atova01 wrote:
BrushMyQuant wrote:
Find the sum of all three digit numbers which leave a remainder 2 when divided by either 7 or5
Three digit numbers which will satisfy the condition will be 2 + Multiple ofLCM(5,7)
= 2 + 35k (where k is aninteger)
Firstnumber = 107 (As 107 = 35*3 +2)
LastNumber =982
Common difference,d =35
Number of terms,n = (Last term - First Term)/d + 1 =\(\frac{ 982 -107}{35}\) + 1 = 25 + 1 =26
Sum = n * (First Term + Last Term)/2 = 26 *\(\frac{ (107 +982)}{2}\) = 14,157
So, Answer will beB
Hope ithelps!
Watch the following video to MASTER Sequenceproblems
Iframe
Three digit numbers which will satisfy the condition will be 2 + Multiple ofLCM(5,7)
= 2 + 35k (where k is aninteger)
Firstnumber = 107 (As 107 = 35*3 +2)
LastNumber =982
Common difference,d =35
Number of terms,n = (Last term - First Term)/d + 1 =\(\frac{ 982 -107}{35}\) + 1 = 25 + 1 =26
Sum = n * (First Term + Last Term)/2 = 26 *\(\frac{ (107 +982)}{2}\) = 14,157
So, Answer will beB
Hope ithelps!
Watch the following video to MASTER Sequenceproblems
Iframe
Hello - what is the fastest/ easiest way to find the last number aka the982?
This is what I did to get the last number quickly.
Once we establish that it's an AP, we know that a(n) = a + (n-1)*d
Now we know the series will end at 999.
So we can find a(n) <= 999 => a + (n-1)*d < 999
a = 107, d = 35, substituting the values we get
107 + (n-1)*35 <= 999 => (n-1)*35 <= 999 - 107 =>
...
Statistics : Posted by siddhantvarma • on 10 Dec 2021, 03:15 • Replies 10 • Views 2672
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