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Subjects
Class 10MathematicsReal NumbersEuclid’s Division LemmaHardLevel 1
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If a number gives quotient 31 when divided by 21, what is the least possible value of that number?

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Answer and explanation

Correct answer: 651

Step 1: Number (=21\times31+r), where (0\le r<21). Step 2: For the least value, take (r=0), so the number is (21\times31=651). Step 3: For the least possible number, use remainder zero.

Related tags

Euclids-Division-LemmaMinimum-NumberQuotient

Frequently asked questions

What is the correct answer to this question?

651

Why is this the correct answer?

Step 1: Number (=21\times31+r), where (0\le r<21). Step 2: For the least value, take (r=0), so the number is (21\times31=651). Step 3: For the least possible number, use remainder zero.

Which subject and chapter does this question cover?

This is a Class 10 Mathematics question. Chapter: Real Numbers. Topic: Euclid’s Division Lemma.

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10 more MCQs
  1. 1In Euclid’s Division Lemma, if (a) and (b) are positive integers, which condition is correct for the remainder (r)?
  2. 2If (35=6q+r) and (0 \le r < 6), what are the values of (q) and (r)?
  3. 3Which is the standard form of Euclid’s Division Lemma?
  4. 4If a number leaves remainder (0) when divided by (9), in which form can it be written?
  5. 5What possible remainders can occur when an integer is divided by (5)?
  6. 6What is the remainder when (52) is divided by (7)?
  7. 7Which condition is necessary for (b) in Euclid’s Division Lemma?
  8. 8If (a=17) and (b=4), which is the correct Euclidean form (a=bq+r)?
  9. 9How many possible remainders are there when a number is divided by (3)?
  10. 10If a number is odd, what remainder will it leave when divided by (2)?
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