For how many values of x from 1 to 300, inclusive, is the sum of the

Question

For how many values of x from 1 to 300, inclusive, is the sum of the integers from 1 to x, inclusive, divisible by 3 ?

(A) 30

(B) 99

(C) 100

(D) 199

(E) 200

DZ3 · Accepted Answer

I would like to share another method.

The question asks for the following calculation "If we add 1+2+3+...+x, is this number divisible by 3?"

So if we simply do so we can notice a pattern. The first sequence not divisible by 3 is x=1 followed by x=4 i.e. 1+2+3+4=10. We can conclude, that every third increment of the sequence is not divisible by 3 and therefore there are 100 values of x for which the sequnce is not divisible by 3.

IMO E

gmatophobia · Answer

The sum of x integers from 1 to x is given by \frac{x * (x+1) }{ 2} So we can translate the question as - " For how many values of x from 1 to 300, inclusive, is \frac{x * (x+1) }{ 2} divisible by 3 " OR " For how many values of x from 1 to 300, inclusive, is \frac{x * (x+1) }{ (2 * 3)} is an integer " Simplifying further we finally visualize the question as, For how many values of x from 1 to 300, inclusive, is \frac{x * (x+1) }{ 6} is an integer The information is a bit more manageable to work with

Let's proceed - For \frac{x * (x+1) }{ 6} to be an integer, one of the following should be true - Case 1 - x is a multiple of 6 Case 2 - (x+1) is a multiple of 6 Case 3 - Neither x nor (x+1) is a multiple of 6, but the product is multiple. When can this happen ? For ex. 2 and 3. Neither 2 is a multiple of 6, nor is 3. But (2*3) is a multiple of 6. Solving these cases - Case 1 x is a multiple of 6 Between 1 to 300, there are 50 numbers which are multiple of 6 Case 2 (x+1) is a multiple of 6 Between 1 to 301, there are 50 numbers which are multiple of 6 Case 3 Neither x nor (x+1) is a multiple of 6, but the product is multiple. Note: For the product to be multiple of the number, the product of remainder when x is divided by 6 and when (x+1) is divided by 6 should be a multiple. Taking the same example - 2 is not a multiple of 6; Remainder when 2 is divide by 6 = 2. 3 is not a multiple of 6; Remainder when 3 is divide by 6 = 3 Now the product of the remainders (2 * 3) is a multiple of (via the Remainder theorem) There can just be two combinations when the product of remainders of x and x+1 is divisible by 6 while x and (x+1) are not. x = 2 & (x+1) = 3 OR x = 3 & (x+1) = 4 Hence x can be represented as x = 6 * (some factor) + 2 OR x = 6 * (some factor) + 3 Therefore the number of possible values satisfying this condition will be same for both. Assume, x = 6 * (some factor) + 2 = \frac{300 - 2 }{ 6} + 1 = 50 Similarly for x = 6 * (some factor) + 3 = 50 Total possible values of x = (Case 1) + (Case 2) + (Case 3) Total possible values of x = (50) + (50) + (50 + 50) = 200 IMO - E

versewriter · Answer

According to the question we need to find total such consecutive series starting from 1 till 300, that if we add all numbers of such series resultant should be divisible by 3 i.e 1 to 2 sum equals to 3 which is divisible by 3 or 1 to 5 sum equals to 15 which is also divisible by 3. So such series are -

1+2=3,
1+2+3=6,
1+2+3+4+5=15,
1+2+3+4+5+6=21
1+2+3+4+5+6+7+8=36

From the above we can conclude two things -
1) If the last no. of the series divisible by 3, the sum will automatically divisible by 3
2) The series, 1+2 =3 or 1+2+3+4+5=15 or so on i.e. the series having last no as 2 or 5 (2+3=5) or 8 (5+3) or so on, the such series is also divisible by 3

So total such series are -
1) According to option 1 Total such numbers are 300/3 =100
2) According to option 2. Total such series are 300/3 = 100
So total such numbers are 200

hughng92 · Answer

Why does not 99 work?

samarpan.g28 · Answer

Ofcourse 99 is valid.
See, we have to select x so that 1+....+x is divisible by 3.
We know that the sum of the first x terms is x(x+1)/2, additionally, it shall be divisible by 3 so we have to find values of x s.t. it satisfies x(x+1)/(2*3) = x(x+1)/6.

We need to find out how many terms are divisible by 2, divisible by 3 and divisible by 6. But notice, when we are finding the number of terms that are divisible by 6, they will repeat with the number of terms that are either divisible by 2 or divisible by 3.
For eg. 60 is a term that comes when we are finding divisible by 2, divisible by 3, as well as divisible by 6.
Therefore we need only one 60 in the set, which is why we will do (number of terms div by 2) + (number of terms div by 3) - (number of terms div by 6) so that if 60 is present thrice then two of them will cancel out and only one will exist.

Now find the number of terms div by 2:
2,4,....,300 - apply AP 300=2+(n-1)*2 or, n=150.
The number of terms div by 3:
3,6,....,300 or, 300=3+(n-1)*3 or, n=100.
The number of terms div by 6:
6,12,....,300 or, 300=6+(n-1)*6 or, n=50.

Finally, 150+100-50=200. Option (E) is correct.

Adarsh_24 · Answer

If we start the series we will get
1,3,6,10,15,21,28,36,45,55,66,78,...

Every 2 out of 3 is divisible by 3. So 2/3*300=200

Akshaynandurkar · Answer

The question is not tricky calculation wise but it takes lot of time if you are not aware of the right direction. Whenever you are stucked with which path to take, start looking for patterns that can give some insights on how to proceed with such questions.
when x = 1, sum = 1 (x cannot be 1)
when x = 2, sum = 3 ( x can be 2)
when x = 3, sum = 6 ( x can be 2)
when x = 4, sum = 10 (x cannot be 4)
when x = 5, sum = 15 ( x can be 5)
when x = 6, sum = 21 ( x can be 6)
when x = 7, sum = 28 (x cannot be 7)
when x = 8, sum = 36 ( x can be 8)
when x = 9, sum = 45 ( x can be 9)

you will see that we need to eliminate numbers = 1,4,7,10,...
but it is difficult to count all this number. we need to check some pattern of recurrence
write down the numbers till 1 is repeated in unit place
1,4,7,10,13,16,19,22,25,28,31(repeat). so upto 30 we have 10 numbers
so up to 300 we have 10x10 = 100 numbers that needs to be eliminated. so total number = 200
Search for the patterns and try to formulate some formula to get to the answer. This is good strategy in such questions

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For how many values of x from 1 to 300, inclusive, is the sum of the

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