Question : If the number formed by the last two digits of a three-digit integer is an integral multiple of 6, the original integer itself will always be divisible by:
Option 1: 6
Option 2: 3
Option 3: 2
Option 4: 4
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Correct Answer: 2
Solution : Given: The last two digits of a three-digit integer is an integral multiple of 6. Let the 3-digit integer be $100x+10y+z$. $10y+z =6m$ ($m$ is any integer) The 3-digit integer is $100x+10y+z = 100x+6m = 2(50x+3m)$. So, the original integer itself will always be divisible by 2. Hence, the correct answer is 2.
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Question : Which is the smallest three-digit number which when increased by 5 becomes divisible by both 2 and 3?
Option 1: 102
Option 2: 105
Option 3: 103
Option 4: 108
Question : The ratio of a two-digit number and the sum of the digits of that number is 4 : 1. If the digit at the unit's place is 3 more than the digit at the ten's place, then the number is:
Option 1: $47$
Option 2: $69$
Option 3: $36$
Option 4: $25$
Question : The least number of five digits exactly divisible by 88 is:
Option 1: 10032
Option 2: 10132
Option 3: 10088
Option 4: 10023
Question : Which is the greatest number of four-digit divisible by 12, 25, and 45?
Option 1: 9890
Option 2: 9999
Option 3: 9800
Option 4: 9900
Question : Which smallest number must be added to 100, so that the resulting number is completely divisible by 6?
Option 1: 2
Option 3: 5
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