Question

Replace A and B in the number 2A769B so that the number is divisible by 3,5 and 11. Show that they are two solutions.​

Answer 1

Answer:

277695

Step-by-step explanation:

Lets start with B. It can be replaced by 0 or 5 as per divisibility rule by 5.

Then, lets find value of A for each case considering divisibility rules by 3 and 11.

So starting point:

1. B=0 and number is 2A7690

Sum of numbers= 24+A

To be divisible by 3, A must be any one of following: 0, 3, 6 or 9

As per divisibility rule by 11 the difference in sums of 2+7+9=18 and A+6 must be divisible by 11.  So, here A must be = 1

In this case we have no solution as A=1 doesn't mach with A= 0, 3, 6 or 9

2. B=5 and number is 2A7695

Sum of numbers=29+A

To be divisible by 3, A must be 1, 4 or 7

As per divisibility rule by 11 the difference in sums of 2+7+9=18 and A+11 must be divisible by 11.  So, here A = 7

So we found one number divisible by all 3, 5 and 11: 277695

Answer 2

Answer: A = 7 and B = 5

Step-by-step explanation:

Divisibility of 5:- Last digit of the no. should be either 5 or 0

Divisibility of 3:- Sum of all the digits should be divisible by 3

Divisibility of 11:- Difference of sum of alternative digits should

                           Be either 0 or divisible by 11

Hit and trial Method:-

Let's put B = 5,      2A7695

2+A+7+6+9+5 = 29+A = 30 (nearest no. which is divisible by 3)

A=1  , 217695

(2+7+9) - (1+6+5) = 6 (Not divisible by 11)

A=1+3=4   , 247695

(2+7+9) - (4+6+5) = 3 (Not divisible by 11)

A = 4+3 = 7     ,     277695

(2+7+9) - (7+6+5) = 0 (Divisible by 11)

Hence, our required no. is 277695

A= 7 and B= 5

I hope u got ur answer....