Chapter 13: Playing with Numbers

Solution: The correct answer is option (c) 1000 a + 100 b + 10 d + c

Explanation:

We know that, the numbers are expressed as the sum of the product of it digits with the respective place value. So the generalised form of abdc is 1000 a + 100 b + 10 d + c

Solution: The correct answer is option (b) 10x + y

Explanation:

The numbers are expressed as the sum of the product of it digits with the respective place value. So the generalised form of xy is 10x+y

Solution: The correct answer is option (c) aobc

Explanation:

The usual form to represent the aobc is 1000xa+100xo+ 10xb+1xc, which is equal 1000a+10b+c

Solution: The correct answer is option (c) 18

Explanation:

The general form of abc is 100a+10b+c

The genera form of cba is

Now (abc- cba) = (100a+10b+c ) – (100c+10b+a) =99a-99c

= 99(a – c)

Now, abc – cba is divisible by 99, because 99 is the factor of abc – cba

So, all the numbers which are the factors of 99 will also be divisible by abc-cba

Here, 9, 11 and 33 are the factors of 99. But 18 is not a factor of 99. Hence abc – cba is not divisible by 18.

Solution: The correct answer is option (c) 37

Explanation:

It is known that, three numbers which can be formed by using the digits x, y and z are xyz, yzx and zxy.

The general form of xyz = 100x + 10y + z.

The general form of yzx = 100y + 10z + x.

The general form of zxy = 100z + 10x + y.

Add the three numbers, we will get

xyz + yzx + zxy = (100x + 10y + z) + (100y + 10z + x) + (100z + 10x + y).

xyz + yzx + zxy = 100x + 10x + x + 100y + 10y + y + 100z + 10z + z

xyz + yzx + zxy = 111x + 111y + 111z.

Now, simplify the expression, we will get

xyz + yzx + zxy = 111(x + y + z).

Now, it is clear that 111 is the common factor. So,

xyz + yzx + zxy is divisible by 111

Also xyz + yzx + zxy will be divisible by factors of 111.

From the given options, 11, 33, and 74 are not factors of 11, whereas 37 is a factor of 111.

Solution: The correct answer is option (c) 0 and 5

Explanation:

It is known that if a number is divisible by 55, then the number should be divisible by the factors of 55

It means that the number aabb is divisible by 5. By using the divisibility test of 5, it must be 0 or 5.

Solution: The correct answer is option (d) 9

Explanation:

By simplifying the general form of abc, bca and cab, we will get

= abc+bca+cab = 111(a+b+c)

Hence, abc+bca+cab is divisible by 111, and also it is divisible by the factors of 111.

Here, 3 and 7 are the factors of 111, and a+b+c is also a factor of 111(a+b+c).

But 9 is not the factor of 111.

Solution: The correct answer is option (b) 1

Explanation:

We know that the four-digit number 4ab5, which is divisible by 55 is also divisible by 11 and also the factors of it. By using the divisibility test of 11, the difference between the sum of the alternate digits should be a multiple of 11.

It means that

(4+b) – (a+5) =0, 11, 22, …it becomes

b – a – 1 = 0

Hence, b – a = 1

Solution: The correct answer is option (a) 9

Explanation:

We know that the general form of abc is 100a+10b+c

Then the given number is abc – a – b – c =100a+10b+c – a – b- c

By simplifying the above expression, we will get

abc – a – b – c = 9(11a+b)

Hence, the number abc – a – b – c is divisible by 9.

Solution: The correct answer is option (d) 1001

Explanation:

From the given question, the number should be of the form abcabc

So the general form of abcabc is 1000000a+100000b+1000c+100a+10b+c

Now, abcabc = is 1000000a+100000b+1000c+100a+10b+c

By simplifying the above expression, we will get

abcabc = 1001(100a+10b+c)

Hence, the six digit number should be divisible by 1001.