Playing with Numbers Test - 1

If a is a factor of b, then b is divisible by a. Here, 2 and 5 are factors of the required number. Thus, the number is divisible by both 2 and 5.

A number is divisible by 2 if its last digit is even (0, 2, 4, 6, or 8).

A number is divisible by 5 if its last digit is either 0 or 5.

Therefore, a number is divisible by 2 and 5 if its last digit is 0.

Among the options, only the number 30 has 0 as its last digit. Thus, 2 and 5 are both factors of 30.

We know that if a three-digit number has the same digit in its ones and hundreds places, then the difference between the number and the number formed by reversing its digits is always 0.
Also, 0 is an additive identity.
So, the difference between the number and the number formed by reversing its digits is always an additive identity.

It can be seen that the four digit number abcd has a thousands, b hundreds, c tens, and d ones.

Thus, the number abcd can be written in the general form as

abcd =1000 × a + 100 × b + 10 × c + d.

A number is divisible by 2 if its units digit is even (0, 2, 4, 6, 8).

The units digit of 10,302 is 2. Therefore, 10,302 is divisible by 2.

A number is divisible by 3 if the sum of its digits is also divisible by 3.

For the number 10,302, sum of the digits = 1 + 0 + 3 + 0 + 2 = 6, which is divisible by 3.

Therefore, 10,302 is divisible by 3.

A number is divisible by 4 if the number formed from its last digits is 00 or is divisible by 4.

For the number 10,302, the number formed from its last 2 digits is 02, which is not divisible by 4. Thus, the number 10,302 is not divisible by 4.

A number is divisible by 5 if its units digit is 0 or 5.

The number 10,302 has the digit 2 at units place. Thus, it is not divisible by 5.

Thus, the given number is divisible by 2 and 3.

A number is divisible by 11 if the difference between the sum of digits at odd places (from right) and even places is either 0 or divisible by 11.

The given number is 15Δ936.

Sum of digits at odd places from right = 6 + 9 + 5 = 20

Sum of the digits at even places except Δ = 1 + 3 = 4

∴ Difference = 20 − 4 = 16 = 11 + 5

Therefore, if Δ = 5, then the given number will be divisible by 11.

Thus, the least value of Δ is 5.

It is known that if a number ends in 0, 2, 4, 6 or 8, then it is divisible by 2.

The number 96152 ends in 2. Therefore, it is divisible by 2.

The correct answer is A.

Why alternative B is wrong:

9 + 6 + 1 + 5 + 2 = 23 is not divisible by 3. Therefore, 96152 is not divisible by 3.

Why alternative C is wrong:

The number 96152 does not end in 0 or 5. Therefore, the given number is not divisible by 5.

Why alternative D is wrong:

The number 96152 is not divisible by 3. Hence, it is not divisible by 6.

Let the original three-digit number be of the form 100x + 10y + z, where x, y, and z represent the digits at hundreds, tens, and units places respectively.

Then, the new four-digit number formed is of the form 1000x + 100y + 10z + 4.

It is given that the difference between the two numbers is 1111.

∴ 1000x + 100y + 10z + 4 − (100x + 10y + z) = 1111

⇒ 900x + 90y + 9z + 4 = 1111

⇒ 900x + 90y + 9z = 1107

⇒ 9 (100x + 10y + z) = 1107

⇒ 100x + 10y + z = 123

Thus, the original three-digit number i.e., 100x + 10y + z is 123.