What is the least natural number you have to add to 1024 to get a multiple of 181?

Question

Philoid · Accepted Answer

We know If s → (q, r) denotes the quotient q and r denotes the remainder r

Then s = (b × q ) + r where s and b > 0 and 0 ≤ r < b

When s is divided by 181 then

s = 1024

1024 = 181 × 5 + 119

Here the quotient = 5 and r = 119

If we subtract the reminder from the divisor we will get the required number which should be added to get the multiple of 181

⇒ 181 – 119 = 62

The required number is 62 which when added to 1024 gives the multiple of 181