Practice30Create account

The greatest number that divides 62, 132 and 237 leaving the same remainder in each case is used as the divisor. That common remainder is:

Competition practiceNumber & Arithmetic
  • A35
  • B2
  • C27
  • D6

Step-by-Step Solution

The divisor must divide all the differences: 132 − 62 = 70, 237 − 132 = 105 and 237 − 62 = 175. HCF(70, 105, 175) = 35, so the divisor is 35. Now 62 = 35 + 27, 132 = 105 + 27 and 237 = 210 + 27, so the common remainder is 27. Answering 35 gives the divisor instead of the remainder.
Answer: C) 27

Master this topic in 30 minutes a day.

Don't just memorise the answer — build a daily practice habit.

Practice 5 Similar Questions Free →

All Number & Arithmetic practice questions

More Number & Arithmetic practice

If x² − y² = 91 and x + y = 13, where x and y are positive integers, then x − y equals:The value of 0.444… + 0.222… − 0.111… (each of the digits repeating for ever) is:The value of 1/(1 + 1/(1 + 1/(1 + 1))) is:How many perfect squares between 1 and 500 end in the digit 5?On rationalising the denominator, (√7 + √3)/(√7 − √3) becomes: