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Cryptarithmetic Multiplication 11

 1. Value of H + S + P ? (a) 10 (b) 11 (c) 12 (d) 13
 2. Which of the following forms a triangle ? (a) H, S, P (b) H, T, O (c) D, U, O (d) U, T, P
 3. Value of 2D + P ? (a) 14 (b) 15 (c) 16 (d) 17

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1. Vinay

Plz let us know what happens if we take case 2 while solving….how can we know which case is best to solve.

1. satheesh

if we take case2 then p=5 and p+p=10 ie O=0 WHICH IS NOT TRUE BCOZ IF O=0 THEN EVERYTHING MULTIPLIED BY ZERO WILL BE ZERO

2. Shujayat Ali1

@Sandeep…. actually we are not using here pythagoras theorem to check formation of triangle.

We are applying here…. (“”in a triangle the sum of two sides must be greater than the third side””)

a) H=3,S=8,P=2 (not forming triangle because H+P is not greater than S)
b) H=3,T=5,O=4 (here it is following the rule..in a triangle the sum of two sides must be greater than the
third side)
c) D=6,U=1,O=4 (it is not following the rule…in a triangle the sum of two sides must be greater than the third
side)
d) U=1,T=5,P=2 ((it is not following the rule…in a triangle the sum of two sides must be greater than the third
side)

3. vikas solanki

we can proceed like this also AND BEST WAY I THINK TO LEARN THESE MULTIPLICATION IS DONOT LOOK FOR OPTION and just follow unit digit method

as we can easily see U is =1
and d*p=p
possible of values for P and D is
P{5} D{3,7,9,}
P{2,4,8} D{6}
WE CAN EASILY REJECT FIRST SET OF VALUES IF YOU WILL SEE
P+P=0
AND IF P=5 THEN O SHOULD BEE ZERO. BUT O CANT BE ZERO IF YOU OBSERVE THE MULTIPLIACTION
SO PROCEED WITH SECOND SET OF VALUES AND HIT ANS TRIAL WITH
P{2,4,6}
D{6}