Question:
Find the least number which when divided by 35, 45, and 55 leaves the remainder 18, 28, and 38 respectively.
Find the least number which when divided by 35, 45, and 55 leaves the remainder 18, 28, and 38 respectively.
Show 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.
Updated On: Jul 13, 2026
- 3465
- 17
- 3449
- 3448
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Spot the pattern in the remainders.
Compare each divisor to its remainder:
\[ 35 - 18 = 17 \]
\[ 45 - 28 = 17 \]
\[ 55 - 38 = 17 \]
Every divisor is exactly 17 more than its remainder. This is the key pattern to use.
Step 2: Use the shortcut for this pattern.
When a number N, divided by several divisors, always falls short of each divisor by the same amount k, it means N is exactly k less than a common multiple of all the divisors:
\[ N = \text{LCM(divisors)} - k \]
Here \(k = 17\).
Step 3: Find the LCM of 35, 45, and 55.
Break each into prime factors:
\[ 35 = 5 \times 7 \]
\[ 45 = 3^2 \times 5 \]
\[ 55 = 5 \times 11 \]
The LCM takes the highest power of every prime that appears, \(3^2\), \(5^1\), \(7^1\), \(11^1\).
\[ \text{LCM} = 9 \times 5 \times 7 \times 11 \]
\[ 9 \times 5 = 45, \quad 45 \times 7 = 315, \quad 315 \times 11 = 3465 \]
So the LCM of 35, 45, and 55 is 3465.
Step 4: Subtract to get the least number.
\[ N = 3465 - 17 = 3448 \]
Step 5: Verify.
\[ 3448 \div 35 = 98 \text{ remainder } 18 \quad (35 \times 98 = 3430,\ 3448 - 3430 = 18) \]
\[ 3448 \div 45 = 76 \text{ remainder } 28 \quad (45 \times 76 = 3420,\ 3448 - 3420 = 28) \]
\[ 3448 \div 55 = 62 \text{ remainder } 38 \quad (55 \times 62 = 3410,\ 3448 - 3410 = 38) \]
All three remainders check out exactly as given.
Step 6: Why the other options fail.
Option (1), 3465, is just the LCM itself, it would leave remainder 0 with every divisor, not the remainders given.
Option (2), 17, is only the constant gap we found, not the actual least number satisfying all three conditions.
Option (3), 3449, is one more than the correct answer and does not divide out to the right remainders.
Final Answer:
The least such number is 3448. \[ \boxed{3448} \]
Compare each divisor to its remainder:
\[ 35 - 18 = 17 \]
\[ 45 - 28 = 17 \]
\[ 55 - 38 = 17 \]
Every divisor is exactly 17 more than its remainder. This is the key pattern to use.
Step 2: Use the shortcut for this pattern.
When a number N, divided by several divisors, always falls short of each divisor by the same amount k, it means N is exactly k less than a common multiple of all the divisors:
\[ N = \text{LCM(divisors)} - k \]
Here \(k = 17\).
Step 3: Find the LCM of 35, 45, and 55.
Break each into prime factors:
\[ 35 = 5 \times 7 \]
\[ 45 = 3^2 \times 5 \]
\[ 55 = 5 \times 11 \]
The LCM takes the highest power of every prime that appears, \(3^2\), \(5^1\), \(7^1\), \(11^1\).
\[ \text{LCM} = 9 \times 5 \times 7 \times 11 \]
\[ 9 \times 5 = 45, \quad 45 \times 7 = 315, \quad 315 \times 11 = 3465 \]
So the LCM of 35, 45, and 55 is 3465.
Step 4: Subtract to get the least number.
\[ N = 3465 - 17 = 3448 \]
Step 5: Verify.
\[ 3448 \div 35 = 98 \text{ remainder } 18 \quad (35 \times 98 = 3430,\ 3448 - 3430 = 18) \]
\[ 3448 \div 45 = 76 \text{ remainder } 28 \quad (45 \times 76 = 3420,\ 3448 - 3420 = 28) \]
\[ 3448 \div 55 = 62 \text{ remainder } 38 \quad (55 \times 62 = 3410,\ 3448 - 3410 = 38) \]
All three remainders check out exactly as given.
Step 6: Why the other options fail.
Option (1), 3465, is just the LCM itself, it would leave remainder 0 with every divisor, not the remainders given.
Option (2), 17, is only the constant gap we found, not the actual least number satisfying all three conditions.
Option (3), 3449, is one more than the correct answer and does not divide out to the right remainders.
Final Answer:
The least such number is 3448. \[ \boxed{3448} \]
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