If abc, cab, bca are three digit numbers formed by the digits a, b, and c then the sum of these numbers is always divisible by 37.

Solution:

The generalised form of abc, cab and bca can be written as follows:

(i) 100a + 10b + c

(ii) 10a + b + 100c

(iii) a + 100b + 10c

Sowhen add all the three numbers i.e. (i) + (ii) + (iii) we get

111a + 111b + 111c = 111(a+ b + c) = 37 × 3 × (a + b + c)

Therefore the sum of the numbers abc, cab and bca is divisible by 37

Hence the statement is True.

✦ Try This: If ab and ba are two digit numbers formed by digits a and b then the sum of these two numbers will be divisible by 11.

The generalised form of ab, ba can be written as follows:

(i) ab = 10a + b

(ii) ba = 10b + a

Adding (i) and (ii) we have

11a + 11b

= 11(a + b)

Hence the sum of numbers ab and ba is divisible by 11.

☛ Also Check: NCERT Solutions for Class 8 Maths Chapter 16

NCERT Exemplar Class 8 Maths Chapter 13 Sample Problem 8

If abc, cab, bca are three digit numbers formed by the digits a, b, and c then the sum of these numbers is always divisible by 37.

Summary:

If abc, cab, bca are three digit numbers formed by the digits a, b, and c then the sum of these numbers is always divisible by 37 is a True(T) statement.

☛ Related Questions:

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