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If a number leaves remainder (15) when divided by (40), (56), and (88), what is the smallest such number?
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Answer and explanation
Correct answer: (3095)
Step 1: Subtracting (15) makes the number divisible by all three numbers. Step 2: (40=2^3\times5), (56=2^3\times7), and (88=2^3\times11), so LCM (=2^3\times5\times7\times11=3080). Hence the number is (3080+15=3095). Step 3: Add the common remainder at the end.
Frequently asked questions
What is the correct answer to this question?
(3095)
Why is this the correct answer?
Step 1: Subtracting (15) makes the number divisible by all three numbers. Step 2: (40=2^3\times5), (56=2^3\times7), and (88=2^3\times11), so LCM (=2^3\times5\times7\times11=3080). Hence the number is (3080+15=3095). Step 3: Add the common remainder at the end.
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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