the cube of the first no exceed their product by 2,the cube of the second no is smaller than their product by 3,and the cube of the third no exceeds their product by 3

A)3^(1/3) B)9^(1/3) C)2 D)any of these

The number of persons who booked ticket for the New Year’s concert is a perfect square. If 100 more persons booked ticket then the number of spectators would be a perfect square plus 1. If still 100 more persons booked ticket then the number of spectators would be again a perfect square. How many persons booked ticket for the concert?

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Thank you, Sir

For how many non-negative integers, n, 2^{4}+2^{7}+2^{n} results in a perfect square?

2^{4} + 2^{7} + 2^{n}

= 2^{4} [ 1 + 2^{3} + 2 ^{(n -4)} ]

= 2^{4 }[ 9 + 2^{(n -4) }]

9 + 2^{(n -4) }has to be a perfect square .

n = 8 is only possibility.

9 + 2^{(n -4) }has to be a perfect square

Sir, From this how do we conclude that n=8 is the only possibility from all the natural numbers, unless range is not given to us in the question?

9 + 2^{(n-4)} is a perfect square

Two cases :

(i)When ( n -4) is a perfect square

9 + x^{2} = y^{2}

9 = y^{2} – x^{2}

(x-y) (x+y) = 1 x 9 or 3 x 3

Solving this we get x = 4 is a possible value .

(ii) When ( n -4) is not a perfect square

9 + 2^{(n-4) }is of the form 3k + 2 Hence , Never a perfect square .

**[All perfect squares are of the form 3n or 3****n + 1.]**

**Find the sum of divisors of 544 which are perfect squares.**

**Find the sum of the sum of even divisors of 96 and the sum of odd divisors of 3600**

96 = 2^{5}x3; Sum of even divisors of 96 = (2^{1}+2^{2}+2^{3}+2^{4}+2^{5})(3^{0}+3^{1}) = 248=

3600=2^{4}x3^{2}x5^{2}; Sum of odd divisors of 3600 = (3^{0}+3^{1}+3^{2})(5^{0}+5^{1}+5^{2}) = 403

Ans = 403+248 = 651

544 =2^{5}x17. Perfect squares divisors of 544 are: 1, 4, 16 Sum = 21

Let a, b, c be the numbers.

a^{3} = (abc + 2), b^{3} = (abc - 3), c^{3} = (abc + 3). Multiply all three equations and keep abc = x.

x^{3} = (x + 2)(x - 3)(x + 3) --> 2x^{2} - 9x - 18 = 0 --> (2x + 3)(x - 6) = 0 --> x = -3/2 or x = 6. Taking abc = 6 and noting that b is the smallest number b^{3} = 6 - 3 = 3 --> b = 3^{1/3}

How many divisors of the number N = 2^7 x 3^5 x 5^4 have unit digit equal to 5?

Hello Ritika ,

We know , Product of an odd natural number and 5 always ends with the unit digit 5.

So we can select

2^0 ( 1 way )

3^0 , 3^1 , 3^2 , 3^3 , 3^4 , 3^5 ( in 6 ways )

similarly powers of 5 in 4 ways

hence 1 * 6 * 4 = 24 factors of N have unit digit equal to 5.

Hello Raman,

Shouldn't there be at least one times 5 present?

5^1 5^2 5^3 5^4 = 4 ways. Hence, 1*6*4 = 24 factors.

Hi sir,

I have a confusion. 5^0 is equal to 1 which can never create a number with unit digit 5.

Hello Apoorva,

5^0=1 => multiple of 5^0 gives unit digit 5 when multiply by 5^n

Where n>0

Find the sum of the prime divisors of 1² + 2² + 3² +.......+ 2005²

Hello Vidisha , we know sum of squares of first n natural number =

n ( n +1 )( 2n +1 )/6

1² + 2² + 3² + ... + 2005²

= 2005 × 2006 × 4011/6

= 2005 × 1003 × 1337

= 5 × 401 × 17 × 59 × 7 × 191

= 5 × 7 × 17 × 59 × 191 × 401

Hence, required sum of primes

= 5 + 7 + 17 + 59 + 191 + 401 = 680.

How many three digit natural number has less than 5 prime factors.Please explain in detail.

Hello Apoorva !

Smallest number which has five distinct prime factors is 2 × 3 × 5 × 7 × 11 = 2310 .

All three digit natural numbers are smaller than 2310 .

Hence there are 900 three digit natural numbers which have less than 5 prime factors. 🙂

A number N has 5 factors between 1 and root N . How many factors does the number have if

1. N is a perfect square

2. N is not a perfect square.

Hello Apoorva !

For a natural number N the number of factors of N below √N is equal to number of factors of N above √N .

So if N has five factors between √N and 1 [ 6 below √N ] it must have 6 factors above it's square root.

If it is a perfect square then it's square root is also a factor .

Hence , 6 + 1 + 6 = 13 for first question and 6 + 6 = 12 is the answer for the second question. 🙂