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How many even integers n ; 13<=n<=313 are of the form 3k+4,

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honchos
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How many even integers n ; 13<=n<=313 are of the form 3k+4, [#permalink] New post   14 Sep 2013, 00:01
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How many even integers n ; 13<=n<=313 are of the form 3k+4, where k is any natural number ?

A. 101
B. 51
C. 50
D. 96
E. None of these
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C
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Bunuel
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Re: How many even integers n ; 13<=n<=313 are of the form 3k+4, [#permalink] New post   14 Sep 2013, 03:15
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honchos wrote:
How many even integers n ; 13<=n<=313 are of the form 3k+4, where k is any natural number ?

A. 101
B. 51
C. 50
D. 96
E. None of these


Given that n=3k+4. So, n is 4 greater than a multiple of 3: 4, 7, 10, 13, 16, ... Since also given that n is even, then n is 4 greater than a multiple of 6: n=6x+4.

Thus we have that \(13\leq{6x+4}\leq{313}\) --> \(9\leq{6x}\leq{309}\) --> \(1.5\leq{x}\leq{51.5}\) --> x can take 50 values from 2 to 51, inclusive.

Answer: C.
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Re: HOW MANY EVEN INTEGERS N; 13<=n<=313 [#permalink] New post   14 Sep 2013, 01:50
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honchos wrote:
HOW MANY EVEN INTEGERS N; 13<=n<=313 are of the form 3k+4 where k is any natural Number
1. 101
2. 51
3. 50
4. 96
5. None of these


The first term (13<=n<=313) that satisfies 3k+4 is 13
2nd term will 16

So the sequence would like 13, 16, 19, 22 .........310, 313
Since we want only even numbers
16, 22......310
The number of terms would be= [(310-16)/6] +1
= 49 + 1
=50

Answer C
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Re: HOW MANY EVEN INTEGERS N; 13<=n<=313 [#permalink] New post   14 Sep 2013, 02:14
fameatop wrote:
honchos wrote:
HOW MANY EVEN INTEGERS N; 13<=n<=313 are of the form 3k+4 where k is any natural Number
1. 101
2. 51
3. 50
4. 96
5. None of these


The first term (13<=n<=313) that satisfies 3k+4 is 13
2nd term will 16

So the sequence would like 13, 16, 19, 22 .........310, 313
Since we want only even numbers
16, 22......310
The number of terms would be= [(310-16)/6] +1
= 49 + 1
=50

Answer C


Thanks, Can some one think of any other method.
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Re: How many even integers n ; 13<=n<=313 are of the form 3k+4, [#permalink] New post   14 Sep 2013, 03:28
Bunuel wrote:
honchos wrote:
How many even integers n ; 13<=n<=313 are of the form 3k+4, where k is any natural number ?

A. 101
B. 51
C. 50
D. 96
E. None of these


Given that n=3k+4. So, n is 4 greater than a multiple of 3: 4, 7, 10, 13, 16, ... Since also given that n is even, then n is 4 greater than a multiple of 6: n=6x+4.

Thus we have that \(13\leq{6x+4}\leq{313}\) --> \(9\leq{6x}\leq{309}\) --> \(1.5\leq{x}\leq{51.5}\) --> x can take 50 values from 2 to 51, inclusive.

Answer: C.



Wow. Excellent approach, You are the best, we can use this technique elsewhere also.
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Re: How many even integers n ; 13<=n<=313 are of the form 3k+4, [#permalink] New post   30 Apr 2014, 06:27
I did the following

3k + 4, first term in expression falling in the given range is 13, then we have that each number will be of the form 13 + 3k <=313

Therefore, 13+3k <=313, k<=100

There are 101 numbers in this range. Since we are asked for even numbers and since range begins with 13 = odd number, then we have 50 even numbers in the range

Answer: C

Hope this helps
Cheers
J :)
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Re: How many even integers n ; 13<=n<=313 are of the form 3k+4, [#permalink] New post   30 Apr 2014, 14:43
I tried making a series:

first number:16 . last number 310. difference between numbers :3
formula: 310 = 16+6d (there is a jump of 6 between every two even numbers).
The solution for this was a little over 49.
How should I have known to pick 50 and not 49 (if 49 were a valid answer)?

Can someone help?
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Re: How many even integers n ; 13<=n<=313 are of the form 3k+4, [#permalink] New post   30 Apr 2014, 22:14
First Value of K that will satisfy the expression is 4.
And the last value is 103

Since we want only even numbers, so in our case last value will be 102

You know the first number and the last number..
Boom, put them in the AP formula..

An = A1 + (n-1)d

102 = 4 + (n-1)2

solve for n.

n = 50

Option C is right. :)
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Re: How many even integers n ; 13<=n<=313 are of the form 3k+4, [#permalink] New post   30 Apr 2014, 22:58
DesiGmat wrote:
First Value of K that will satisfy the expression is 4.
And the last value is 103

Since we want only even numbers, so in our case last value will be 102

You know the first number and the last number..
Boom, put them in the AP formula..

An = A1 + (n-1)d

102 = 4 + (n-1)2

solve for n.

n = 50

Option C is right. :)



This way I already did it and succeeded.
I had a problem with the way I did it, but I can't figure out what it is.
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Re: How many even integers n ; 13<=n<=313 are of the form 3k+4, [#permalink] New post   01 May 2014, 03:55
Hi ronr34

As per your approach,

First number is 16 and last number is 310.

The number of terms will be given by

310 = 16 + (n-1)6
n = (310-16)/6 + 1

Solve for n

n = 50

I think you forgot to add 1.
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Re: How many even integers n ; 13<=n<=313 are of the form 3k+4, [#permalink] New post   17 Jan 2016, 04:11
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How many even integers n ; 13<=n<=313 are of the form 3k+4, where k is any natural number ?

A. 101
B. 51
C. 50
D. 96
E. None of these[/quote]

for n to be between 13 & 313(inclusive) having form 3K+4
we have least value of K=3
3*3+4=13
.
.
.3*103+4=313(so last value of k= 103)

Now 3*(odd)+4=odd term
so we have to look for K even between range 3 to 103
or our aim is to find K even in our defined range

no. of even terms= (102-4)/2+1
98/2+1
=50
hence ans D

Good ques
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Re: How many even integers n ; 13<=n<=313 are of the form 3k+4, [#permalink] New post   02 Nov 2020, 08:43
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Re: How many even integers n ; 13<=n<=313 are of the form 3k+4, [#permalink]
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