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Question
The greatest number which divides 134 and 167 leaving 2 as remainder in each case is
Options
- 14
- 17
- 19
- 33
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Solution 1
33
Since we need 2 as the remainder, we will subtract 2 from each of the numbers.
167 − 2 = 165
134 − 2 = 132
Now, any of the common factors of 165 and 132 will be the required divisor.
On factorising:
165 = 3 × 5 × 11
132 = 2 × 2 × 3 × 11
Their common factors are 11 and 3.
So, 3 × 11 = 33 is the required divisor.
Solution 2
33
Subtract the required remainder from 134 and 167.
So we get 132 and 165.
132 = 2 × 2 × 3 × 11 = 22 × 3 × 11
165 = 3 × 5 × 11 = 31 × 5 × 11
HCF = 3 × 11 = 33
Hence, the greatest number which divides 134 and 167 leaving 2 as remainder in each case is 33
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