A number r when divided by 8 leaves remainder 3. What will be the remainder when (r 2 +6r+7) is divided by 8?

Correct — d. Only the remainder matters, so replace r by the smallest number that fits.

r leaves 3 on division by 8, so take r = 3. Any r = 8k + 3 behaves identically.

r² + 6r + 7 = 9 + 18 + 7 = 34
34 = 8 × 4 + 2 → option (d).

The algebra agrees: r² + 6r + 7 = (r + 3)² − 2, and r + 3 ≡ 6 (mod 8), so (r + 3)² ≡ 36 ≡ 4, leaving 4 − 2 = 2.

Why the others are wrong

• (a)
  4 — 4 is the remainder of (r + 3)² alone — 36 mod 8. The expression is (r + 3)² − 2, so two more must come off.
• (b)
  3 — 3 is the remainder r itself leaves, carried over unchanged. Squaring and adding shifts it: 3² + 6×3 + 7 = 34, and 34 leaves 2.
• (c)
  1 — 1 would need the expression to land on 33 or 41. With r = 3 it lands on 34, and every r of the form 8k + 3 gives the same remainder.

Remainders survive both addition and multiplication, so the remainder of a polynomial in r depends only on r's own remainder — never on which r you picked.

Here r ≡ 3 (mod 8), so testing r = 3 settles the whole family: 3, 11, 19, 27 and the rest all give the same answer.

The general form proves it. With r = 8k + 3, r² + 6r + 7 = 64k² + 48k + 9 + 48k + 18 + 7 = 64k² + 96k + 34. Both 64k² and 96k are multiples of 8, so the remainder is 34 mod 8 = 2.

Key facts

• If r ≡ 3 (mod 8), any polynomial in r has a remainder fixed by 3 alone.
• 3² + 6×3 + 7 = 34, and 34 = 8×4 + 2.
• r² + 6r + 7 can be rewritten as (r + 3)² − 2.

Common traps

• Choosing a large r such as 27 and then dividing a four-figure number by 8 by hand
• Reporting r's own remainder of 3 instead of the expression's
• Losing the +7 while copying the expression out of the stem

No directly related past PYQ was found.