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

(a) M, S, H | (b) Z, S, X | (c) Z, M ,U | (d) U, E, R |

2. | Which of the following set contains odd numbers only ? | |||

(a) Z, M, U | (b) S, K, R | (c) R, K, H | (d) Z, U, X |

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

(a) Z, S, U | (b) U, R, H | (c) E, H, X | (d) E, U, M |

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● Cryptarithmetic Problem 30 |

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● Cryptarithmetic Multiplication Problem |

743*674

there is no clue…

split into 3 part 3*1 matrix.

then hit and trial for S={0-9}

case 1:

S=0,then U=even,M=5;

case 2:

s=1, U,M={3,7 or 7,3}

s=2, U,M={3,4},{4,3}………..{8,9},{9,8}

it will satisfy at S=2, and U=3,M=4……

7 4 3

6 7 4

……………………….

2 9 7 2

5 2 0 1

4 4 5 8

………………………

5 0 0 7 8 2

If i take U = 6, M = 7, then Z = 3, K = 6 & S = 2, then also it matches. It becomes 3 7 6 * 7 = 2 6 3 2

In this problem

one clue

m+1=x

if m=4

then

m+1=x=5 put the value then observe and find best option

z*z=x_

means

7*7=49

put z=7

m=4

E=6 e*z=mm 6*7=42

u=3

7 4 3

6 7 4

……………………….

2 9 7 2

5 2 0 1

4 4 5 8

………………………

5 0 0 7 8 2

actually there is a clue, X+M is generating a carry so possible values of M & X can only be (4,5) or (5,6) or (6,7) as X cannot be greater than 7(guess why..?)

now for set (5,6) M cannot be 5 as it can be seen in the multiplication which leaves only 2 possible values of M and X

either M=4 & X=5

or M=6 & X=7

now use hit and trial

One can answer atleast 2 questions from options only.

Let’s consider 3rd question where options a) b) d) has U common so Let’s assume ‘U’ is prime number( 2 or odd number-3,7).

Case 1- if we take U => 2 then, in question 1 option c) and d) must be correct so now consider option C) the probable values for ‘M’ could be 4,6,8 thus (M*M+1) will always give an odd value for Z but Z is even according to our assumption. Therefore option c) is incorrect

and let’s take option d) and it will be cancelled out.

Case 2- if ‘U’ is odd i.e either 3 or 7, thus this leads to another conclusion that ‘U’ is odd now we will check question 2 and option a) and d) are correct becoz they contain U as well as Z in common. So Z is again odd that is sure now. Now again move to question 1 and as Z is odd so it can’t be even hence option b) is also wrong, so the only option left is a). and now coming to question 2 back we can conclude that option d) is correct.

So it’s a kindof calculated risk but then it finally works.