A number leaves remainders (63), (81), and (117) when divided by (72), (90), and (126) respectively. What is the smallest such number?
Answer and explanation
Correct answer: (2511)
Step 1: In each case, divisor minus remainder is (9), so adding (9) to the number makes it exactly divisible by all three divisors. Step 2: The LCM of (72), (90), and (126) is (2520), so the number is (2520-9=2511). Step 3: When each remainder is equally less than the divisor, subtract that common difference from the LCM.
Frequently asked questions
What is the correct answer to this question?
(2511)
Why is this the correct answer?
Step 1: In each case, divisor minus remainder is (9), so adding (9) to the number makes it exactly divisible by all three divisors. Step 2: The LCM of (72), (90), and (126) is (2520), so the number is (2520-9=2511). Step 3: When each remainder is equally less than the divisor, subtract that common difference from the LCM.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Real Numbers. Topic: HCF and LCM using prime factorisation.
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