Answer
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Hint: We first describe the theorems of the divisibility. The example helps to understand how the theorem works. We find the sum of the digits and the last two numbers to find out the divisibility of 3 and 4.
Complete step-by-step solution:
We use the divisibility theorem for 3 and 4 to find out if the given numbers are divisible by 3 and 4.
For the divisibility of 3, we know that if the sum of the digits in a number is divisible by 3 then the number itself is divisible by 3.
For example, we take a number (abc) where a, b, c are the digits in that number in the hundredth, tenth, unit places. So, we find $a+b+c$. If the sum is divisible by 3 then (abc) is divisible by 3. Take 4737. We add up the digits and get $4+7+3+7=21$ which is divisible by 3. So, 4737 is divisible by 3 where $\dfrac{4737}{3}=1579$.
For the divisibility of 4, we know that if the last two digits i.e. the tenth place and unit place digits together in a number are divisible by 4 then the number itself is divisible by 4.
For example, we take a number (abc) where a, b, c are the digits in that number in the hundredth, tenth, unit places. So, we find $bc$. If the term $bc$ is divisible by 4 then (abc) is divisible by 4. Take 4738. We take digits 38 and it’s not divisible by 4. So, 4738 is not divisible by 4 where $\dfrac{4738}{4}=1184.5$. Decimal appears which means 4738 is not fully divisible by 4.
Now we apply the theorems for the given numbers. We check the numbers’ divisibility for 3 and 4 simultaneously.
(i). For 63712, the digits add up $6+3+7+1+2=19$ which is not divisible by 3. So, 63712 is not divisible by 3.
And its last two digits are 12 which is divisible by 4. So, 63712 is divisible by 4.
(ii). For 2314, the digits add up $2+3+1+4=10$ which is not divisible by 3. So, 2314 is not divisible by 3.
And its last two digits are 14 which is not divisible by 4. So, 2314 is not divisible by 4.
(iii). For 78962, the digits add up $7+8+9+6+2=32$ which is not divisible by 3. So, 78962 is not divisible by 3.
And its last two digits are 62 which is not divisible by 4. So, 78962 is not divisible by 4.
(iv). For 10038, the digits add up $1+0+0+3+8=12$ which is divisible by 3. So, 10038 is divisible by 3.
And its last two digits are 38 which is not divisible by 4. So, 10038 is not divisible by 4.
(v). For 20701, the digits add up $2+0+7+0+1=10$ which is not divisible by 3. So, 20701 is not divisible by 3.
And its last two digits are 01 which is not divisible by 4. So, 20701 is not divisible by 4.
Note: We need to remember the theorems for the divisibility, If we can’t then there will be no way to find out the answer other than doing long division, But dividing the numbers will be a lengthy approach and there are more chances to do calculation mistakes so using the theorem saves a lot of time.
Complete step-by-step solution:
We use the divisibility theorem for 3 and 4 to find out if the given numbers are divisible by 3 and 4.
For the divisibility of 3, we know that if the sum of the digits in a number is divisible by 3 then the number itself is divisible by 3.
For example, we take a number (abc) where a, b, c are the digits in that number in the hundredth, tenth, unit places. So, we find $a+b+c$. If the sum is divisible by 3 then (abc) is divisible by 3. Take 4737. We add up the digits and get $4+7+3+7=21$ which is divisible by 3. So, 4737 is divisible by 3 where $\dfrac{4737}{3}=1579$.
For the divisibility of 4, we know that if the last two digits i.e. the tenth place and unit place digits together in a number are divisible by 4 then the number itself is divisible by 4.
For example, we take a number (abc) where a, b, c are the digits in that number in the hundredth, tenth, unit places. So, we find $bc$. If the term $bc$ is divisible by 4 then (abc) is divisible by 4. Take 4738. We take digits 38 and it’s not divisible by 4. So, 4738 is not divisible by 4 where $\dfrac{4738}{4}=1184.5$. Decimal appears which means 4738 is not fully divisible by 4.
Now we apply the theorems for the given numbers. We check the numbers’ divisibility for 3 and 4 simultaneously.
(i). For 63712, the digits add up $6+3+7+1+2=19$ which is not divisible by 3. So, 63712 is not divisible by 3.
And its last two digits are 12 which is divisible by 4. So, 63712 is divisible by 4.
(ii). For 2314, the digits add up $2+3+1+4=10$ which is not divisible by 3. So, 2314 is not divisible by 3.
And its last two digits are 14 which is not divisible by 4. So, 2314 is not divisible by 4.
(iii). For 78962, the digits add up $7+8+9+6+2=32$ which is not divisible by 3. So, 78962 is not divisible by 3.
And its last two digits are 62 which is not divisible by 4. So, 78962 is not divisible by 4.
(iv). For 10038, the digits add up $1+0+0+3+8=12$ which is divisible by 3. So, 10038 is divisible by 3.
And its last two digits are 38 which is not divisible by 4. So, 10038 is not divisible by 4.
(v). For 20701, the digits add up $2+0+7+0+1=10$ which is not divisible by 3. So, 20701 is not divisible by 3.
And its last two digits are 01 which is not divisible by 4. So, 20701 is not divisible by 4.
Note: We need to remember the theorems for the divisibility, If we can’t then there will be no way to find out the answer other than doing long division, But dividing the numbers will be a lengthy approach and there are more chances to do calculation mistakes so using the theorem saves a lot of time.
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