Playing with Numbers Test - 1

(10ⁿ - 1) is divisible by 9 for n ∈ N as 10¹ - 1 = 9, 10² -1 = 99, 10³ - 1 = 999 and so on, are divisible by 9.

28345x4 is divisible by 8, if its last three digits are divisible by 8.

So, we have 5 x 4 to be divisible by 8.

Therefore, least value of x is 0 (because 504 is divisible by 8).

It is given that when X is divided by 5, the remainder is 3.
So, X can be 3,8, 13,18, 23, 28, 33, 38, 43, 48, ...

Also, it is given that when X is divided by 2, the remainder is 1.
So, X can be 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, ...

Seeing the common elements in both the cases, we see that numbers common in both the cases have 3 as the units digit.

The number 43476p2086 is divisible by 3, if the sum of all the digits is divisible by 3.
So, sum of the digits = 4 + 3 + 4 + 7 + 6 + p + 2 + 0 + 8 + 6 = 40 + p

So, the possible values of p are 2, 5 and 8.
(Because only 42, 45 and 48 are divisible by 3)

If the number is xy, then its reverse is yx. Since the number remains unchanged, 10x + y = 10y + x.
The possible number can be 11, 22, 33, or 44.

The required prime number is 11.
The square of 11 is 121.

The prime factors of 6 are 2 and 3. So, if a number is divisible by 2 and 3, it must also be divisible by 6.

As the number has an even digit at its unit place, so the number would be divisible by 2 and also, it is divisible by 3.

So, the number would be divisible by 6 as well.

A number is divisible by 3 if the sum of the digits of the number is divisible by 3.
ab736 is divisible by 3 if the (a ＋ b ＋7 ＋ 3 ＋ 6) is divisible by 3.

i.e. 16 + a + b is divisible by 3.
Or a + b = 2, 5, 8
Since a + b < 4;
a + b = 2
So, the only possible values of (a, b) are (1, 1) or (2, 0).