HSBC Divisibility Questions and Answers Quiz-1

(x³ + 4x² + 6x - 2) ÷ (x + 5), find 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

• “Remainder when 17 power 23 is divided by 16\" is a bit challenging question for the people who are getting prepared for competitive exams.

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\"

For example, if we take exponent \"0\" for 17, we get 17? = 1

Here 1 is less than the divisor 16 and 1 can not be divided by 16.

If the dividend is less than the divisor, then that dividend itself to be considered as \"Remainder\"

So, if 17? is divided by 16, we get the remainder \"1\"

If the dividend is greater than the divisor, then we have to divide the dividend by the divisor and get remainder.

Let us deal our problem in this way.

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

17^5 ends with 7....

So from here on it follows a cyclicty of 4. after that it follows a pattern.

Now 23 can be written as 4*5 +3

So if u find the remainder of 17 ^3 , it will be same as 17^23.

remainder of 17^ 3 when divided by 16 is 1 .

So for 17^23 also it will be 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 Define x: Let the number be x Solve x: x ÷ 999 = 366 remainder 103 x = 999 x 366 + 103 x = 365737

let larger no.is x and smaller no.is y. now we get from question is that (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*3 In the given options , 3125736 is divisible by both 8 and 3... So its also divisible by 24... Ans : (4) 3125736

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 check ...15411/467=33

7936 => 2^2 * 2^2 * 2^2 * 2^2 * 31^1 To make it as a perfect square, we have to multiply 7936 with 31... Hence the reqd no. is 7936*31 = 246016 Ans : 246016

rem=16,divisor=16*5=80,quotient=80/20=4; no.=divisor*quotient+rem=80*4+16=336 is the right answer