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A right angled triangle has its sides in Arithmetic progression and be

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A right angled triangle has its sides in Arithmetic progression and be  [#permalink]

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New post Updated on: 01 Aug 2018, 10:43
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A right angled triangle has its sides in Arithmetic progression and being integers. At most one of its sides is a multiple of 10. Its perimeter is less than 115. How many such triangles are possible?
A. 9
B. 8
C. 7
D. 6
E. 5

Edit: corrected answer options. Thank you pushpitkc @chetan2u

Originally posted by PKN on 31 Jul 2018, 12:48.
Last edited by PKN on 01 Aug 2018, 10:43, edited 2 times in total.
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Re: A right angled triangle has its sides in Arithmetic progression and be  [#permalink]

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New post 31 Jul 2018, 18:48
PKN wrote:
A right angled triangle has its sides in Arithmetic progression and being integers. At most one of its sides is a multiple of 10. Its perimeter is less than 115. How many such triangles are possible?
A. 8
B. 7
C. 6
D. 5
E. 4


Algebraically let's how many are there..
Let sides be a-d,a,a+d
\(a^2-2ad+d^2+a^2=a^2+2ad+d^2......a^2-4ad=0\)
So a=4d or multiple of 4..
So 3:4:5 or 6:8:10 and so on

Otherwise also we know 3:4:5 is a right angled triangle..

Atmost one is multiple of 10, means it may not have any multiple of 10 or just one at the max.
Least perimeter = 3+4+5=12
Next bigger perimeter will be 12*2 when sides are 6:8:10
So total possible perimeter will be the greatest multiple of 12 just less than 115
12*9=108, so 9 such perimeter

All 9 will have atmost one multiple of 10..
Ans 9 but not in choices

But if you are asking for atmost one multiple of 5..
Then only Ans will be 9-1=8 as sides 15:20:25 will not be counted
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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A right angled triangle has its sides in Arithmetic progression and be  [#permalink]

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New post Updated on: 02 Aug 2018, 00:16
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PKN wrote:
A right angled triangle has its sides in Arithmetic progression and being integers. At most one of its sides is a multiple of 10. Its perimeter is less than 115. How many such triangles are possible?
A. 8
B. 7
C. 6
D. 5
E. 4


From the question stem, we can deduce the following information

1. The sides of the right-angled triangle are in an AP.
2. At most one of the sides is a multiple of 10
3. The perimeter is less than 115

We need a set of numbers which is both a Pythagorean triplet and in an AP
One such set of numbers that is both in an AP and sides of a right-angled triangle are 3x-4x-5x

The triangle possible are
3,4,5(when x = 1) | 6,8,10(when x = 2) | 9,12,15(when x = 3) |
12,16,20(when x = 4) | 15,20,25(when x = 5) | 18,24,30(when x = 6) |
21,28,35(when x = 7) | 24,32,40(when x = 8) | 27,36,45(when x = 9) |
At x = 10, the perimeter of the triangle is 12x or 120(and this fails condition 3)

Therefore, there are at least 9(Option A) possibilities where right-angled triangles can be formed.
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Originally posted by pushpitkc on 01 Aug 2018, 00:36.
Last edited by pushpitkc on 02 Aug 2018, 00:16, edited 2 times in total.
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Re: A right angled triangle has its sides in Arithmetic progression and be  [#permalink]

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New post 01 Aug 2018, 01:32
atmost 9 options are possible out of which option with d= 2,4,5,6,8 will have atleast one side in multiple of 10
so, one of its kind is allowed and hence 9-4=5 , 5 such triangle are possible
pls correct if i m wrong
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Re: A right angled triangle has its sides in Arithmetic progression and be  [#permalink]

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New post 01 Aug 2018, 10:46
PKN wrote:
A right angled triangle has its sides in Arithmetic progression and being integers. At most one of its sides is a multiple of 10. Its perimeter is less than 115. How many such triangles are possible?
A. 8
B. 7
C. 6
D. 5
E. 4


Let the sides be x, x+y, x+2y.

Then, by Pythagorean theom. (x+2y)^2 = x^2 + (x+2y)^2

Simplifying the above equation we get 3*(y/x)^2 + 2*(y/x) -1 = 0, solving the quadratic equation we get y=2x/3

Thus, the three sides are x, 5x/3, 7x/3 and the perimeter is 5x.

Now as the perimeter is less than 115 and the sides are integers both x and y will also be integers. Hence the possible values of the perimeter (5x) are 110, 105, 100......5. so the possible values for x are 22, 21, 20 ...... 1.

However, as y=2x/3 and y is an integer x should be a factor of 3.

Therefore, x can take the value of 3, 6, 9, 12, 15, 18, 21 i.e 7 possible values and hence 7 possible triangles.

Please help me identify where i went wrong.
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A right angled triangle has its sides in Arithmetic progression and be  [#permalink]

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New post 01 Aug 2018, 11:13
prashant17p wrote:
PKN wrote:
A right angled triangle has its sides in Arithmetic progression and being integers. At most one of its sides is a multiple of 10. Its perimeter is less than 115. How many such triangles are possible?
A. 8
B. 7
C. 6
D. 5
E. 4


Let the sides be x, x+y, x+2y.

Then, by Pythagorean theom. \((x+2y)^2\) = \(x^2\) + \((x+2y)^2\)
Simplifying the above equation we get 3*(y/x)^2 + 2*(y/x) -1 = 0, solving the quadratic equation we get y=2x/3

Thus, the three sides are x, 5x/3, 7x/3 and the perimeter is 5x.

Now as the perimeter is less than 115 and the sides are integers both x and y will also be integers. Hence the possible values of the perimeter (5x) are 110, 105, 100......5. so the possible values for x are 22, 21, 20 ...... 1.

However, as y=2x/3 and y is an integer x should be a factor of 3.

Therefore, x can take the value of 3, 6, 9, 12, 15, 18, 21 i.e 7 possible values and hence 7 possible triangles.

Please help me identify where i went wrong.


Hi prashant17p,
Welcome to GMAT Club.
Highlighted portion is wrong.
It should be \((x+2y)^2\)=\(x^2\)+\((x+y)^2\)
Also, At most one of its sides is a multiple of 10 implies NOT MORE THAN ONE of the sides of the right angled triangle is a multiple of 10. You need to take care of this statement while determining possible combinations of 3 sides of the right angled triangles.
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PKN

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Re: A right angled triangle has its sides in Arithmetic progression and be  [#permalink]

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New post 03 Aug 2018, 12:51
1
PKN wrote:
A right angled triangle has its sides in Arithmetic progression and being integers. At most one of its sides is a multiple of 10. Its perimeter is less than 115. How many such triangles are possible?
A. 9
B. 8
C. 7
D. 6
E. 5



Let the sides be \((a - d)\), \(a\), \((a + d)\), where d is the common difference for the sides to form an AP.

Given the perimeter \((a - d) + a + (a + d) < 115\)

Hence we have \(a < 38.3\), since a is an integer we get \(a = 1,2,......38\)

Also, the given triangle is a right triangle, hence \((a -d)^2 + a^2 = (a + d)^2\), the longest side as the hypotenuse.

we get \(a = 4d\), hence \(a\) is a multiple of 4 & has a value between 1 & 38.

we get 9 such values for \(a\) to satisfy above criterion.


Answer A.



Thanks,
GyM
Re: A right angled triangle has its sides in Arithmetic progression and be &nbs [#permalink] 03 Aug 2018, 12:51
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