Question

Find the least number which when divided by 35, 45, and 55 leaves the remainder 18, 28, and 38 respectively.

• A. 3465
• B. 17
• C. 3449
• D. 3448

Hint:

Notice that each divisor minus its remainder gives the same value (17). The answer is the LCM of the divisors minus that common difference.

Answer 1

The Correct Option is D

A different way to attack this is to build up the answer step by step from the remainder condition, instead of jumping straight to the LCM shortcut.

First notice something useful: $35 - 18 = 17$, $45 - 28 = 17$, and $55 - 38 = 17$. So in every case, the remainder is exactly 17 less than the divisor. In other words, if we added 17 to our mystery number, it would divide evenly (with no remainder) by all three numbers 35, 45, and 55.

1. Turn the problem around: Let $M = N + 17$. Then $M$ must be exactly divisible by 35, by 45, and by 55, with no remainder left over.
2. Find the smallest such $M$: the smallest positive number divisible by all three at once is their LCM. Factor each: $35 = 5 \times 7$, $45 = 3^2 \times 5$, $55 = 5 \times 11$. Taking the highest power of each prime gives $\text{LCM} = 3^2 \times 5 \times 7 \times 11 = 3465$.
3. Recover N: since $M = N + 17$, we get $N = 3465 - 17 = 3448$.

Now check this actually works: $3448 = 35 \times 98 + 18$, $3448 = 45 \times 76 + 28$, and $3448 = 55 \times 62 + 38$. All three remainders line up exactly with what the question asks for.

Let's summarize:

• The trick is spotting that divisor minus remainder gives the same constant (17) in every case.
• Adding that constant back turns the problem into finding an LCM.
• Subtracting the constant from the LCM gives the smallest number that fits all three conditions.

So the least number satisfying all the given conditions is 3448, which is option (4).

\[ \boxed{3448} \]