When 416, 888, 1537 and 2245 are divided by the greatest number x, the remainder in each case is the same. The remainder is :

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MPPKVVCL Jabalpur JE Electrical 16 August 2018 Official Paper
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  1. 1
  2. 3
  3. 11
  4. 13

Answer (Detailed Solution Below)

Option 2 : 3
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Detailed Solution

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Given:

When 416, 888, 1537 and 2245 are divided by the greatest number x, the remainder in each case is the same.

Concept used:

For this type of question we operate on the HCF of difference between the numbers.

Calculation:

The greatest number when dividing 416, 888, 1537, and 2245 leaves the same remainder each time is HCF of (888 − 416), (1537 − 888), (2245 − 1537), (1537 − 416), (2245 − 416), (2245 − 888)

⇒ HCF of 472, 649, 708, 1121, 1357, 1829

Now,

The factors of 472 are: 1, 2, 4, 8, 59, 118, 236, 472

The factors of 649 are: 1, 11, 59, 649

The factors of 708 are: 1, 2, 3, 4, 6, 12, 59, 118, 177, 236, 354, 708

The factors of 1121 are: 1, 19, 59, 1121

The factors of 1357 are: 1, 23, 59, 1357

The factors of 1829 are: 1, 31, 59, 1829

So, the x is 59

Now,

If we divided 416 by 59 we get 3 as the remainder

∴ The required answer is 3.

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