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P Pankaj Sanodiya
The addition of A and B is giving a number whose ones digit is 9. The sum can only be 9 not 10 as a sum of two single digits cannot exceed 18. hence there will not be any carry for the next step  2 + A = 0  implies A = 8  2 + 8 = 10 and 1 is the carry for next step. 1 + 1 + 6 = A = 8  it satisfies hence A = 8 and B = 1 is the correct answer.

P Pankaj Sanodiya
The additiion of B and 1 is 8  is giving a number whose ones digit is 8. this means digit B is 7 .  B + 1 = 8 and no carry for next step. next step : Now, A + B = 1 => A + 7 which implies A = 4  A  + B = 11 and 1 is carry for next step  1 + 2 + A  = B 1 + 2 + 4 = 7  Hence A = 4 and B = 7 is correct answer.

P Pankaj Sanodiya
The addition of 1 and B gives a number whose ones digit is 0. this is possible when digit B = 9 .  1 + B  = 10  and 1 is the Carry for the next step  Now, A + 1 + 1 = B => 9 Implies  A = 7. Hence A = 7 and B = 9 is Correct answer

P Pankaj Sanodiya
The product of 6 & B gives a number whose unit digit is B again. possible value of B = 0. 2 , 4, 6 or 8  If B = o then our product will be zero. hence this value of B is not possible. If B = 2, then B x 6 = 12 . Carry for next step = 1. 6A + 1 = BB = 22 implies A = 21/6 = not any integer value hence this case is also not possible. If B = 6  then B *6 = 36 and 3 will be carry for next step. 6A +...

P Pankaj Sanodiya
Here , multiplication of B and 5 gives a number whose ones digit is B. this is possible when B = 0 or 5 let B = 5  B * 5 = 5 * 5 = 25 Carry = 2 5*A + 2 = CA  , This os possible only when A = 2 or 7 when A = 2 , 5*2 + 2 = 12 which implies C = 1 when A = 7  5*7 + 2 = 37 which implies C = 3   now  B = 0  B*5 = 0*5 = 0  Carry = 0 so  5 * A = CA which is possible when A = 0 or 5 Howerver A cannot...

P Pankaj Sanodiya
Here multiplication of 3 and B gives  a number whose unit place digit is B . Possible value of B = 0 and 5 let B = 5  3 * A  + 1 = CA this is not possible for any value of A.  Hence B = 0 now  A * 3 = CA ( a number whose unit place digit is A itself when multiplied by 3) hence  possible value of A = 5 and 0  since AB is a two digit number A can not equal to 0. hence A = 5  A * 3 + 1 = CA 5 * 3...

P Pankaj Sanodiya
There can be two cases 1. when the addition of unit place digit doesn't  produce Carry A + 3 = 6 A = 3 However, to get 3 in unit place of our answer our B has to be 6 and that would produce carry hence this case is not possible. 2. when the addition of unit place digit produces Carry A + 3 +1 = 6 A = 2 for getting 2 in unit place of answer we need the sum of unit digit of numbers = 12 B + 7 =...

P Pankaj Sanodiya
Here first clue : we have A = a number which when multiplied by itself gives the same number in the unit digit.  possible numbers = 1 and 6  Second Clue: number when multiplied with 1 and added with the reminder of previous multiplication( A*A) = 9  both first and second clue implies that A = 6.

P Pankaj Sanodiya
Here answer's unit place = 3 , possible addition of unit places digit = 13  A + 8 = 13    A = 13 - 8 = 5 remainder = 1 Ten's place of answer  = 4 + 9 + 1 =  14  B = 4  remainder = 1  100's place = C = 1   Hence value of  A = 5 , B = 4 and C = 1.

P Pankaj Sanodiya
Here we are adding two numbers and unit place of the first number and the second number is A and  5 respectively. unit place of the answer is 2 so the way we can get this result is when we get 12 on adding unit places of both number i.e.                                                             A + 5 = 12                                 which implies      A = 12 - 5 = 7. Ten's digit of both...

P Pankaj Sanodiya
The detailed solution for the above-written question is as follows N is odd; so it's unit digit is odd. Therefore, the unit digit must be 1, 3, 5, 7 or 9.

P Pankaj Sanodiya
The detailed solution for the above-written problem is as follows, The unit digit, when divided by 5, must be leaving a remainder of 3. So the unit digit must be either 3 or 8.

A Amit Singh
If the number abcd is divisible by 11 then [ (b + d) - (a + c) ] also must be divisible by 11. let the number be 1089 here a = 1, b = 0, c = 8 and d = 9 1089 = 1000*1 + 100*0 + 10*8 + 9          = (1001*1 + 99*0 + 11*8) + [(0 + 9) - (1 + 8)]          = 11(91*1 + 9*0 + 8) + [ 9 - 9 ] here  [ (b + d) - (a + c) ] = [9 - 9 ] = 0 which is divisible by 11. hence If the number abcd is divisible by 11...

P Pankaj Sanodiya
let the number abc be 132 Here a = 1, b = 3 and c = 2 132= 100*1 + 10*3 + 2 = 99 + 11*3 + (1 - 3 + 2)     = 11(9*1+3) + (1 - 3 + 2 )  if number is divisible by 11 then (a - b + c )  must be divisible by 11. as in above case of number 132 the a - b + c = 1 -3 + 2 = 0 whichis divisible by 11. Hence we conclude ( a - b + c ) should be divisible by 11 if abc is divisible by 11.

P Pankaj Sanodiya
Yes, it has been prooved that if a number is divisible by any number m, then it will also be divisible by each of the factors of m. Let's Assume n is divisible by m, and m is divisible by k.This means n=pm and m = qk where all are integers Now, n = p(qk) =( pq)k which means n is divisible by k. Hence a number is divisible by any number m, then it will also be divisible by each of the factors of m.

P Pankaj Sanodiya
Any number will be divisible by 3 only if the sum of all the digits in that number will be divisible by 3. Sum of digits of number 927 = 9 + 2 + 7 = 18 which is divisible by 3.  Hence we conclude number 108 is divisible by 3.

P Pankaj Sanodiya
Any number will be divisible by 3 only if the sum of all the digits in that number will be divisible by 3. Sum of digits of number 432 = 4 + 3 + 2 = 9 which is divisible by 3.  Hence we conclude number 432 is divisible by 3.

P Pankaj Sanodiya
Any number will be divisible by 3 only if the sum of all the digits in that number will be divisible by 3. Sum of digits of number 294= 2 + 9 + 4 = 15 which is divisible by 3.  Hence we conclude number 294 is divisible by 3.