# Cryptarithmetic Division Problem 3 First we will convert this question into multiple, Cryparithmetic Multiplication Problems –

Before that in 4th row you see ANML and the divisor is also ANML, so value of D = 1 clearly.

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`As, A N M L x     D-------- A N M Lthus, D = 1.         C 1 N K B          _______________A N M L |P K N L B C 1 E         P 1 B A         --------------           L C B B           A N M L           --------------           C M M M C           C C B K N           -----------------             B K C L 1             B M N B M             -----------------               1 L B N E               1 B L E 1               -----------------                 B M A PCheck Unit digit method here.Now, Find the multiplication with most substituted values,it is -  A N M L  x     B--------1 B L E 1Now, using Unit digit's method possible ways of getting 1 are(check the link above for Unit digit method to find all rules)3 x 7 = _1 (21 only last digit considered)7 x 3 = _1 (21 only last digit considered)Thus, values are either L = 3, B = 7 and L = 7, B = 3using case 1 first,   A N M 3  x     7---------1 7 3 E 1Now, if you consider A x 7, this results in 17.There is only one possibility, for A x 7 + carry = 17when A = 2 and carry = 3( A = 1 not possible as7 + 6(max carry ie 7 x 9 =63) results only in 13 which is <17.So now eqn is   2 N M 3  x     7---------1 7 3 E 1Now, we know what carry from 7 x N was 3, this is only possible when7 x N = _3 when N = 4( 28 + some carry >=2) or 5( 35 + some carry less equal to 4)Taking N = 4,  2 4 M 3  x     7---------1 7 3 E 1Now, at unit's digit 3 value and 3 as carry.It is possible for 7 x 4 + 5(carry) = 33Now to get 5 carry from last step you have to multiply 7 with either 7 or 8Now, take M = 8  2 4 8 3 x      7---------1 7 3 E 1Now carry from unit's 7 x 3 will be 2and 7 x 8 + 2 carry = give value of E = 8. and M also 8. So, not possible.Sim. M = 7 is not possible(try yourself)Now try L = 7 and B = 3  A N M 7  x     3---------1 3 7 E 1Now A = 4 to get 13 with 1 carry from prev. multiplication,  1  4 N M 7  x     3---------1 3 7 E 1Now, to give 1 carry and 7 at unit's placevalue of N = 5 with 2 carry from previous operation  1 2  4 5 M 7  x     3---------1 3 7 E 1Now, 3 x M can give 2 carry for M={6, 8, 9}Let's do hit and trial. Take M = 6.  1 2 2  4 5 6 7  x     3---------1 3 7 E 1Now, 3 x 7 gives 2 carry to next mult.and 3 x 6 + 2 as carry = 20 so E = 0.  1 2 2  4 5 6 7  x     3---------1 3 7 0 1E = 0, D = 1, A = 4, N = 5, M = 6 its easy to find other values too. `

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