let n =1
a)15. when divided by 5, remainder=0
b)27 when divided by 5, remainder =2
c)83 remainder=5 when divided by 6
d)92 remainder=2 when divided by 5
E)123 remainder=3 when divided by 4,5,6
IMO E

dell2 wrote:if n is a positive integer which one of the following numbers must have a remainder of 3 when divided by any of the numbers 4,5 & 6 ?

A- 12n + 3
B- 24n + 3
C- 80n + 3
D- 90n + 2
E- 120n + 3

Hi,
(k-3) is divisible by 4,5&6. So, (k-3) is divisible by LCM of(4,5,6).
So, (k-3) is divisible by 60.
So, k is of the form 60*p +3.
Hence, E

hello,

120 is the common multiple of 4, 5 and 6 so answer would be E

regards nafi

It is good to know that when the test talks of remainders it is often looking at some version of factoring. you can thus factor out the coeficients of each of the numbers to find that 120 has 4, 5, and 6 as a factor and thus would be the answer.

you could also use plugging in a number if you don't recognize the factor pattern - pick a number for n and divde the results of each answer choice and then divide - you quickly realize that the units digit must be 3 for the remainder to be 3 when divided by 5.