NCERT Solutions for Class 8 Maths Ganita Prakash Chapter 5 — Figure it Out (Page 131)#

The digital root behaves like the remainder on division by 9. Adding 10 adds $9 + 1$, so the remainder goes up by 1:

New digital root $= 5 + 1$
New digital root $= 6$

Example: $40000001$ has digital root $4 + 1 = 5$, and $40000011$ has digital root $4 + 1 + 1 = 6$.

∴ The digital root will be 6.

Let us start with 10 and keep adding 11:

Observations:

• Each time, the digital root goes up by 2, because $11 = 9 + 2$. After 9 it wraps round: 9 → 2, then 8 → 1.
• The digital roots run through all nine values 1 to 9 before repeating, so the pattern repeats every 9 numbers.
• This happens whatever number we start with.

We separate the part that is a multiple of 9:

So the number is 4 more than a multiple of 9, and its remainder on division by 9 is 4.

∴ The digital root is 4, whatever whole numbers $a$ and $b$ are. For example, with $a = b = 1$: $9 + 36 + 13 = 58$, and $5 + 8 = 13 \to 1 + 3 = 4$.

There is no consistent pattern. Even numbers can have odd or even digital roots, and so can odd numbers:

• 10 is even, with digital root 1 (odd); 12 is even, with digital root 3 (odd); 22 is even, with digital root 4 (even).
• 11 is odd, with digital root 2 (even); 13 is odd, with digital root 4 (even); 15 is odd, with digital root 6 (even).

This is because adding 9 changes the parity of a number but not its digital root.

Division by 9. If the digital root is 9, the remainder is 0. Otherwise, the remainder equals the digital root.

Division by 3. The remainder depends on the digital root like this:

That is, the remainder on division by 3 is the remainder of the digital root itself on division by 3. Both facts come from the place values $10, 100, 1000, \ldots$ all being 1 more than a multiple of 9 (and so of 3).