In the trick given above, instead of finding the difference of the two 2-digit numbers, find their sum. What will happen? For example : Class 8

In the trick given above, instead of finding the difference of the two 2-digit numbers, find their sum. What will happen? For example : Class 8

Question 1.

In the trick given above, instead of finding the difference of the two 2-digit numbers, find their sum. What will happen? For example : Class 8

We start with 31. After reversing we get 13. Adding 31 and 13, we get 44.

We start with 28. After reversing we get 82. Adding 28 and 82, we get 110.

We start with 12. After reversing we get 21. Adding 12 and 21, we get 33.

Observe that all these numbers are divisible by 11. Is this always true? Can we justify this claim using algebra?

Solution:

Yes. Let the number be ab.

So, we have :

ab + ba = 10a + b + 10b + a

= 11 (a + b)

Hence, the sum will be divisible by 11.

Question 2.

Consider any 3-digit number, say abc (100a + 10b + c). Make two other 3-digit numbers from these digits by cycling these digits around, yielding bca and cab. Now add the three numbers. Using algebra, justify that the sum is always divisible by 37. Will it also always be divisible by 3? [Hint: Look at some multiples of 37.] Class 8

Solution:

abc + bca + cab = (100a + 10b + c) + (100b + 10c + a) + (100c + 10a + b)

= 111a + 111b + 111c

= 111 (a + b + c)

= 37 × 3 (a + b + c)

Hence, the sum of such three digit numbers will be divisible by 37 as well as 3.