Question

When a heap of pebbles is grouped in 32, 40, or 72, it leaves remainders 10, 18, and 50 respectively. What is the minimum number of pebbles in the heap?

• A. 1416
• B. 1418
• C. 1412
• D. 1420

Hint:

When remainders differ across moduli, first combine two congruences into one using $N = r_1 + m_1 a$, then solve the reduced congruence with the third modulus. Always reduce coefficients and divide by $\gcd$ when possible.

Answer 1

The Correct Option is B

Step 1: Translate to congruences.
Let $N$ be the number of pebbles. Then \[ N \equiv 10 \pmod{32},\qquad N \equiv 18 \pmod{40},\qquad N \equiv 50 \pmod{72}. \] Step 2: Solve the first two congruences (combine mod $32$ and $40$).
Write $N = 10 + 32a$. Impose $N \equiv 18 \pmod{40}$: \[ 10+32a \equiv 18 \ (\bmod\ 40)\ ⇒\ 32a \equiv 8 \ (\bmod\ 40). \] Since $32 \equiv -8\ (\bmod\ 40)$, \[ -8a \equiv 8 \ (\bmod\ 40)\ ⇒\ 8a \equiv 32 \ (\bmod\ 40)\ ⇒\ a \equiv 4 \ (\bmod\ 5). \] Hence $a=4+5t$ and \[ N=10+32(4+5t)=138+160t\ ⇒\ N \equiv 138 \pmod{160}. \] Step 3: Bring in the third congruence (mod $72$).
We need $138+160t \equiv 50 \ (\bmod\ 72)$. Reduce: \[ 138 \equiv 66\ (\bmod\ 72), 160 \equiv 16\ (\bmod\ 72). \] So \[ 66+16t \equiv 50\ (\bmod\ 72)\ ⇒\ 16t \equiv -16 \equiv 56\ (\bmod\ 72). \] Divide by $\gcd(16,72)=8$: \[ 2t \equiv 7\ (\bmod\ 9). \] The inverse of $2$ modulo $9$ is $5$ (since $2 5 \equiv 1$), hence \[ t \equiv 5 7 \equiv 35 \equiv 8\ (\bmod\ 9) ⇒ t=8+9k. \] Step 4: Smallest positive solution.
Take $t=8$: \[ N=138+160 8 = 138+1280 = 1418. \] This satisfies all three congruences and is minimal. \[ \boxed{1418} \]