**INDUSTRY AND COMPANY AWARENESS (ICA)**

Question

:

C

The correct answer can be marked in 30 seconds if you know that this is an application of "number of terms in an Arithmetic Progression”. If we distribute 500 consecutive numbers with a common difference of 9, number of terms can only be 5009= 55 or (55+1). You can directly mark option (c)

Why a common difference of 9? Because the first number with a property of its digit sum=5 is 05, the next term is 14 (1+4=5), the next is 23(2+3=5) and so on

The difference between any two consecutive terms with this property is 9.

Thus, the above question is nothing but finding the number of terms in an Arithmetic Progression with the terms as 5,14.... 500.

Notice another interesting thing

When 5 is divided by 9,the remainder is 5

When 14 is divided by 9, the remainder is 5

When 23 is divided by 9, the remainder is 5

3 different ways of asking the same question

1) How many numbers less than or equal to 500 will leave a remainder of 5 when divided by 9?

2) Find the number of terms in the AP 5,14,23,32...........500 ?

3) If B = sum of digits of A, C = sum of digits of B, .........X = X itself. If X = 5 find the different possible values of A given that 0<A <500

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Suppose the seed of any positive integer n is defined as follows:

Seed (n) = n, if n < 10

= seed(s(n)), otherwise,

Where s(n) indicated the sum of digits of n. For example, seed(7)=7, seed(248) = 2+4+8 = seed(14) = seed(1+4) = seed(5) = 5, etc.

How many positive integers n, such that n< 500, will have seed(n)=5?

Seed (n) = n, if n < 10

= seed(s(n)), otherwise,

Where s(n) indicated the sum of digits of n. For example, seed(7)=7, seed(248) = 2+4+8 = seed(14) = seed(1+4) = seed(5) = 5, etc.

How many positive integers n, such that n< 500, will have seed(n)=5?

**Answer: Option C**

:

C

The correct answer can be marked in 30 seconds if you know that this is an application of "number of terms in an Arithmetic Progression”. If we distribute 500 consecutive numbers with a common difference of 9, number of terms can only be 5009= 55 or (55+1). You can directly mark option (c)

Why a common difference of 9? Because the first number with a property of its digit sum=5 is 05, the next term is 14 (1+4=5), the next is 23(2+3=5) and so on

The difference between any two consecutive terms with this property is 9.

Thus, the above question is nothing but finding the number of terms in an Arithmetic Progression with the terms as 5,14.... 500.

Notice another interesting thing

When 5 is divided by 9,the remainder is 5

When 14 is divided by 9, the remainder is 5

When 23 is divided by 9, the remainder is 5

3 different ways of asking the same question

1) How many numbers less than or equal to 500 will leave a remainder of 5 when divided by 9?

2) Find the number of terms in the AP 5,14,23,32...........500 ?

3) If B = sum of digits of A, C = sum of digits of B, .........X = X itself. If X = 5 find the different possible values of A given that 0<A <500

Was this answer helpful ?

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