Question

An odd three digit number is divisible by 5 and 11 and sum of its digit is 16. Find the number​

Answer 1

Solution :

Let the required number be pqr .

Now , this is divisible by 5 and 11 .

For a number to be divisible by 5, the last digit can be either 0 or 5 .

Thus , the value of r can be either 0 or 5 .

Case 1 :

r = 0

Thus , the number becomes pq0

This is divisible by 11 .

So >

p + 0 - q = 11k

=> p - q = 11k .

Also ,

The sum of the digits is 16 .

So, p + q + r = 16

=> p + q + 0 = 16

=> p + q = 16 & p - q = 11k

Adding the two equations :

2p = 16 + 11k

Now, as p is an integer between 0 - 9, k € Even .

Let k = 2n

=> 2p = 16 + 22n

=> p = 8 + 11n .

But, we know that :

P_min = 0 & P_max = 9 .

Here, p = 8 + 11n , n € R+

So , p ≥ 8 .

The only value of n such that p ≤ p_max is n = 0

So, p = 8

now , p + q = 16 ( mentioned above )

=> q = 8.

Thus , the number becomes 880 .

Case 2 :

This case is quite interesting ... Observe carefully

Here, r = 5 .

So , the number becomes pq5 .

This is divisible be 11 .

So,

p + 5 - q = 11k

=> p - q = 11k - 5 . .. (1)

Now , the sum of the digits is 16

=> p + q + 5 = 16

=> p + q = 11 . . . . (2)

Adding ;

2p = 11k - 5 + 11

=> 2p = 11 ( k + 1 ) - 5

=> 2p = 11a - 5 [ when a = k + 1 , a € N ]

Now, as p € I, 11a - 5 has to be even .

Let 11a - 5 = b & b = 2c

=> 11a - 5 = 2c

=> 11a = 2c + 5

=> 11 ( k + 1 ) = 2c + 5

=> 11k + 11 = 2c + 5

=> 11k = 2c - 6 .

So,

2c - 6 ≡ 0 ( mod 11 )

=> 2c ≡ 6 ( mod 11 )

=> c ≡ 3 ( mod 11 ) [ as ( 2, 11 ) = 1 or 2 and 11 are coprime ]

c = 3, c = 14, etc .

c = 3 is the only possible value .

Now , as before mentioned , b = 2c = 6

So,

11a - 5 = 6

=> 11a = 11

=> a = 1 .

2p = 11a - 5

=> 2p = 11 - 5 = 6

=> p = 3 .

Now , p + q = 16

=> q = 13 .

But , this is not possible as q ≤ 9 .

So , this case is discarded when r = 5 .

Answer :

The only number possible is 880 .

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