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# Quantitative Aptitude Objective Questions and Answers for placement

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 1 Find the sum of even divisors of 4096. A. 8192 B. 6142 C. 6144 D. 8190
See Answer & Explanation Lets Discuss
Explanation

4096= 212 .

So sum of all the factors= (20+ 21+22+23+24+25 + 26+27+28+29+210+211+212).

Now that we need only the even numbers we can remove 20 from the equation. That will leave only the even numbers as 20=1 was the only odd number, all the other powers of 2 is even and we know that

Odd x even = even. Hence the resultant has to be even if there is even 1 even numbers in the whole product.

So, The sum is = 21+22+23+24+25 + 26+27+28+29+210+211+212 = 8190

 2 Find the number of zeroes in 58! (58 factorial) A. 11 B. 12 C. 13 D. 14
See Answer & Explanation Lets Discuss
Explanation

We know that  a pair of 2 x 5 makes 10. So we need to find the number of occurrences of this pair.  In any factorial value,  the number of 5’s will always be lesser than the number of 2’s. Hence, we just need to count the number of 5’s.

Now, Lets get back to this Question.

To find the number of 5’s just  count the number less than 58 which are divisible by 5.

In this case, 55,50,45,40,35,30,25,20,15,10,5.

Now  there are 11 numbers which are divisible by 5 where 5 occurs only once, and there are two 5’s in 50 and 25 out of which  one 5 has already been included earlier, hence you  need to add 2 to the number of factors divisible by 5 i.e. , 11. Hence, 11+2 = 13.

Now that you know the concept, here is a shortcut :

If you have to count the number of zero’s in n!.

n/5 + n/(52) + n/(53)….

Hence in this case: 58/5+ 58/25 = 11+2 = 13.

 3 Find the number of zero’s in 173! (173 Factorial) A. 39 B. 40 C. 41 D. 42
See Answer & Explanation Lets Discuss
Explanation

If you have to count the number of zero’s in n!.

n/5 + n/(52) + n/(53)….

Using the formula given above:

173/5 + 173/25 + 173/125 = 34 + 6 + 1 = 41.

Now lets verify this to grasp the concept,

All the numbers less than 173 which are divisible by 5 are:

170,165,160,155,150, 145, 140 , 135, 130, 125, 120,115,110,100,…..5

5x34, 5x33, 5x32, 5 x 31, 52x6, 5x29, 5x28,5x27,5x26,53,5x24,5x23,5x22,5x21,52x4,….,5x3,5x2,5

If you count the number of 5’s in the above equation, you will find out that the answer is 41!

 4 Find the remainder of 1421 x 1423 x 1425 when divided by 12. A. 3 B. 11 C. 13 D. 5
See Answer & Explanation Lets Discuss
Explanation

1421 x 1423 x 1425/12

= 5x7x9/12

= 35x9/12

=11x9/12

which gives a remainder of 3.

 5 The largest number amongst the following that will perfectly divide 101100 - 1 is? A. 10000 B. 100 C. 100000 D. 100100
See Answer & Explanation Lets Discuss
Explanation

Lets use hit and trial method:

1012 = 10201.
1012 - 1 = 10200. This is divisible by 100.

On trying for 1013 - 1 = 1030301 - 1 = 1030300.

So we can conclude that (1011 - 1) to (1019 - 1) will be divisible by 100.
(10110 - 1) to (10199 - 1) will be divisible by 1000.
Therefore, (101100 - 1) will be divisible by 10,000.

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