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(#M40166209) M4MATHS Numerical Ability question #PnC Keep an EYE

Find the number of triangles with Exactly one side odd and perimeter = 203

Asked In M4maths chakravorty devopriya devashish (5 years ago)

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Devendra Marghade  (5 years ago)

Ans: 101

We can presume that, one side has odd length and sum of the lengths of other two sides is even formed by sum of two even lengths.
Combinations of the sides A & (B+C) can be (1, 202), (3,200), (5,198), ..... (201,2)
Now to calculate number of triangles, we can consider the series of increasing length 1, 3, 5, ...., 201 which has 'n' terms, where first term a=1, last term=201 & difference d=2
Therefore, Last term=a + (n-1)d
=> 201 = 1+(n-1)2
201 = 2n - 1
n = 101
Hence, required number of triangles=101

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chakravorty devopriya devashish  (5 years ago)

total soln.
a + b + c = 203
a + b > c
b + c > a
c + a > b


203/2 = 101.5

so we can have the max value of the sides = 101

(101 - a) + (101 - b) + (101 - c) = 203
a + b + c = 100
102c2
we want distinct soln.

a b c
0 0 100
1 1 98
....
50 50 0

51 solutions will be of type aab
and repeated 3 times.
(102c2 - 51*3)/3! + 51 = 884

all odd we need to exclude that...
2a + 1 + 2b + 1 + 2c + 1 = 203
a + b + c = 100
100/2 = 50
(50 - a) + (50 - b) + (50 - c) = 100
a + b + c = 50
52c2

0 0 47
...
25 25 0
26 sol
26 * 3 = 78

(52c2 - 26 * 3)/3! + 26 = 234

884 - 234 = 650

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