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 |

```
H S P
x D U O
U T P S
H S P
P P A P
P H O T O S
You can easily predict the value of U=1 as,
H S P
x U
H S P
i.e. after multiplication you are getting the same results which is only possible when U=1
For example.
A B C 3 4 5
x 1 x 1
A B C 3 4 5
[If you are sure about value of any variable, then replace it with digit and rewrite the problem again. It will help you in solving the problem.]
Further, you have one more clue
H S P
x D U O
U T P S
H S P
P P A P
P H O T O S
D x P = _P [ P P A P]
Hence, Possible values of D and P are
Case I : When D={6} and P={2, 4, 8} Rule 3
Case I : When P={5} and D={3, 7, 9}
Now Proceed further...
```

```
1. D
2. B
3. A
```

```
H=3, S=8, P=2, D=6, U=1, O=4, T=5, A=9
3 8 2
x 6 1 4
1 5 2 8
3 8 2
2 2 9 2
2 3 4 5 4 8
```

H S P x D U O U T P S H S P P P A P P H O T O S You can easily predict the value of U=1 as, H S P x U H S P i.e. after multiplication you are getting the same results which is only possible when U=1 For example. A B C 3 4 5 x 1 x 1 A B C 3 4 5 Further, You have one more clue H S P x D U O U T P S H S P P P A P P H O T O S D x P= _P [ P P A P ] Hence, Possible values of D and P are Case I : When D= {6} and P={2, 4, 8} Rule 3 Case I : When P= {5} and D={3, 7, 9} Now, You have to collect some more clues for solving the problem, Such as H S P x D U O U T P S H S P P P A P P H O T O S P + P = O and P x O =_S Now, take case I to solve the problem Case I - When D={6} and P={2, 4, 8} Now, you have to start hit and trial with the value of P={2, 4, 8} Let's take P=2 then O=4 [P + P = O] and as P x O=_S Hence S=8 i.e. U=1, O=4, S=8, P=2 and D=6 Put these values in the main problem and rewrite the problem. H 8 2 x 6 1 4 1 T 2 8 H 8 2 2 2 A 2 2 H 4 T 4 S As, H 8 2 x 6 2 2 A 2 Now you can easily predict the value of H=3 and it also satisfies Basic Cryptarithmetic Rules. 18 + 4(carry)= 33[6 x 3 =18] Now, 3 8 2 x 6 2 2 9 2 [ 2 2 A 2] Hence, A=9 3 8 2 x 6 1 4 1 5 2 8 3 8 2 2 2 9 2 2 3 4 5 4 8 [Relax - it's going to take some time to understand the whole concept. If you are facing any difficulty. Please go through the Cryptarithmetic Tutorial.]

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

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

You can not predict it which case is write Go on your luck bro.

How do we get and. 2.

How triangle is formed by HTO

Is it because 16+9=25?

Ans 2

@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)

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}