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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#?

Answer 1

Jameson Allen

The integers meeting the criteria are 13, 25, 37, 49, 61, 73, 85, 97. Their sum is 430.

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Answer 2

Abigail Allen

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

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#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.