CoCubes Divisibility Question and Answers Quiz-1

(x³ + 4x² + 6x - 2) ÷ (x + 5), To find the remainder

Find the value of x: x + 5 = 0 x = -5

Find the remainder: Sub x = -5

to find the remainder: x³ + 4x² + 6x - 2 = (-5)³ + 4(-5)² + 6(-5) - 2

= -125 + 4(25) + (-30) - 2

= -125 + 100 - 30 - 2

= -57

Assuming that you know binomial theorem… The solution is as follows

Actually it is not a difficult question and it can be answered easily once we know the stuff.

Let us take exponents 0, 1, 2, 3, ....one by one to "17"

When we look at the above table carefully, 17? is divided by 16, we get the remainder "1".

Again we get remainder "1" for the exponent "1".

Next we get remainder "1" for the exponent "2" and so on.

So, we get remainder "1", for all the exponents we take for 17.

Since we get the remainder 1 for all the exponents we take for 17,

when we divide 17²³ by 16, the remainder will be "1 ".

• OTHER METHOD :

17 ^1 ends with 7 in unit's digit
17 ^2 ends with 9
17^3 ends with 3
17 ^4 ends with 1

13^{36} = (13^{3})^{12} = 2197^{12} = (2196+1)^{12}

when ((2196+1)^{12} is divided by 2196, all terms in the expansion are divisible by 2196 except 1^{12}, so the remainder will be 1

Divisor = 999 quotient = 366 Remainder = 103

Number = 999 x 366 + 103 = 365737

Let larger no.is x and smaller number be y.

Now as per the question,

(x-y)=1365...(1),

6y+15=x...(2).

Now from equation (1)-(2)

we get y=270,x=1635.

Therefore the smaller no. is y=270.

For a no. to be divisible by 8, the last 3-digits of that no. should be divisible by 8... Here 6x2 should be divisible by 8... Try for all possible value from 0 to 9... x = 3

=> 632 which is perfectly divisible by 8...

Ans : 3

For a no. to be divisible by 24, it should be divisible by both 8 nd 3... since 24 = 8 x 3

In the given options , 3125736 is divisible by both 8 and 3... So its also divisible by 24...

15411 when we Divide 15207 by 467, we get 263 as remainder 467-263=204. Either subtract 263 from 15207 or add 204 in 15207 to get a number divisible by 467. The number nearest to 15207, which is divisible by 467 = 15207+ 204=15411

let's check: \frac{15411}{467} = 33

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

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

Hence the required number is 7936 x 31 = 246016

remainder =16

divisor=16 x 5=80

quotient=\frac{80}{20}=4

no.=divisor x quotient+rem=80 x 4+16=336