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Which of the following is divisible by 2, 3, and 5?
2240
54332
3150
14175
3150
Since a number divisible by 2 and 5 must end in 0, we can immediately eliminate 54332 and 14175. For 2240 and 3150, check the sum of digits: 2240 (sum=8, not div by 3) vs 3150 (sum=9, div by 3).
A list of four numbers: 2240, 54332, 3150, 14175. Question asks for the number divisible by 2, 3, and 5.
Divisibilityrulefor2:UnitdigitmustbeevenDivisibilityrulefor5:Unitdigitmustbe0or5Divisibilityrulefor3:Sumofdigitsmustbedivisibleby3
Since a number divisible by 2 and 5 must end in 0, we can immediately eliminate 54332 and 14175. For 2240 and 3150, check the sum of digits: 2240 (sum=8, not div by 3) vs 3150 (sum=9, div by 3).
Students often check divisibility by 2 and 5 first and forget to verify the sum of digits rule for 3.
Filter by 2 and 5
A number is divisible by 2 and 5 if it is divisible by 10, meaning it must end in 0. Looking at the options: 2240 and 3150 end in 0, while 54332 and 14175 do not.
Unitdigit=0
Apply divisibility rule for 3
A number is divisible by 3 if the sum of its digits is a multiple of 3. For 2240, the sum is 2+2+4+0=8, which is not divisible by 3. For 3150, the sum is 3+1+5+0=9, which is divisible by 3.
3+1+5+0=9→9÷3=3
Conclusion
Since 3150 is divisible by 2 (ends in 0), 5 (ends in 0), and 3 (sum of digits is 9), it satisfies all conditions.
3150
C is correct because 3150 is the only option that satisfies the divisibility rules for 2, 3, and 5 simultaneously.
This concept of divisibility by 2, 3, and 5 is equivalent to testing divisibility by LCM(2,3,5)=30. Any number ending in 0 whose digits sum to a multiple of 3 is divisible by 30.
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