HackerRank Number System Quiz- 1

For divisibility by 10,the unit digit must be zero.

But here the number at once place is 5,

So,874575 is not completely divisible by 5

A number is divisible by 9 if the sum of its digits is completely divisible by 9.

Here, 9+3+5+8+1+9+1=36,which is completely divisible by 9.

Hence,9358191 is also completely divisible by 9

Given: a + b = 15 a - b = 3

Using the identity (a + b)(a - b) = a^2 - b^2, we can find the difference of their squares as follows:

(a + b)(a - b) = a^2 - b^2 (15)(3) = a^2 - b^2 45 = a^2 - b^2

Therefore, the difference of their squares is 45.

Since the given number is divisible by 5, so 0 or 5 must come in place of $.

But, a number ending with 5 is never divisible by 8.

So,0 will replace $.

Now, the number formed by the last three digits is 6*0, which becomes divisible by 8, if * is replaced by 4. Hence, digits in place of * and $ are 4 and 0 respectively.

For a number to be divisible by 11, the sum of the numbers that occupy the even places minus the sum of the numbers that occupy the odd places is equal to 0 .

(0+1+5+A)-(3+2+6+4)=0

6+A-15=0

A-9=0

A=9

Let the total number of apples be x.

Then,Apples sold to 1st customer= \frac{x}{2}+1

Remaining apples =  x-\frac{x}{2}+1=\frac{x}{2}-1

Apples sold to 2nd customer = \frac{1}{3}(\frac{x}{2}-1)+1<br /> <br /> =\frac{x}{6}-\frac{1}{3}+1

= \frac{x}{6}+\frac{2}{3}

Remaining apples = (\frac{x}{2}-1)-(\frac{x}{6}+\frac{2}{3})

= \frac{x}{3}-\frac{5}{3}

Apples sold to 3rd customer = \frac{1}{5} (\frac{x}{3}-\frac{5}{3}

= (\frac{x}{15}+\frac{2}{3})

 

Remaining Apple= \frac{x}{3}-\frac{5}{3}-  (\frac{x}{15}+\frac{2}{3})

= \frac{4x}{15}-\frac{7}{3}

Therefore,

\frac{4x}{15}-\frac{7}{3} =3

Solving above equation, we get

x=20

987= 3 \times 7 \times 47

So, the required number must be divisible by each one of 3,7,47.

None of the numbers in (a) and (b) are divisible by 3, while (d) is not divisible by 7

By actual division, we find that 999999 is exactly divisible by 13. The quotient 76923 is the required number.

He wants to write from l to 999.

He has to write 9 numbers of one digit, 90 numbers of two digits and 900 numbers of three digits.

Total number of times= 1 \times 9 + 2 \times 90 + 3 \times 900 =2889

\frac{\sqrt{28}+\sqrt{7}}{\sqrt{28}-\sqrt{7}}

= \frac{2\sqrt{7}+\sqrt{7}}{2\sqrt{7}-\sqrt{7}}

= \frac{3\sqrt{7}}{\sqrt{7}}

= 3