Theory of Numbers Test - 1

Given number x01y is divisible by both 8 and 9.
For divisibility by 8: The number formed by the last three digits should be evenly divisible by 8.
For divisibility by 9: The sum of all the digits should be divisible by 9.
⇒ x + y + 1 = 9, and y can only be 6 (as 016 is divisible by 8).
⇒ x + 6 + 1 = 0 or Multiple of 9
⇒ x + 7 = 9
⇒ x has to be 2.
So, x + y = 2 + 6 = 8

(56 - 1) = (53)2 - (1)2
= (125)2 - (1)2 = (125 + 1) (125 - 1)
= 126 × 124
= 31 × 4 × 126
Therefore, it is clear that the expression is divisible by 31.

For the number to be divisible by 18, it should be divisible by both 2 and 9.
Now, sum of the digits of the number = x + y + 39
So, x + y should be 6 or 15.
Next value should be 24, but since x and y are distinct digits, it is not possible.
The possible pairs of (x, y) are (0, 6), (6, 0), (1, 5), (5, 1), (2, 4), (4, 2), (7, 8), (8, 7), (9, 6) (6, 9) and (3, 3).
Since the number has to be divisible by 2, the last digit should be even.
Then, there are 6 favourable cases. But (0, 6) is not possible, as then the given number will not be an eight-digit number.
Hence, the number of possible pairs is 5.

When a digit is written six times together, i.e. aaaaaa,

Then factors of aaaaaa is = 1001 × (aaa).

or, aaaaaa = 1001 × (aaa) = 7 × 11 × 13 × (aaa)

Hence, the given number is divisible by 7, 11, 13 and 1001.

The first 11 digits of the number are from 1 to 10.
Now we have another 90 digits to write.
Each of these 90 digits will come in pairs from a 2-digit number.
Thus, 45 numbers more will be required as 45 x 2 = 90
Now, the last number to be filled in will be 10 + 45 = 55
Before this, 54 shall be filled in.
Thus, the last 3 digits of the number will be 455. Thus, the number so obtained will be of the form 1000y + 455.
Now, 1000y = 8a
Thus, the only remainder term will come from 455.
Now, 455 = 448 + 7 = 56 x 8 + 7
Therefore, remainder = 7

Let the two numbers be 21x and 21y.

21x + 21y = 105

x + y = 5 where x, y are co-prime numbers

x = 1, y = 4 or x = 2, y = 3

⇒ Numbers = (21 × 1 and 21 × 4) or (21 × 2 and 21 × 3) = (21 and 84) or (42 and 63)

LCM of 21 and 84 is 84 and LCM of 42 and 63 is 126.

Product of 'n' numbers = [(HCF for each pair)n-1 × (LCM of 'n' numbers)]

= 33 × 126 = 27 × 126 = 3,402

x136y is a five digit number divisible by 15, means number should be divisible by both 3 and 5.
Number divisible by 5 means y should be 0 or 5.

Case 1: y = 0
x1360 is divisible by 3, so sum of digits should be multiple of 3.
Sum of digits = 10 + x
Possible value of x = 2, 5 and 8.

Case 2: y = 5
x1365 is divisible by 3, so sum of digits should be multiple of 3.
Sum of digits = 15 + x
Possible value of x = 0, 3, 6 and 9.

For least value of x + y, take y = 0 and x = 2
So, x + y = 2

For divisibility by 8, the last three digits should be divisible by 8.
∴ A can be 1, 3, 5, 7 or 9.
∴ There are 5 possible numbers, i.e. 7128, 7328, 7528, 7728 and 7928.