#### ElitmusZone

ElitmusZone » Cryptarithmetic Problem 10

# Cryptarithmetic Problem 10

 1. Value of N ? (a) 2 (b) 4 (c) 3 (d) 8
 2. Value of T + 2E ? (a) 15 (b) 17 (c) 16 (d) 11
 3. Which of the following forms Right Angled Triangle ? (a) N, P, E (b) T, P, E (c) T, H, A (d) B, N, S

1. ElitmusZone Post author

Hi Ishani,

The Pythagorean theorem states that:

In any right triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle).

This can be stated in equation form as

a^{2}+b^{2}=c^{2} i.e.
a^2+b^2=c^2

where c is the length of the hypotenuse, and a and b are the lengths of the remaining two sides.

Pythagorean triples are integer values of a, b, c satisfying this equation.

So, you have to check with the values of alphabets, whether it satisfies the above condition or not.

(a) N, P, E
(b) T, P, E
(c) T, H, A
(d) B, N, S

Best Regards,
ElitmusZone

1. vikas solanki

we can also approach like this.
we have following clues
1. E*E=E
2. P*E=E
POSSIBLE SET OF VALUES OF P AND E ARE=
P{6} E{2,4,8,}
P{3,7,9} E{5}
FROM ABOVE 2 SETS YOU CAN ELIMINATE 1 SET BECAUSE
T H E
*E
_______
P I A E AND HERE P CANT BE 6 BECAUSE MAXIMUM 5*9=45 AND MAX REMAINDERS ARE{0,1,2,3,4,5} SO ________ NONE OF THEM WILL GIVE TENTH DIGIT AS 6.

SP PROCEED WITH SECOND SET ::i.e P{3,7,9} E{5}
AS E IS 5 E*ODD NUMBER = 5 AS UNIT DIGIT AND E*EVEN NUMBER = 0 AS UNIT DIGIT .
5 CANT BE UNIT DIGIT IN
(T H E) *N SO VALUE OF I WILL BE 0 FOR SURE AND N MUST BE A EVEN INTEGER SO POSSIBLE VALUES OF N ARE N{2,4,6,8}

NOW PROCEED IN THE SAME DIRECTION WE HAVE ENOUGH CLUES TO SOLVE