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A
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Difficulty:
95%
(hard)
Question Stats:
43% (02:21) correct
57%
(02:47)
wrong
based on 157
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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
A. 9
B. 8
C. 7
D. 6
E. 5
Moderator Note: corrected answer options. Thank you pushpitkc @chetan2u
31 Jul 2018, 18:48
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PKN
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
PKN
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.
