3 × 7²² + 2 × 10²² - 44 when divided by 18 leaves the remainder __________.

Solution

Maths sir · 4 steps

Write 7 as 1 + 6 and 10 as 1 + 9, and almost everything turns into a multiple of 18.

1. First term. Expand 7²² = (1+6)²² = 1 + 6k for some integer k, because every term after the first has a factor 6. So 3 × 7²² = 3 + 18k.
2. Second term. In the same way, 10²² = (1+9)²² = 1 + 9m for some integer m. So 2 × 10²² = 2 + 18m.

   3 × 7²² + 2 × 10²² - 44 = 18(k+m) + 3 + 2 - 44 = 18(k+m) - 39

3. Fix the sign. A remainder cannot be negative, so borrow 54 = 18 × 3: 18(k+m) - 39 = 18(k+m-3) + 15. Since 0 ≤ 15 < 18, this is the true remainder.
4. So. The remainder on division by 18 is 15.

The common slip: 3. Treating -39 as +39 and taking 39 - 36 = 3 gives 3, not 15. The correct step is -39 + 54 = 15.

The idea to keep. For remainders of big powers, write the base as 1 plus a multiple of the divisor's factor and expand. If the leftover is negative, add multiples of the divisor until it lies between 0 and the divisor.

Answer from the NTA final answer key. Solution by PrepWiser.