## Divisor question

How to approach the following question ?
Please tell me in details .

A number and its 17th multiple leave respective reminders of 8 and 4 when divided by a divisor. Another multiple of the same number when divided by the same divisor leaves the reminder 3.
which of the following is a possible value of the divisor
a. 22
b. 33
c. 44
d. 66

## In mid way

I tried but it seems all the options are possible Let the number be N 17th multiple of N is N'=17N when we divide N by d we get 8 reminder Thus N = kd + 8 and 17N =k'd + 4 = 17kd + 136 SO d got to be a factor of 136-4 = 132 and possible values are 11, 22, 33, 44, 66 . . . The second condition is "Another multiple of the same number when divided by the same divisor leaves the reminder 3" so any of the above numbers are possible .... ???

## you are correct

you are correct as per your equation N = kd + 8 and its given that Another multiple of the same number when divided by the same divisor leaves the reminder 3 let another multiple of N be n' then Nn'=kn'd + 8n' when divided by d leaves reminder 3 that means 8n'/d leaves a reminder of 3 then d must be odd thus the answer is 33.

## please anyone can make it

please anyone can make it more clear

## Ok let me try to explain it

Ok let me try to explain it in a very simple way. First Statement:

A number and its 17th multiple leave respective reminders of 8 and 4 when divided by a divisor.
Let the number be N When N is divided by d a divisor it gives a reminder 8 So N can be written as N=ad + 8 The 17th multiple of N is 17ad+136 and when divided by d it should give a reminder of 136 but as per the question it gives a reminder of 4 => when 136 is divided by d we get a reminder 4 Now which are the numbers when divided by 136 gives reminder 4 132, 66, 44, 33, 22 Any of the above can be d Second Statement:
Another multiple of the same number when divided by the same divisor leaves the reminder 3.
Let us consider another multiple (say 6) of the number N Then 6N = 6ad + 48 But the reminder is given as 3 (odd number) Now lets check for which of the above numbers we are getting an odd divisor 48/132 reminder 48 48/66 reminder 48 48/44 reminder 4 48/33 reminder 15 48/22 reminder 4 Ok none of the reminder is 3 but there should be some other number (not 6) when multiplied by N shall a give a reminder 3 But from the above its clear to get an odd reminder we need to dived the number by an odd number (48/33 gives 15 as reminder) The only odd number which satisfies stament1 is 33 And that’s the answer. Hopefully it must be clear now.

## thanks.do u know of any

thanks.do u know of any methods to solve large number problems faster eg:faster multiplication of 53x91etc.so that we can arrive at answers quicklyor can u mention any sites for that?

## My suggestion would be to

My suggestion would be to stay away from such short cuts and fast method It’s just a misconception that those methods help in exams like CAT. You don’t need to waste your time in learning those methods. Any how you can check the Vedic math chapter on this site http://cat4mba.com/math-e-boom/vedic-mathematcis-1

## let me try to solve the

let me try to solve the problem as SecondHarsa did only more formally :) The idea is of the congruent modulo ! first the notation a=bmod(m) means m divides a-b => ca = cb mod(m) where c is a +ve integer. now the problem N=8 mod(m) => 17N = 136 mod(m) ------------ (1) also xN = 8x mod (m) ------------------(2) but 17N = 4 mod(m) ------------------- (3) also xN = 3 mod(m) ------------------------(4) another property is that we can add ot substart to modulos ie. if a = b mod(m) and c = d mod(m) then a+- c = b+-d mod(m) thus (3) - (1) gives 0 = 4 - 136 mod(m) => 132 = 0 mod(m) thus m divides 132 = 4*3*11 also (4) - (2) gives 8x-3 = 0 mod(m) thus m divides 8x - 3 => m is ODD thus possible solutions for m = 11 , 33 11 is not there in the options => the correct answer is 33.