Problem Solving

What is the third term in a sequence of numbers that leave remainders of 1, 2 and 3 when divided by 2, 3 and 4 respectively?

1.11

2.17

3.19

4.23

5.35

I believe the answer is 3) 19


19 /4 = 4 + remainder 3

17 /3 = 5 + 2

15/ 2 = 7 + 1

danjuma wrote:What is the third term in a sequence of numbers that leave remainders of 1, 2 and 3 when divided by 2, 3 and 4 respectively?

1.11

2.17

3.19

4.23

5.35

There can be Many value for the third term in the series. Because we don't know what is the first term of the sequence.

it can be 11, 23 , 35,

It can be -11, -23, -35, and with the difference of 12 there can be n number of values.

I know as per the options 35 seems to be first choice. But What if negative values.

danjuma wrote:What is the third term in a sequence of numbers that leave remainders of 1, 2 and 3 when divided by 2, 3 and 4 respectively?

1.11

2.17

3.19

4.23

5.35

Please post the OA and the explanation if you have with you?

danjuma wrote:What is the third term in a sequence of numbers that leave remainders of 1, 2 and 3 when divided by 2, 3 and 4 respectively?

1.11

2.17

3.19

4.23

5.35

i think the answer is 11, because 11 : 2= 5 with remainder 1
11:3 =3 with the remainder 2
and 11:4 =2 with remainder 3
in case the number is the same
in case the number are different from the sequence, 19 or 23 also be the answers
there's something not right with this PS
experts please!

Instead of 'numbers', the question should read ' positive integers'
Also, it should be made clear that this is an increasing sequence.