First Naukri Divisibility Quiz -1

Solution:

For a number to be divisible by 3, the sum of the digits should be divisible by 3,

7+8+2+3+0+9+3=32,

Since 32 is not divisible by 3,

Hence, 7823093 is also not divisible by 3

Solution:

For a number to be divisible by 8,the last three digits of the number must be divisible by 8.

Since, 344 is divisible by 8,

So,57479344 is also divisible by 8.

Solution:

7936 => 2^2 \times 2^2 \times 2^2 \times 2^2 \times 31^1

To make it a perfect square, we have to multiply 7936 with 31

Hence the reqd no. is 7936\times 31 = 246016

Solution:

p-7= 11\times a (as it is multiple of 11)

p=11\times (a+7)

so (p+4)(p+15)= (11a+7+4)(11a+7+15);

= (11a+11)(11a+22);

= 11\times 11(a+1)(a+2) ;

= 121\times 2

=242

Solution:

If a − b is divisible by 3, then a − b = 3k, for some integer k

(a-b)^2=(3k)^2

a^2-2ab+b^2=9k^2

a^3-b^3=(a-b)(a^2+ab+b^2)

= (a-b)(a^2-2ab+b^2+3ab)

=3k(9k+3ab)

= 3k\times 3(3k+ab)

=9k(3k+ab)

Since k(3k+ab) is an integer, then 9k(3k+ab) is divisible by 9

Solution:

For a number to be divisible by 9, the sum of the digits should also be divisible by 9,

3+6+3+x+2+9+5=28+x

The required number should be 8, which makes 363x295 divisible by 9.

Hence, x=8

Solution:

L.C.M. of 6, 7, 8 and 9.

Prime factorization of 6 = 2\times 3

Prime factorization of 7 = 7

Prime factorization of 8 = 2\times 2\times 2

Prime factorization of 9 = 3\times 3

L.C.M. = 2\times 2\times 2\times 3\times 3\times 7

= 504

The L.C.M. of 6 seconds, 7 seconds, 8 seconds and 9 seconds is 504

seconds.

Now, 1 hour = 3600 seconds

So, 2 hours = 3600\times 2 = 7200 seconds

The number of times the four bells will toll together in the next 2 hour

= \frac{7200}{504}

= 14.28 or 14 times

They will toll together 14 times in the next 2 hours

Solution:

The remainders when 1998, 1999, and 2000 are divided by 7 are 3, 4, and 5 respectively.

Hence the final remainder is the remainder when the product 3 \times 4 \times 5 = 60 is divided by 7. Therefore, remainder = 4

Solution:

The common factor between 2^{96} and 96 is 32 = 2^{5} .

Removing 32 from the dividend and the divisor we get the numbers 2^{91} and 3 respectively.

The remainder when 2^{91} is divided by 3 is 2.

Hence the real remainder will be 2 multiplied by common factor 32.

Remainder = 64

Solution:

Since 7 is prime, n^7 - n is divisible by 7.

n^7 - n= n(n^6 – 1) = n (n + 1)(n – 1)(n^4 + n^2 + 1)

Now (n – 1)(n)(n + 1) is divisible by 3! = 6. Hence n^7 - n is divisible by 6 \times 7 = 42. Hence the remainder is 0.