1. | Value of C ? | |||

(a) 2 | (b) 4 | (c) 6 | (d) 8 |

2. | Value of E + A + R + T ? | |||

(a) 10 | (b) 11 | (c) 12 | (d) 13 |

3. | Value of 2E + H ? | |||

(a) 7 | (b) 5 | (c) 6 | (d) 2 |

H A T x C U P E I U I E A R T E U P I H I E E E I Here, T x U=_T [E A R T] Therefore, Possible value of T and U are Case I - When T=[5] AND U=[3, 7, 9] or Case II - When U=[6] AND T=[2, 4, 8] Detailed Explanation- Rule 3 Let's take T=5 H A T x C U P E I U I E A R T E U P I H I E E E I Further as, T x P=_I [ E I U I ] T x C=_I [ E U P I ] i.e. 5 x P = _I [E I U I ] and 5 x C=_I [ E U P I] Therefore, I=0 and P and C are even numbers. [If you multiply 5 to any number, you will only get [0,5] as their last digit)[(5 x Even=_0 and 5 x Odd=_5)] Put I=0 and T=5 and rewrite the problem, H A 5 x C U P E 0 U 0 E A R 5 E U P 0 H 0 E E E 0 Now Proceed Further...

1. D 2. C 3. A H=3, A=4, T=5, C=8, U=7, P=6, E=2, I=0, R=1 3 4 5 x 8 7 6 2 0 7 0 2 4 1 5 2 7 6 0 3 0 2 2 2 0

(The solution has been given considering you as beginner in Cryptarithmetic.) H A T x C U P E I U I E A R T E U P I H I E E E I Here, T x U=_T [E A R T] Therefore, Possible value of T and U are, Case I - When T=[5] AND U=[3, 7, 9] or Case II - When U=[6] AND T=[2, 4, 8] Detailed Explanation- Rule 3 Let's take T=5 H A T x C U P E I U I E A R T E U P I H I E E E I Further as, T x P=_I [ E I U I ] and T x C=_I [ E U P I ] i.e. 5 x P=_I [ E I U I ] and 5 x C=_I [ E U P I ] Therefore, I=0 and P and C are even numbers. [If you multiply 5 to any number, you will only get [0,5] as their last digit) [(5*even=_0 and 5*odd=_5)] Put I=0 and T=5 and rewrite the problem, H A 5 x C U P E 0 U 0 E A R 5 E U P 0 H 0 E E E 0 At this stage, Possible values of variable U, P and C U= {3, 7, 9} P= {2, 4, 6, 8} C= {2, 4, 6, 8} At this stage, you have one more clue, H A 5 x C U P E 0 U 0 E A R 5 E U P 0 H 0 E E E 0 U + 5 = E Now, start hit and trial with the possible values of U={3, 7, 9} Firstly take U=3 You have U + 5 = E therefore E=8 Now, put E=8 and check further whether it satisfies the Basic Cryptarithmetic Rules. H A 5 x C U P 8 0 3 0 8 A R 5 8 U P 0 H 0 8 8 8 0 i.e. 0 + R + 0 = E Therefore R=8 Rejected as E=8 and R=8 [In Cryptarithmetic each variable should haveuniqueanddistinctvalues] Now, check with U=7 You have U + 5 = E [7 + 5 =_2(last digit)] therefore E=2 and as, 0 + R + 0 = 2, Therefore, value of R = 1 [R + 1(carry)=2] Put R=1, E=2 and rewrite the problem, H A 5 x C 7 P 2 0 7 0 2 A 1 5 2 7 P 0 H 0 2 2 2 0 [2+1(carry)=3(H)-- As value of E=2] therefore H=3 put E=2 H=3 T=5, R=1 in main problem 3 A 5 x C 7 P 2 0 7 0 2 A 1 5 2 7 P 0 3 0 2 2 2 0 2. 3 A 5 x 7 2 A 1 5 Now You can easily predict the value of A=4 [(28+3(carry)=_1(last digit)] [(28(7*4(A))+3(carry 7*5=35)=_1[31](last digit)] 3 4 5 x 7 2 4 1 5 Hence, 3 4 5 x 8 7 6 2 0 7 0 2 4 1 5 2 7 6 0 3 0 2 2 2 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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why you haven’t taken case 2 into consideration??

Case II – When U=[6] AND T=[2, 4, 8]

Hello Ravi,

You can also proceed with case II.

Please read below article and try to solve firstly try with case II.

Unit Digit Method- Cryptarithmetic

Best Regards,

ElitmusZone Team

wrong answer 8*5=40

carry 4 ?

thanks for such a good collection

can u plzz give the solution using case 2.Will the problm gets solved by using case 2 or not??

No ,it won’t solve the problem . also time will be consumed on that alot.