Find the sum of all three digit numbers which leave a remainder 2 when

Question

Find the sum of all three digit numbers which leave a remainder 2 when divided by either 7 or 5.

(A) 14179
(B) 14157
(C) 15011
(D) 14171
(E) 14000

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anish777 · Accepted Answer

The question should be "and" instead of "or". If we consider "or", we should not only look at the values divisible by 5 or 7 but also the LCM. If we consider "and", then we should consider 35 only. Isn't it ? Or is my understanding incorrect

BrushMyQuant · Answer

Find the sum of all three digit numbers which leave a remainder 2 when divided by either 7 or 5

Three digit numbers which will satisfy the condition will be  2 + Multiple of LCM(5,7) 
= 2 + 35k (where k is an integer)

First number  = 107 (As 107 = 35*3 + 2)
 Last Number  = 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 be B 
Hope it helps!

Watch the following video to MASTER Sequence problems

https://www.youtube.com/watch?v=EfNT2JoE5CM

beeblebrox · Answer

refer attached fro detailed solution

RanonBanerjee · Answer

B.

Those numbers should be divisible by 5*7 =35 too.

First number is 107 and last number is 982.

Difference in the series of all these 3 digit number should also be 35. There fore the series would be, 107, 142, 167,...,982.
Number of terms= (982-107)/35 +1= 26

Sum would be= (107+982)*26/2 =14157

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rajshreeasati · Answer

At first I took just by 5, which gave a LOT of No. and a large sum not even in the option. But then the best way was to take LCM of 7 and 5, ie 35 and then look at the no. divisible by 35 and add 2. Thus such no. will make an AP as follows -
107, 142, 177,.....982. Now apply formulas of AP to find the no. of terms.
982 = 107 + (n-1)*35
n = 26.
Sum = Sn = 26/2{(2*107 + (26-1)*35)}
Thus Sn = 7 is the last digit, thus B is correct! no need to solve fully.

VLantoyMC · Answer

yeah that's what i thinking all-day

KarishmaB · Answer

Though I didn't think twice about what it is asking and was sure that it is looking for numbers that leave remainder 2 when divided by 35 (perhaps because this is how questions are phrased in remainders topic), upon further evaluation I do see how the question could be ambiguous. 
Perhaps it could be re-written to clarify that the numbers should leave the remainder 2 in both cases - when divided by 5 and when divided by 7.

VLantoyMC

atova01 · Answer

Hello - what is the fastest/ easiest way to find the last number aka the 982?

KarishmaB · Answer

Multiple ways of doing this. n = 35a + 2 If a is 10, you get 350, if a = 20, you get 700 and if a = 30, you get 1050. This means 1015 and 980 are multiples of 35 too. So the greatest 3 digit value of n is 980+2 = 982 or Divide 1000 by 35 to get 20 as remainder. So you know that 1000 - 20 = 980 is the greatest 3 digit number divisible by 35. Division and remainders are discussed here: Division: https://youtu.be/A5abKfUBFSc https://anaprep.com/number-properties-b ... isibility/ Remainders: https://anaprep.com/number-properties-m ... emainders/ https://anaprep.com/number-properties-a ... -shortcut/ https://anaprep.com/number-properties-t ... emainders/

siddhantvarma · Answer

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 => (n-1)*35 <= 892 n-1 <= 892/35 => n - 1<= 25 => n <=26 You can then verify that a(26) < 999 and a(27) > 999 hence n = 26 is the last term in this sequence so we can then easily find a(26).

BrushMyQuant · Answer

Start with 1000. Number will leave a reminder of 2 when divided by 5 and 7

Looking at multiples of 5 is easier so we know that units digit will be 0 or 5.
So, we need to look at numbers which are multiples of 7 and which have a units digit of 0 or 5.

Start with 995 and then try 990, 985, 980 to eventually land at 980. 
980 is divisible by both 5 and 7. So , add the remainder of 2 to get 982 as the number.

Other method  is using 35k + 2. We are looking at 1000 as the max number.
35*30 will be around 1000, 1050 to be exact

But 1050 + 2 (1052) > 1000
So start subtracting 35 from it

1052 - 35 > 1000
=> 1052 - 70 = 982 is the number

bumpbot · Answer

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Find the sum of all three digit numbers which leave a remainder 2 when

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