1. | Which of the following set contains only even numbers ? | |||

(a) M, X, A | (b) A, V, L | (c) X, L, W | (d) P, W, S |

2. | Which of the following set forms a right angle triangle ? | |||

(a) P, A, X | (b) A, V, L | (c) L, G, P | (d) P, K, W |

3. | Which of the following set contains prime numbers only ? | |||

(a) A, V, P | (b) M, X, A | (c) A, V, L | (d) S, K, W |

You might be interested in this article |

● Cryptarithmetic Problem 27 |

● Cryptarithmetic Problem 25 |

● Cryptarithmetic Tutorial |

● Cryptarithmetic Multiplication Problem |

945*476

Answers of question

1) C

2) A

3) A

im not getting the solution pls any one explain this one

V*A=A So,A can be from either the set of {5} when V={3,7,9} or {2,4,8} when V={6}

but X*A=K and L*A=K .So, A can only be 5 and K=0 while X and L are from the set of even numbers.

P+1=X

V+A=W

V * A = A which means either

A = 5 and V = (3,7,9) or A = (2,4,8) and V = 6

Multiplication of A with the number gives K and A only. This thing is possible in case of 5 where it gives 5 and 0 when multiplied by anyone .

Therefore A can be assuned to be 5 and K is 0.

Now from M X A * L = A L V K, we have M X 5 * L = 5 L V 0.

Now we can have L = any odd numbers (2, 4, 6, 8) Now L = 2, 4 cant be possible as multiplication of L with any value of M will yield maximum of 1 and 3 respectively as the value of A in the 1000th place.

So now we have

A = 5

K = 0

V = 3,7,9

X = 2,4,6,8

L = 6,8.

Now we have got two case

Case 1: L = 6 & X = 2,4,8.

Case 2: L = 8 & X = 2,4,6.

Now by hit and trial for Case 1 and Case 2 in M X 5 * 6 = 5 L V 0 , we can get it satisfies case 1 only for X = 4 and dont satisfies Case 2.

So we get

L=6

X=4

M=9

V=7

K=0

A=5.

Now put this vues and proceed further

945

*476

———-