Find the sum of all such integers less than #100#, which leave a remainder #1# when divided by #3# and leave a remainder #2#, when divided by #4#?

1 Answer
Jan 30, 2017

Answer:

Sum of all such integers less than #100# would be #416#.

Explanation:

Integers which leave a remainder #1# when divided by #3# are

#{4,7,10,13,16,19,22,......}#

and of these, those which leave a remainder #2# when divided by #4# are

#{10,22,34......}#

this is an arithmetic sequence, with first term as #a_1=10# and common difference as #d=12# (note it is LCM of #3# and #4#). As #(100-10)/12=7+....#, the last number in the sequence, up to #100# should be #7+1# or #8^(th)# term.

and sum of the arithmetic sequence given by #S_n=n/2(2a(n-1)d))# would be

#8/2(2xx10+(8-1)xx12)=4(20+84)=4xx104=416#