1. | Value of S ? | |||

(a) 6 | (b) 7 | (c) 8 | (d) 9 |

2. | Which of the following follows the Pythagoras theorem ? | |||

(a) H, A, D | (b) H, A, B | (c) T, E, A | (d) T, E, D |

3. | Value of H + R + S + A ? | |||

(a) 14 | (b) 15 | (c) 16 | (d) 17 |

T E A x H A D L D T R H R S A E W D A L E S S E R Here, A x A = _A [ H R S A ] Therefore, Possible values of A= {5, 6} Detailed Explanation- Rule 2 Further, T E A x H A D L D T R H R S A E W D A L E S S E R Here, H x A= _A [E W D A] Therefore, two possible cases for the values of H and A Case I - When A={5} then H={3, 7, 9} Case II - When H={6} then A={2, 4, 8} Detailed Explanation-Rule 3Now proceed further...

1. B 2. A 3. B T=6, E=1, A=5, H=3, D=4, L=2, S=7, R=0, W=8 6 1 5 x 3 5 4 2 4 6 0 3 0 7 5 1 8 4 5 2 1 7 7 1 0

(Solution has been given considering you as beginner in Cryptarithmetic.)

T E A x H A D L D T R H R S A E W D A L E S S E R Here, A x A = _ A [ H R S A ] Therefore, Possible values of A= {5, 6} Detailed Explanation- Rule 2 Further, T E A x H A D L D T R H R S A E W D A L E S S E R Here, H x A= _A [ E W D A ] Therefore, two possible cases for the values of H and A Case I - when A={5} then H={3, 7, 9} Case II - when A={2, 4, 8} then H={6} Detailed Explanation-Rule 3 Firstly taking case - I Taking A=5 rewrite the problem again, T E 5 x H 5 D L D T R H R S 5 E W D 5 L E S S E R Further, T E 5 x H 5 D L D T R H R S 5 E W D 5 L E S S E R Here, 5 x D = _ R [ L D T R ] Now, you can easily predict the value of R = 0 and possible values of D= {2, 4, 6, 8} [If you multiply 5 to a number, you will only get[0,5] as their unit digit.] 5 x Even =_0 [2, 4, 6, 8] 5 x Odd =_5 [3, 5, 7, 9] [Relax it's going to take some to understand the concept. Please read... again!] Put R=0 and write the problem again, T E 5 x H 5 D L D T 0 H 0 S 5 E W D 5 L E S S E 0 At this stage, divide the problem into 3 parts, (1) T E 5 x D L D T 0 (2) T E 5 x 5 H 0 S 5 (3) T E 5 x H E W D 5 Now, take (2) [ As it has less number of variables. 5 is repeated three times.] (2) T E 5 x 5 H 0 S 5 Now you have to start hit and trial with the possible values of E Firstly take E=1 Put E=1 in (2) (2) T 1 5 x 5 H 0 S 5 [ H 0 7 5] If you compare side by side, then you will get S=7 Put S=7 and E=1 in the main problem. [It needs to be checked further whether these values satisfies the Basic Cryptarithmetic Rules] T 1 5 x H 5 D L D T 0 H 0 7 5 1 W D 5 L 1 7 7 1 0 At this stage you can easily predict all the values as You can see T + 5 =_1 (which is only possible when the value of the T=6) L=2 (As, 1 + 1(carry) = L) Hence T=6, L=2. Now you can easily solve the problem. 6 1 5 x 3 5 4 2 4 6 0 3 0 7 5 1 8 4 5 2 1 7 7 1 0 [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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In the first case that is predicting ‘A’, the value of A can be 1 as well apart from being 5 or 6. Why 1 is not considered for A?

Suppose A={1, 5, 6} Reference: A x A= _A [H R S A ]

Let’s take A=1

then,

T E A

x A

H R S A

T E 1

x 1

H R S 1 [ T E 1 ]

If you will multiple any number with 1, result will be number itself.

A B C

x 1

A B C

Detailed Explanation( Rule 2)

sir How L=2 is not explained in detail

when E=1 it a trick that the carry of two digits will always be 1…so from there the value of L =2 is evaluated.

Why E=1 not any other value initially we take?

Hello Tushar,

This is just an hit and trial approach with the values of E.

So, you have to make a set of possible values of E.

Here E can be {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

Now, as you have already considered R=0 and A=5

So, the possible values of E= {1, 2, 3, 4, 6, 7, 8, 9}

Now you have to start hit and trial with the possible values of E from above set.

So, firstly consider E=1 proceed further ……

Please let us know if more information is required.

Best Regards,

ElitmusZone Team

‘A’ shouldn’t be considered in {2, 4, 8} in step-2. {2, 4, 8} for A, don’t satisfy A X A= _A (in step 1). Hence there is only one case, i.e, A=5