Question

when number is divided by any divisor from 2 to 7 it leaves a remainder 1 less than the divisor. Which of the following cannot be the number? (a.419 b.839 c.2099 d.1249)

Answer 1

You want me to explain it or just solve it for ya @debpriya ?

Answer 2

explain please :)

Answer 3

Okay ,as far as I am know,we will have to do trial and error with all the operators.Do you understand the condition that has been put forward in the question?

Answer 4

yes I understood the question

Answer 5

Do one thing,take all the numbers separately,that is your options and check for the condition that is given above by diving it by all the numbers from 2 to 7.
For instance,take up 419.Divide it by 2,3,4 and so on until 7 .If you get the remainder that does not satisfy the condition above,that will be you answer
Does this make sense to ya?

Answer 6

@kity

Answer 7

@kiity

Answer 8

@debpriya ?

Answer 9

@KamiBug

Answer 10

@sweetburger

Answer 11

@bournville I understood what you said. But is there any way to do it fast ?

Answer 12

None of which I am aware of

Answer 13

oh okay thank you so much for your help :)

Answer 14

no:)

Answer 15

Heyy

Answer 16

heyyy !!!

Answer 17

Let the required number be \(x\)

Answer 18

\(x\) leaves a remainder \(a-1\) when divided by \(a = 2, 3, 4, 5, 6, 7\)

Answer 19

then would you agree that \(x+1\) is divisible by \(a = 2, 3, 4, 5, 6, 7\)

Answer 20

yes !!!!!!

Answer 21

just add 1 to the options and use the divisibility tests to check the correct option

Answer 22

wow this would make it so fast! thank you ! thank you so much !! :)

Answer 23

Np :)