =64
a = 3, n=729.
a = 4, n=4096.

Now coming back to the question, we have to find out all the possible three digit numbers which are of the form a^{6}.
For a = 1 , n =1 No. of possible 3 digit number =1 i.e. 111
a = 2, n =64 possible digit combinations are (1,8,8 ), (2,4,8) and(4,4,4)
with 1,8,8 possible no. of 3 digit numbers are 3
with 2,4,8 possible no. of 3 digit numbers are 6
with 4,4,4 possible no. of 3 digit numbers are 1
a = 3, n=729. Possible digit combinations is 9,9,9 . so we have only 1 three
digit number
a = 4, n=4096. No there digit number is possible.

So the total no. Of possible 3 digit numbers are 12.
Q1. Possible series are
500
98 + 99 +100+101+102
59 + 60 + 61. . . . . + 66
8 + 9 + 10 + 11 + . .+ 32

## Thanks for expelaining in

Thanks for expelaining in details abt the second question
but how to approach questions like the first one.

## Re: Thanks for explaining in

Yes, How can a question like Q1 be approached? Even i had the doubt!!!
Thank in advance.

## explanation.....

nice explanatin of 2nd questn bt wot shud b da approch f 1st questn?

## How many sets containing one

How many sets containing one or more consecutive natural numbers are there, such that the sum of the elements in the set is 500?
a. 1
b. 2
c. 3
d. 4
1. The set can contain any number of natural numbers
2. Sum of the consecutive natural numbers is 500
Case 1: Set contain 500 it self
Case2: "sum of two consecutive number ll be an even number."
Case3: Let’s see if we can have three consecutive numbers whose sum is 500.
500 is not divisible by 3
So add 1 to it and divide by 3 i.e. 501/3 = 167
If there is any set of 3 numbers whose sum is 500 then it shd contain 167
Possible sets are
1. 165, 166, 167 – Sum is 498
2. 166, 167, 168 - Sum is 501
3. 167, 168, 169
So there is no any such set with three consecutive numbers whose sum is 500
Case4: 500/4 = 125 not possible
Case5: Lets check for a set with 5 consecutive numbers
Divide 500 by 5
So the set shd contain 100
And the possible set is 98, 99, 100, 101, 102
Similarly we ll have two other sets containing 8 and 25numbers.
Point to remember :
odd + odd = even
Odd + even = odd
Even + even = even
At first it might look very confusing and time taking but once you know the way you have to check it is quite easier.
Always do check the option here we have max numbers is 4
Is there any other method – I don’t know but would love to know if there is any.

## 1st one

Q1. How many sets containing one or more consecutive natural numbers are there, such that the sum of the elements in the set is 500?
a. 1
b. 2
c. 3
d. 4
Ans.
Here let us assume n is the 1st term and m is the number of terms in the set. So the terms in the set are -> n,(n+1),(n+2),.....,(n+m-1)
Sum of all the terms = n + (n+1) + (n+2) + ..... + (n+m-1) = m.n + {1+2+3+.....+(m-1)} = m.n + (m-1).m/2
Now, this sum has to be 500
i.e. m.n + (m-1).m/2 = 500
=> n = 500/m - (m-1)/2
The solution to the above problem will be number of possible solution of above equation for n.
Now, as n is natural number both (500/m) and (m-1)/2 has to be natural numbers. But (m-1)/2 will be a natural number only if m is odd. Also, 500 = 5^3 x 2^2 i.e. odd factors of 500 are 1,5,25 and 125.
So, for 500/m to be a natural number there are 4 possible values of m (1,5,25,125).
Note that only +ve integer values of m is possible because as per the question there is at least one term in the set; so m>0. Corresponding n values are (500,98,8 and -58).
So, answer to the question will be -
d. 4

## Small Mistake!!!

In my above reply n=-58 is not acceptable as n is a natural number.
But then there exists one solution for even m. (m>0)
If m is even then decimal fraction part of (m-1)/2=.5
So, to make n a natural number 500/m will also have decimal fraction part =.5
For this only possible m=8 (as 500 = 5^3 + 2^2). Corresponding n=59
But even after this correction - the answer remains the same... :-)
d. 4

## Q2

Q2. How many three-digit numbers have the product of their digits as a perfect square as well as a perfect cube greater than 0?
a. 11
b. 12
c. 15
d. 17
Ans.
To be both a square & cube number has to be of the form n=a^6.
Now, maximum 3-digit number = 999.
Product of its terms = 9x9x9 = 3^6
So, possible values of a = 1,2 and 3
For a=1 (n=1^6)-> possible digit = 1; possible 3-digit numbers =1 {111}
For a=2 (n=2^6)-> possible digits = 1,2,4,8; possible 3-digit numbers = 3+3x2+1 = 10 {(188,818,881),(248,284,428,482,824,842),444}
For a=3 (n=3^6)-> possible digit = 9; possible 3-digit numbers =1 {999}
So, total possible 3-digit numbers = 1+10+1 = 12
Hence, answer to Q2 is ->
b. 12

## No, not by trial and

No, not by trial and error...
500/m has to be a natural numeber for odd m. (as n is natural & m-1/2 is natural for odd m)
But 500 = 5^3 X 2^2; so odd factors of 500 = 1,5,5^2,5^3 = 1,5,25,125
125 is not possible as for m=125 => n=-58 which is not acceptable
When m (m>0) is even - fraction part of (m-1)/2=.5 to cancel this .5 -> 500/m should have fraction part = .5 But 500 is divisible by 2^2=4 hence for m=8 500/m will have fraction part=.5
So, possible m = (1,5,25,8) and n = (500,98,8,59)
Hope it is understandable now... :-)

## for problem 1 i am getting

i got an easier sol i guess:
Heres how.
the sum of any consecutive natural nos can be written as :
n(n+1)/2 - k(k+1)/2 = 500 in this case.
on simplifying:
(n-k)(n+k+1) = 1000 = 2^3 * 5^3
clearly one factor is even and the other is odd.
So clearly one factor must CONTAIN 2^3 or else the other factor becomes even too!
so the we can create new factors by changing the power of 5 in one factor and
putting the remaining 5's with 2^3
since the power of 5 is 3, we can do it in 4 = (0,1,2,3) ways.

## what is n and k

what is n and k here.....????
i mean wot does this signify here...??

## I think

what Rasjorshi trying to do
let
k = is the first number
n = is the last number
its like 1, 2, 3. . . . . k, , , , n. . . ..
Then sum of all the numbers from 1 to k = k(k+1)/2
Then sum of all the numbers from 1 to n = n(n+1)/2
As per the question sum of the numbers with in the set is 500
so n(n+1)/2 - k(k+1)/2 = 500
Am i correct Rasjorshi ???

## yeps..anita u r

yeps..anita u r rite......!!!!
i also got dat

## ya ! only that if k is

ya ! only that if k is supposed to be in the set{k,...,n} the sum should be till k-1. Not that it makes much of a difference, I mean k is supposed to be a general number , but its good to get it right all the way.