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Which of the following integers can be written as both the s
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13 Apr 2020, 09:40
13 Apr 2020, 09:40
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Which of the following integers can be written as both the sum of 5 consecutive odd integers and 7 consecutive odd integers?
A. 49
B. 70
C. 140
D. 215
E. 525
Kudos for the right answer and explanation
A. 49
B. 70
C. 140
D. 215
E. 525
Kudos for the right answer and explanation
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: Which of the following integers can be written as both the s
[#permalink]
13 Apr 2020, 10:52
13 Apr 2020, 10:52
Carcass wrote:
Which of the following integers can be written as both the sum of 5 consecutive odd integers and 7 consecutive odd integers?
A. 49
B. 70
C. 140
D. 215
E. 525
Kudos for the right answer and explanation
A. 49
B. 70
C. 140
D. 215
E. 525
Kudos for the right answer and explanation
If we have consecutive odd integers, and each value is 2 greater than the value before it
So for example, if we let x = the smallest odd integer in the set of five consecutive odd integers then...
The sum of the 5 integers = x + (x + 2) + (x + 4) + (x + 6) + (x + 8) = 5x + 20 = 5(x + 4)
NOTE: since x is ODD, the sum of the 5 integers = 5(ODD + 4) = 5(some odd integer)
In other words, the sum must be an ODD MULTIPLE OF 5 (e.g., 5, 15, 25, 35, 45, etc)
Check the answer choices....
We can ELIMINATE A, B and C since they are NOT odd multiples of 5
Likewise, if we let k = the smallest odd integer in the set of seven consecutive odd integers then...
The sum of the 7 integers = k + (k + 2) + (k + 4) + (k + 6) + (k + 8) + (k + 10) + (k + 12) = 7k + 42 = 7(k + 6)
NOTE: since k is ODD, the sum of the 7 integers = 7(ODD + 6) = 7(some odd integer)
In other words, the sum must be an ODD MULTIPLE OF 7 (e.g., 7, 21, 35, 49, ... etc)
Check the remaining answer choices....
We can ELIMINATE D since it's not even a multiples of 7
By the process of elimination, the correct answer is E
Cheers,
Brent
Re: Which of the following integers can be written as both the s
[#permalink]
26 Jan 2023, 08:26
26 Jan 2023, 08:26
1
Shouldn't the answer be 140?
After some calculations, I could get 140.
Five ODD integers:
(x)+(x + 2)+(x + 4)+(x + 6)+(x + 8)
Seven ODD integers:
(y)+(y + 2)+(y + 4)+(y + 6)+(y + 8)+(y + 10)+(y + 12)
x and y are both odd numbers of course.
x=3
3+5+7+9+11=35
y=9
9+11+13+15+17+19+21=105
35+105=140
I did it by the following logic:
(x)+(x + 2)+(x + 4)+(x + 6)+(x + 8) = 5x + 20
(y)+(y + 2)+(y + 4)+(y + 6)+(y + 8)+(y + 10)+(y + 12) = 7y + 42
5x + 20 + 7y + 42 = z
5x + 7y + 62 = z
5x + 7y = z - 62
Now I can set z with the answer, and let's look at 140.
At this point, we can look only at 70 and 140 because we know the x and y are odd so 5x+7y must be even (odd+odd=even)
5x + 7y = 140 - 62
5x + 7y = 78
Now we need to find what x and y can create 78, which is not too difficult.
x = 3
y = 9
15 + 63 = 78
That's how I found x and y.
Am I wrong about anything here?
After some calculations, I could get 140.
Five ODD integers:
(x)+(x + 2)+(x + 4)+(x + 6)+(x + 8)
Seven ODD integers:
(y)+(y + 2)+(y + 4)+(y + 6)+(y + 8)+(y + 10)+(y + 12)
x and y are both odd numbers of course.
x=3
3+5+7+9+11=35
y=9
9+11+13+15+17+19+21=105
35+105=140
I did it by the following logic:
(x)+(x + 2)+(x + 4)+(x + 6)+(x + 8) = 5x + 20
(y)+(y + 2)+(y + 4)+(y + 6)+(y + 8)+(y + 10)+(y + 12) = 7y + 42
5x + 20 + 7y + 42 = z
5x + 7y + 62 = z
5x + 7y = z - 62
Now I can set z with the answer, and let's look at 140.
At this point, we can look only at 70 and 140 because we know the x and y are odd so 5x+7y must be even (odd+odd=even)
5x + 7y = 140 - 62
5x + 7y = 78
Now we need to find what x and y can create 78, which is not too difficult.
x = 3
y = 9
15 + 63 = 78
That's how I found x and y.
Am I wrong about anything here?
